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[Music] in this video we explore the incredibly important topic

differentiation the derivative is one of the two fundamental concepts in

all of calculus in this video we make the transition from

tangent lines to the derivative function to differentiation rules and then

discuss the question why is the derivative so important so stick around

and get all the calculus you can [Music] okay so welcome to episode three

differentiation the derivative function in this episode we’re going to talk

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about everything about the derivative so everything between in our last

episode we talked about continuous functions

so we’re going to go from continuous functions and we’re going to talk about

everything about the derivative except for and up to the chain rule the

chain will cover the chain rule in the next episode

so yeah we’re going to talk about rates of change in tangent lines the

derivative differentiation rules and how the derivative

is important in the world around us and then

at the end lots of exercises so we have a lot to cover today will be our longest

episode yet um and i’m so excited for it and and let’s get started [Music]

okay we’re going to begin with rates of change and tangent lines

and the r tangent line problem is probably the one of the most important

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problems uh that fermat worked for worked with

it directly led to the idea some ideas that newton and leibniz

you know was responsible for creating calculus for

of course vermont had something the ancients didn’t have

he had the analytic geometry and so he was able to

work out work on the tangent line problem very seriously

um and so newton and leibniz came along and invented calculus so before we begin

i just want to mention these three names every calculus student should see these

three names and you know do some reading on who they were

and what life was like back then for them in terms of mathematics okay

so you know it may not be easy to talk about what a tangent line is in other

words if someone asked you hey what is a tangent line

how would you describe a tangent line to to somebody

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so for example if we just look at a graph right here and ask okay right here

what would a tangent line look like and you say okay it’s a line

and it crosses only once right crosses only once right there

suppose i drew that good enough where it only crosses once

here it only crosses one time so you say okay a tangent line is a line that

crosses the graph only once well we’re tangent right here what if

the graph does this though and the line is extended on it intersects again

so then you would say well it’s not tangent line anymore

well we want to think of it though as a tangent line around this point here

and so the the reason is or or the way to talk about a tangent line

if we put an axis down and you know assign that a number right there call it c

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then what we’ll say is we’re going to be tangent

but around c we have to be able to find an open interval

if we can find an open interval around c where the line crosses the graph once

then we’re going to call this a tangent line

so you know we have to think of tangent line as a local process

so you see these two graphs the graph on the left

the line crosses once the graph on the right the line

crosses twice but we still want to talk about tangent lines so the important

thing is when we talk about a tangent line

it’s a local concept so that’s the key word there

tangent line is a local concept you don’t need to be tangent um globally

it’s locally okay so in our first example here we’re going to

talk about a function and we’re going to be trying to find the

tangent line this is the tangent line problem how do we find

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an equation of the tangent line and we’re going to be doing this the old

school way without knowing what the derivative is and so

we’re going to do a bunch of examples and then we’re going to try to infer

we’re going to try to basically you know use educated guess

on what the equation of the tangent line is so

you know everything we’re about to do here is very important

and in some sense this is kind of a review of some things you’ve learned in

precalculus so for example we have this function here

two x to the third minus two x plus two and we’re given these two points one

half five fourths and we’re given the point one two right

so those are two points on the graph if you were to substitute in here one half

into this function right here so substituting one half you should get out

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five fourths if you substitute in one you should get out two now what we want

to do is to find the equation of the of the line so

you know we have a point we have another point we have a line through that

how do we find the equation of the line so to do that we can find the slope so

let’s say the slope is two minus five five fourths and then one minus a half

so one minus a half so for two let’s say that’s eight fourths minus five fours

and for one let’s say that’s two over two so this will be three fourths over

one-half which we’ll say is what times two over one so six over four so

three over two so three over two is our slope so let’s just say our slope

is three over two okay so now how do we find the equation

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of this line so we can use either point i’ll use this one let’s say y minus two

2 equals the slope 3 over 2 and then x minus 1.

so we have an equation of the line so we have 3 halves x

and then minus 3 halves and then plus 2. so you see the process of given two

points finding the equation of the line is very easy very quick

very easy you just find the slope you just use one of the points

to plug in right and then we have an equation of the line so

if we graph the function 2x to the third minus 2x plus 2

we get a nice curve there nice and smooth and we get this line here the slope is

three over two and then we get the one half for the y intercept there

so as we can see this is a secant line it crosses the graph twice and um

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you know we got so much more to talk about the tangent line but

this is just our first basic example here

so what we’re going to do is we’re going to keep the point one half five fourths

fixed and what we’re going to see what we’re trying to do

is find a tangent line here so if we look at this if we look at this graph here

and we’re at this point right here it’s about right there we’ve got this

nice shape here we’re looking at this point here this is one half five fourths

and then we went over to one two so those points aren’t too close

depending upon your perspective right it’s a difference of one half

we’ve got a secant line right so you might ask

how do we find the equation of the tangent line or crosses one so

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it only crosses once we have one point now how do we find the equation knowing

only one point on the graph right so that’s our goal so the way to do that is to

bump the one half over to a one find equation of this line

but those two points are too far apart so

what we’re going to do now is move that point a little closer

let’s move it a little bit closer i don’t know to something say three four

something three-fourths so it’s a little bit closer so now if

we find the line through there those two points that’ll be a little bit better

a little bit closer to our tangent line let’s do that next

so this time we’re going to look at the same function

the same point one half five fourths but we’re moving

the point one two a little bit closer so i chose three-fourths now if you plug

three-fourths into that function you get out 43

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over 32 and so we have two points again so we can find the equation of the

tangent line and we take the same process we find the slope then we find the y

intercept and we have our we have our line there so our line is the 3 8

as the slope and the y intercept comes out to be 17 over 16.

and so we work this out exactly as we did the numbers have changed

but as you can tell there the secant line

you know the points are closer together right so one-half to three-fourths

that’s only a .25 difference and so now what we’re going to do is do

this not once not twice but we’re going to pull out a

spreadsheet or some type of calculator to help us do all this elementary

operations really quickly so we’re going to repeat this process

several times and we’re going to put this information in a table

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so here are the columns of the table the first one is something we

i’m going to call delta x so in our first example

we went from we went from one-half five-fourths do one two so our delta x is

one minus one-half or 0.5 our next example was one-half five-fourths

and then it was 3 4 and something else 41 i think over something but the delta x

is the three fourths minus the one half so that’s 0.25

so you see i’m getting the delta x the delta x is how much have we changed the

x by so here the change is a lot smaller it’s just 0.25

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so that’s the first column the delta x the second column is just the point the x

and the f of x and that’s going to stay the same all the way through

we’re fixing that point because we’re interested in the tangent line

at that point so that point’s not going to change

it’s the other point that we want to bring closer and closer to it

so to bring it closer and closer my delta x is going to get smaller

so the third column there is the point that that we’re changing

because we’re changing the delta x and then the last one is the equation of the

secant line and so those are the two examples that we’ve done so far

i’ve just put them in decimal notation so at first our delta is 0.5 and then

secondly it’s 0.25 either way we still have

one half five fourths as the fixed point that we’re working on

and then we apply the delta x right so for example in row number one

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we have the second column 0.5 1.25 to get the next column

i add the delta x to the x and i add the delta

and then i and then i get the x which is the one

and then i take that one and i plug it into the function

to get out two we do that again in the second row right that’s exactly what we

already did for those two for those two examples we did previously

and we get those two equations of the line right we find the slope

and then we go find the y intercept [Music]

well now that we’ve done it twice and we put it in a

table right or some spreadsheet or something now using a computing power

we can do the third we can do the fourth and so on

and we can start to look at the equation of the tangent line

and say wow we’re getting really close right our delta x is getting small

0.5 0.25 0.125 0.0625 so delta x is getting smaller and

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smaller getting closer and closer to zero so the point is getting closer and

closer to point five one point two five so in the last one the fourth row

the the new point is 0.56 right that’s very close to 0.5 and

1.23 right that’s really close to 1.25 so for those two points we go and figure

out using some formula what the equation of the secant line is

so you know how many of these rows do we need to do

before we can say i know what the tangent line is going to be

so what are the slopes tending towards 1.5 0.375

we’re getting closer and closer now we’re at negative 0.09 then we’re at

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negative 0.3 can you tell what it’s going towards let’s do another row

so we went from negative for the slope we went from negative 0.3 to negative 0.4

all right now we’re really close .015625 that’s a really small delta x those two

points are really close together and what’s the result for the secant line

negative 0.45 so that’s looking really close to negative 0.5 there

so we’re going to go and stop after that many rows and we’re going to guess

for this table what would you say the equation of the tangent line is

at this point here one-half and one-fourth um should be five-fourths so

tangent line of the function at the point one half

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five fourths so there’s a typo there right it should be

not one fourth it should be one half five fourths

all right so what should we say right so the tangent line is

that’s what my guess is based upon the last column

i’m looking at negative 0.5 and then for the y intercept i’m looking at 1.5

okay so [Music] we’re going to say this is negative one half plus

one point five so we’re going to go with three halves here

so this is what i’m gonna say is the tangent line is my guess for the tangent

line now this took a lot of work to find the equation of the tangent line

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and this is called the tangent line problem how do you find

an equation of the tangent line and if you’re using just the find the slope

and then find the y intercept and do a secant line

and then move that secant line closer and come up with another secant line

when you type try to do that type of process there

you’re going to get a lot of work all right so suppose we have a function

and we’re going to make a change in the variable so this goes back to what our

delta x is when we ever have a delta x a change in x

we also have a corresponding delta y so the delta y is take that change in x

add it to the x and input that into the function and that difference there

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is called the change in y and so we can call this the average rate of change

is the delta y over the delta x this is basically the same thing as the slope

formula so why would that be the slope formula so

we’re given two points here right let’s see here we’re given this point here

x and then f of x and x plus delta x and then plug that point in

we get x plus delta x right so we have these two points here how do we find the

slope of the line through those two points so the slope is f of x plus delta x

minus f of x all over now we subtract the x’s x plus delta x

minus the x and you can see the x’s here cancel

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and so we end up with is x plus delta x [Music] minus f of x

over delta x the x’s just add up to zero so the slope here and this is just

called the average rate of change so we’re going to call this

the average rate of change we’re also later on going to call it the difference

quotient so okay later on as in right now and so we have two names for this

expression right here average rate of change and the difference quotient

so let’s do a quick example here’s a function and we have two values of x

so what’s the change in x what’s our delta x so our delta x is a three

right so our delta x is the three and so let’s go find

our average rate of change here we can do that over here

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so our function is square root of x squared plus nine or minus nine

and our delta x is six minus three or just three right so we can find our

difference quotient our delta x is three so we’re gonna go

and do the x plus delta x that’s our six minus f of the three because we’re

going from three to six so we’re starting at three so x is three here

then we’re going to do six minus three or just three right so we plug in six

here and then we plug in three here so we’re gonna have square root of

36 minus 9 minus and then now we’re going to plug in 3.

so when we plug in 3 we’re going to get 9 minus 9 plus square root of 0

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right and so you know what is that so writing it up the difference quotient is

f of six minus f of three over six minus three

and we calculate that and we get square root of three out of that so

this is also the slope of the secant line as as i was mentioning earlier

so you want to think about this three ways you want to think about it as the

average rate of change the difference quotient and the slope of the secant line

[Music] all right so talk a little bit about the

world around us suppose that object moves along a straight line according to an

equation of motion so s is the displacement or the directed distance of an

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object and we’re going to use t to represent time

so the function that describes the motion is called the position function

we’re going to have a time interval so we need to fix

a moment in time let’s just call it a so from time t equal a to t equals a

plus h so here i’m using my here i’m using an h to represent a change

in the input so now instead of saying average rate of change

we’re going to talk about the average velocity

and so this is the same expression that we had before

the only difference is we’re having this function represent something in the

real world and instead of using a delta x i’m using an

h and so this is the same thing as the slope

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of the secant line through these two points

okay so let’s do an example so let’s say a billiard

is dropped from a height of 500 feet and its height s at some time

is given by position function now later on we’re going to see where we

get that position function from right now this example that’s given to us

so s is measure defeat and t is measured in seconds

so we’re asked to find the average velocity

over these two intervals so let’s work one interval at a time so for this

interval we’re going to calculate s of 2 and s of 2.5 and then we’re going to

find the average velocity so the average velocity is the

change in the outputs divided by the change in the inputs

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and so i calculate the change in the outputs s of 2.5 minus s of 2. i

calculate those two numbers i get 400 and 436 so

when we you know work on all that we’re going to get minus 72

the minus is important the minus tells us that we’re falling downward

and now from 2 to 2.6 so again we have s of 2 and now we calculate s of 2.6

and so we’re going to find the average rate of change or the average velocity

over this interval here so we’re going to get negative 73.6 so

we have a little bit more time you know from 2 to 2.5

and then from 2 to 2.6 so it seems like the average velocity um has gone

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you know the speed the speed is we’re going to talk about the speed as the

absolute value of that um and we’ll do that here in a couple minutes

okay so the average velocities are indicating that the object is moving

downward okay okay so the difference quotient right here

um i put it a little bit differently i put it with x1 and x2

so the delta x here is just x2 minus x1 is an average rate of change so again

the average rate of change is the change in the outputs divided by the change in

the inputs now what happens if we take a limit

though so our delta x here is going to go to zero

and that’s how we’re going to define the derivative so we’re going to input

interpret the limit of the average rate of change

as the interval becomes smaller and smaller which is what we did in our

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example earlier using our table but then we’re going to

call this the instantaneous rate of change

the average rate of change is just the average rate of change

but then if we take the limit then we’re

getting the instantaneous rate of change now if you look throughout the sciences

this has very very many different names this has a lot of different names

so in physics when you’re talking about motion of

objects instantaneous rate of change so the average rate of change approaches

the instantaneous rate of change so the derivative is the limit so at the

top the first equation that’s the average rate of change

that takes place over an interval and the last equation is called the

derivative or the instantaneous rate of change

and it’s the limit as the delta x goes to zero

it’s known as the derivative obviously that’s the main topic

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of this video and you know we have so much more to go still

all right so in this example we’re going to say find the instantaneous rate of

change i’m just putting a function down and i’m going to ask us to find the

instantaneous rate of change so this point what we know is what the

average rate of change is and from our first episode we know how to find limits

but this is the limit of a difference quotient this is the limit of the

average rate of change so it’s not just any just random limit or any random

function this is a specific type of limit the difference quotient

so in order to do that well we can say here’s the derivative okay how do you

find the instantaneous rate of change so the instantaneous rate of change is

found simply by plugging into into the derivative so once i know what

the derivative is there and i wrote it down

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the instantaneous rate of change is just simply found by

substituting in 2 into the derivative and calculating that number and that’s

the instantaneous rate of change so i just want to point out that in

order to find the instantaneous rate of change

you use the derivative function so we were given a

function f we found the derivative function somehow we don’t know how yet we’re

going to spend a great deal of time finding that derivative uh using

a lot of differentiation rules that’s to come

but when we have the derivative function we can substitute in a number and find

the instantaneous rate of change at that point there’s the instantaneous

rate of change so here’s another example find the average rate of change

and then we’re also going to find the instantaneous rate of change

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so the average rate of change takes place over an interval

and this interval is given to us is 2 4.

instantaneous rate of change takes place at a number number two

so to find the average rate of change we find the

change in the outputs over the change in the input

so the change in input the delta x is 4 minus 2.

then we have to substitute in the inputs to make a difference of the outputs

so we substitute in 4 minus then we substitute in 2

and when we calculate all those numbers up we get a 9.

so the average rate of change of f between 2 and 4 is 9.

now to find the instantaneous rate of change we have to somehow

find the derivative which i give to us in this example

the derivative is 4x minus 3 but to find the instantaneous rate of

change we just plug in 2. the instantaneous instantaneous rate of

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change i just plug in 2 and i’m going to get 8 minus 5 uh sorry

8 minus 3 which is 5. so that’s the instantaneous rate of change

so finding the derivative function is something that’s going to you know

take up a considerable a considerable amount of time not only

because not only do we want to know what the derivative is and in terms of a

definition but also in terms of theorems how to

find it but also we want to get skilled at finding the derivative

all right so we’re also going to talk about something called the

relative rate of change so the relative rate of change

is different than say the absolute rate of change and the average rate of change

right the absolute change is not the same thing as the average rate of change

namely the absolute change is just the difference in the y values

so that’s the delta y that we’ve been using for

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delta y is called the absolute change the average rate of change is well you

take the absolute change and you divide it by the size of the

interval the delta x this is the average rate of change

sometimes the average rate of change is more useful okay so

in this example here we’re going to be given a bunch of temperature readings in

celsius and they recorded every hour so from 0 to 24 starting at midnight

one day in april and so we have the x’s we have the temperatures and we want

to find the average rate of change in the temperatures

with respect to time from noon to three and then noon to two

and then noon to one p.m then we’re going to estimate the instantaneous rate

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of change at noon so our time change there is changing

three to two to one but we’re interested in instantaneous rate change at noon

so the average rate of change again is you subtract the outputs

so t of 15 minus t of 12 over the the interval so noon to 3 is the first one

and then noon to 2 is the second one and then noon to

1 o’clock and so we input those numbers or

in this example we don’t have a function so we use the table

we look up those values what is t of 15 we look those values up

right there so t of 15 we can see right there is 18.2

and so we subtract those outputs and we calculate what the average rate of

change is 1.3 1.5 1.7 so we plot the given data we sketch out

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a smooth curve that approximates the graph and we draw the tangent line

and we estimate that the tangent line is approximately 1.9

so we have 18.3 minus 8 over 14 minus 8.5 so we can say here about 1.9

all right very good so what is the relative rate of change

though the relative rate of change is the ratio we have the derivative

divided by the functional value at that number at that fixed number x naught

so here’s an example square root of x find the relative rate of change at five

and at 75 so let’s see why this would be useful so the derivative

is 1 over 2 square roots of x now at this stage i’m still giving you the

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derivative again we’re going to find how to find that derivative

we’re going to work a great deal of time on that

right now i’m just giving that to you illustrating the relative rate of change

right so we have a function and we have the derivative function

what is the relative rate of change at five right

so we plug in five into the derivative and we plug in 5 into the function

we find that ratio we work that out becomes 1 over 10

or 0.1 to find the relative chain rate of change at 75

now use the derivative again plug in 75 plugging 75 into the function find that

ratio and this time we get 1 over 150 so might remember how the square root

function looks something like that so we’re down here at five

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and we’re over here at 75 and so you can see that

you know there’s a huge difference between say an exponential function that just

grows really fast and the square root function

that doesn’t grow as fast it grows fast at the beginning

but then it’s still it’s still increasing but it doesn’t increase as fast

and so one way that you can quantify that is by looking at the relative rate

of change so at five our relative rate of change is 0.1

and then at 75 our relative rate of change is 0.006

okay so sometimes we’re interested in the relative rate of change

instead of the instantaneous rate of change so

for example let’s say you’re earning 25 000 a year and you receive a 5 000 raise

what if you receive a 5 000 raise but instead you’re making a hundred thousand

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dollars a year right so the relative rate of change here is important that’s the

one that’s going to say if you should be happy or not

you want a five percent increase or twenty percent increase [Music] okay so

the average rate of change find the instantaneous rate of change

and find the relative rate of change all together

we got this function here 2x squared minus 3x plus 5

and we’re looking so to find the average rate of change we need an interval

to find the instantaneous rate of change we need a number

find the relative rate of change we use the same number so we can find the

average rate of change from two to four subtract the output values we get 9

use the derivative substitute into the derivative function

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and we get the instantaneous rate of change

and then the relative rate of change at 2. so we looked at all three for this

function right here the only missing part to all this is how

do you find that derivative okay so before we do that we’re going to

work on some more tangent line problems so we’re going to actually solve the

tangent line problem now before we move on so if we have the derivative at some

number a so for example the a in this problem

right here was 2. right so we have the derivative at some number

we can find the equation of the tangent line all we need to do

is to find that derivative at a that’s going to be the slope so

again to find the equation of the tangent line or to find any equation of

the tangent line any equation of the line we need a slope

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and we need a point so we have a point here the point is

a f of a and we have the equation of the line now because we have the slope

all right so we can look at that diagram to see

how all that works perhaps it’s better if i just

reconstruct that for us rather than seeing all that at once

so let’s do that here so we have here the axes let’s say and we have an a

and we’re going to have a let’s say we have a curve going through

here and we want to find the tangent line here so we have

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a we have f of a right here and this line right here it we have the

equation for it is y equals the derivative at a that’s the slope

and we’re going to have an x minus a and then we’re going to have plus f of a

plus f of a right here so the only thing that we’re missing

from all this is how to find that right there

but that right there is defined the derivative at a or the derivative function

the derivative function is the limit as the delta x goes to zero of the

difference quotient this is what we know right now the derivative

is the limit of the difference quotient so we can

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talk about the delta x’s and the delta y’s and find that limit to find the

derivative function once we find the derivative function we plug in a

and that will give us the slope of the tangent line right there

so this completely solves the tangent line problem

assuming this limit exists so remember from episode 1

we’re asked to find the limit over and over again sometimes

our answer was there is no limit the limit doesn’t exist

so it may happen that the derivative doesn’t exist

you will not be able to plug in a which would just simply mean

there would be no tangent line there okay so

find the equations of the tangent lines to the curve

that are parallel to this line right here so we’re given a function x minus one

over x plus one and we’re asked to be parallel to this line right here

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so this line right here has slope one half

right we can find the slope of that line x minus two y equals one

that’s just simply minus two y one minus x divide by minus two

so it’s one half x minus one half right so just simply divide by minus two

so we can see the slope right there is one half that’s the important part

so that line right there we need to be it’s asking us to be parallel to that

line so we need to know the slope of that line it’s one half

and we’re going to use this with the derivative of this function right here

all right so there’s the derivative and i’m giving that to us in this problem

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the derivative is 2 over x plus 1 squared we don’t know yet how to find that

derivative and we’re going to work on that in a great deal in a moment

but what we want to know is what are the x’s where we’re equal to

that slope remember we have to be parallel to that line so we need our

derivative equal to the one half and we need to go solve that for x

so solving that for x we’re going to get one and negative three we just um

you know cross multiply whatever you want to do

you solve for x you get one and minus three those are the x values where the

derivative is going to be one half and that means those are the x values

where tangent line is going to have slope one half

and now i use those x’s to find the y’s so i plot plug each one of those x’s

into the function x minus one over x plus one plug in the 1 for example

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we get out 0 plug in the minus 3 and we get out of 2. so those are the

places where we’re going to have a tangent line and

we can find the equation of the tangent line because we already know the slope

is one half because we’re parallel to that line so

the equation of the line is going to look like

y equals one half x plus b we don’t know what the b

is yet so for each of those tangent lines we have those

points of tangency and we’re going to use each one of those to find the b’s to

find the y intercepts [Music] so using 1 0 1 for the x 0 for the y

we calculate the b using minus 3 for the x and 2 for the y we’ll

find the other b and so now we have our tangent line equations

using the one half and then the minus one half for the b

00:44

and using the plus seven over two and so there you can see in the red is the

function x minus one over x plus one i also put in red the vertical isotope

but there in the blue you can see the tangent lines

you can see the equations you can see the tangent lines there

given by those two equations so those two tangent lines are parallel

to the line x minus two y equals one all right so i hope that example there

was fun for you is interesting let’s do another one like that [Music]

so this time we’re going to be asked how many tangent lines

pass through this point here so just to get a feeling for um what the

00:45

problem is asking let’s look at say the graph the graph will look something like

this so at minus one we have a vertical isotope

and at y equals one we’re going to have a horizontal isotope

the graph is going to look something like this and

something like so it’s going to go through 0 0 right it looks something like

this and you know where’s the point one two is

the point one two even on the graph x over x plus one right

when i substitute in let’s call that f of x

when i plug in one we’re gonna get one half

00:46

and we’re interested in the point one two

so the point one this is right here and this is one half

and where’s one two so so this is y equals one

this let’s say one two is up here so that’s the point here 1 2. [Music]

so we’re looking for the tangent line it has to go through this point and it

has to be tangent to this graph so you know something like this might be tangent

we can put infinitely many um lines through this point

we want to find the one tangent line through this point

that is maybe the tangent line is over here you think

or maybe it’s through here or maybe there’s multiple of them

00:47

so the question is how many right all right so here we go

um and which of these points do the tangent lines touch the curve okay

so in other words don’t just say how many of them

are there but where all right so all tangent lines must pass through the point

so remember to find the equation of the line

we need to know the slope and a point well this time they give us

the point the point is 1 2 and we don’t know the slope of the line though

so i’m telling us though that on this problem for our function x over x plus one

our derivative is one over x plus one square now at this point

maybe you’re just saying how do i actually find

that derivative i mean are we ever going to get to it right

00:48

so for those of us who are impatient let’s go real quick and take a break

from this problem and figure out how to find that derivative right there

how do we know the derivative is one over x plus one squared

starting with our function is x over x plus one

so what is the derivative right so here’s our original function

so let’s recall that the derivative is the limit um as delta x goes to zero

of the difference quotient x plus delta x minus f of x all over delta x

all right and so we can find this limit here let’s go down here equals

the limit as delta x goes to zero and let’s see here we need to plug x

plus delta x into our function everywhere so i’m gonna plug in here

00:49

x plus delta x x plus delta x plus one so that’s for this part right

here i plugged in delta x here i plugged in delta

x plus delta x here and here and then we have minus

and then we have x so we have f of x is x over x plus one all over delta x

all right so very good so there’s our first step

and in order to find this limit delta x is going to zero

so delta x is going to zero here and zero here

and so we’re going to get zero because we’re going to get x

over x plus one minus x over x plus one and then this is also zero so we’re

gonna have to manipulate this and cancel some delta x’s so to do that

00:50

we’re gonna get common denominator so let’s do that perhaps down here on this

area here so i’m going to say this is x plus delta x over

x plus delta x plus 1 times x plus one over x plus one

minus and then this is x over x plus one times x plus delta x plus one

over x plus delta x plus one and then i’m going to say all of that

times one over delta x all right hopefully this will work

okay so now we have a common denominator here and here

so this subtraction right here is separating this term

and this term from each other and so now we have a common denominator

00:51

so the common denominator is x plus one times x plus delta x plus one

and let’s see what we’re getting here so if we multiply this out we’re going to

get x times these two right here so we’re going to get an x squared

plus x times delta x and then 1 times x and 1 times the delta x

and then let’s not forget the we’ll do that in a minute

let’s say now we have minus and now we’re going to say x times

x we’re going to have x times the delta x with a minus

and then the x times the one with the minus and that’s good we just have

x times each of those all right so parentheses delta x all right very good

00:52

so let’s see here what’s happening the x squares add up to zero

the x’s add up to zero and we’re going to get a what’s left

we’re going to get an x plus x times the delta x here

and so that one added up with that one to zero

we’re going to get it delta x here and this was an x times the delta x

so let’s just write all this out here x times delta x

and then all right those two added up those two added up

so we’re getting here left a plus delta x that’s an x make that better so

x delta x and then a delta x and then a minus

oh actually those add up to zero don’t they so here in the top aren’t we just

getting the delta x here delta x here let’s just double check that those

cancel those add up to zero those add up to zero those out we’re just getting

00:53

delta x so delta x over the delta x those are going to cancel out so we end

up with just one over x plus one times x plus delta x plus one

all right so again you can check the top all of it just gives us the delta

x and then we have delta x over delta x so those cancel each other

everything adds up to zero except that one and that one cancels with that one

so we just end up with a one here now the limit’s going to zero here so

we’re going to get one over x plus one square

that simplifies a lot so this was the derivative

all right so the derivative is here let’s go back to our problem

all right so there’s our function right there x over

x plus one and our derivative is one over x plus one squared now

00:54

we’re going to see better ways of finding the derivative later but right now

that’s our derivative and i showed you how to find it using the limit of

a difference quotient using the definition of the derivative

all right so to continue on with our tangent line problem

there’s our derivative so since we’re looking for the intersection

the point of tangency the point that’s on both the graph of the function

and the tangent line we’re going to try to eliminate y

so we’re going to say y is x over x plus 1 that’s our given y

and we’re going to say that you know our equation of the line is y

minus 2 right so i’m going to put the y i’m going to put the plus 2 on the other

side so in other words y is equal to m times x minus 1 plus 2

and then for the m i’m going to use the derivative there

00:55

and solving for x we’re going to be able to say x is minus 2 plus or minus 3.

so there’s going to be two points of tangency one for the

minus two plus square root of three and then one for the minus

two minus square root of three and so those are the two tangencies there

and if we were to sketch the graph there we could see the

function that we drew earlier x over x plus one

and we can see the tangent lines there and we have the exact coordinates for

those right there okay so one last tangent line problem i think

and so we’re going to find equations of both tangent lines at

the point that are tangent to this parabola

so all tangent lines to this point have the form

00:56

right so this is the point that we’re given and we’re going to be looking at the

tangent line we’re going to be looking at a line so all we need is the slope so

i’m going to be giving us the derivative there the derivative is 2x plus 1.

um should we go find that derivative and

make sure that derivative is right first well let’s see here we have y equals

um x squared plus 2 x squared plus x and what is the derivative so the

derivative is if we call this f the limit um last time i used the delta

x but sometimes people want to use an h let’s say h goes to zero and we have

here f of x plus h minus f of x all over h that’s just the definition

00:57

so let’s plug in x plus h into our function

so we’re going to get x plus h squared and then plus the x plus h

right so wherever there’s an x i put x plus h here and i put an x plus h here

now we have the minus sign and then now we’re going to do f of x

here x squared plus x and then all over h so there’s our derivative there

if the limit exists let’s see what we’re going to get

so let’s square that out that’s x squared plus 2 x

h plus h squared and then plus an x and then plus an h and then minus the

x squared and minus the x and then all over h

00:58

all right so we have the limit of all that just expanding it out

with the hopes of it simplifying of course so we have x squared we have x

squared those add up to zero the x’s add up to zero

so let’s see what we’re going to get left we’re going to get a 2xh

plus h squared plus an h and that those combine to zero

those add up to zero and then all over h so this has h h and h so let’s

factor out the h and what do we have two x plus an h plus a one all over h

factor out an h h very good and now we can cancel those h’s

so we’re going to get two x plus h plus one

00:59

and now we can find this limit sorry two x plus

two x plus h plus 1 right cancel those h’s

now we can substitute in the h is zero the h is going to zero

so this is going to 2x plus one so there’s our derivative there

using this simple computation right here there’s our derivative right there

all right so now let’s go on to our problem so all tangent lines um pass through

that point because we were given that point and told

they have to pass through that point so we know some

information about our tangent lines they have to go through that point so

we’re going to use the y minus a negative 3 or y plus 3

and then the m and to find the m we’re going to use our derivative 2x plus 1.

01:00

[Music] okay so we’re looking for the intersection so we’re going to eliminate

the y so we have y equals from two different places

from the line equation and then from the given function equation

x squared plus x solving for x we can find the minus one and the five

so there’s two tangent lines for each of those x’s we plug into the original

function we find the y’s and then we have the tangent line equations

so the minus x minus 1 and the 11x minus the 25. all right so looks good

we can see the parabola there and we can see the point there

um two minus three and we can see where the lines are tangent [Music]

01:01

all right very good so now let’s talk about um [Music]

horizontal tangent lines so if you know the derivative is zero then the

tangent line is horizontal because if the derivative is zero

right so let’s just look at the tangent line equation again

right so for given the point a f of a we’re going to say f minus f of a

the derivative at a and then x minus a so there’s the equation of the tangent

line now if this is zero right here and we have y minus f of a

equals zero in other words f equals f of a

so whatever that number is f of a that output

we’re just gonna have a horizontal line so for example y equals three or y

01:02

equals five it’ll just be a horizontal line so something like this

coming in like this and then like that and then right here we look

and the tangent line is horizontal right there so for example five y equals five

we find that the derivative is zero there and we have a horizontal tangent line

so here’s an example uh for what values of x does this graph

have a horizontal tangent line so to find that horizontal tangent line we’re

going to need the derivative where is the derivative 0

so to find the derivative of that well i’m going to give that to us the

derivative is six x squared minus six x minus six

now to find that derivative using um a limit it’s not gonna be fun

01:03

it’s the third power and then we have a quadratic term

and well there’s going to be a much faster way to find that derivative i’ll

show you very shortly but to find the horizontal tangent lines

we need to know where the derivative is zero

so we’re going to solve where is the derivative equal to zero

we’re gonna find those x values so simplifying that quadratic down a little bit

using the quadratic formula we saw that and we find the x’s

we find two x’s actually now may happen that a function doesn’t

ever have a horizontal tangent line for example something like this

y equals x to the third this has no horizontal tangent line

if the graph curled over and started to decrease

then we would have a horizontal tangent line but this function never does that

so in this example that we’re seeing here that quadratic equation may or may not

01:04

have a solution in this case it does and it has actually two solutions

so there’s two places where we have a horizontal tangent line

so there’s the value of x’s where there’s tangent lines

horizontal tangent lines and you can plug those each of those

x’s into the function and you’ll get the output

and those will be the tangent lines but in this example i just said

for what values of x it didn’t bother to ask you to go find the y’s

all right so here for what values of x does this have a horizontal tangent line

another example so to find the horizontal tangent lines

again we look where the derivative is zero

so we compute the derivative and we need to solve where the derivative is zero

to find the horizontal tangent line so this time the quadratic formula tells

us again we have two values minus one third and one

01:05

and so those are the values where we have a horizontal tangent line

so there’s two examples there quick break [Music]

okay so um now we’re going to talk about the derivative

and we’re going to talk about the derivative as a function

and we’re going to see how to find the derivative um a lot more and

so you know the derivative is the first main idea of calculus um that really has

changed the world limits were put in place to better understand the derivative

at least historically so derivatives derivatives derivatives all right so a

limiting process can be used to study curves in general in other words how to

study the behavior of a function around a number uh but in general speaking the

01:06

derivative is the idea that led to the development of calculus

so we have the difference quotient and we have the derivative

so it’s very important to keep in mind that the difference quotient

does not have a limiting process you put the limit of the difference quotient

and that is the derivative sometimes i write it with the delta x

sometimes i write it with an h of course provided this limit exists if

this limit doesn’t exist then we say the function is not differentiable

now we have lots of ways of writing the derivative sometimes we use dy over dx

sometimes we use y prime sometimes we use df

over dx or sometimes we use the capital d with the subscript the subscript is

denoting the independent variable all right and so sometimes we use dy

01:07

over dx and that’s the limit of the slope or the

limit of the difference quotient so there’s three three ways of writing

it out there and there’s the fourth way so the two outside the d y over d x and

f prime those are obviously notation that is very quick and easy to write

and the limits on the inside are the ways in which you define the limit

and which ways you find the limit the process of finding the derivative is

called differentiation and finally we say a function is differentiable

on an interval or add a number x if the limit exists

if we’re on an interval then it has to the derivative has to exist for every

number in the interval okay so now we’re going to illustrate

01:08

how to find the derivative and we’re going to look at some very not

easy examples right we’ve already done some

examples but now we’re going to do some some some examples here

so let’s do x to the third and we’re looking at the point one one

so let’s find the derivative so the derivative is the limit

i’m going to use an h and so it’s f of x plus h minus f of x all over h

all right so now i’m going to substitute in x plus h

into the function so the function here is x to the third so here we go limit

01:09

h approaches zero we’re gonna have x plus h to the third minus

x to the third all over h okay and let’s see if we can write that out over here

so x to the third um or x plus h to the third what is all that

if we multiply that out maybe you might have to do some

some additional scratch work down here but i think i can get it and it’s going

to be x to the third plus 3 x squared h plus 3 x h squared plus

h to the third and then we still have a minus x to the third all over h

okay so the x to the thirds cancel out so we have limit of three x squared h

01:10

plus three x h squared plus h to the third all over h

now we can factor out the h’s let’s do that over here actually

in the numerator we can factor out the h’s h h and h so we’re going to have an h

and what are we going to have left 3x squared plus 3xh

and then we have h squared all over h so h is going to go to 0 here and we’re

going to have 3x squared plus 3xh plus the h squared

h is going to zero so that’s going to zero that’s going to zero

01:11

so we have three x squared left so there’s our derivative

there’s our function it would be nice to go from here to here

um but to do that we should probably do some more examples first of finding the

derivative using the definition but if you’ve seen the shortcut

if you’ve seen the differentiation rule then what you can say is bring the 3

down and reduce the power by one and there’s the derivative

much quicker much faster so we’ll be able to take the derivative

of this right here when we get to our differentiation rules

we’ll be able to go much faster but for now we need to

understand all these steps here and we’ll see why we need to understand all

these steps here in a second but in this case here’s our function

here’s our derivative and we were asked to look at the point one one

01:12

so if we substitute in one into our function over here we get out one

um so we have the point here one one and what’s the derivative at one so it’s

three so the slope of the tangent line at the point x equals one

is three we could go find easily find the equation of the tangent line

if we wanted to it would be y minus 1 and then 3 and then x minus 1.

so i use this one right here for the y and i use this one right here for the x

and there’s the slope of the tangent line so this is an equation of the

tangent line the equation of the tangent line is easy

once you know what the derivative is okay so let’s look at um this example

there’s our limit that we just computed here when i typed it up i used a delta x

01:13

instead of an h and we get the 3x squared out of all that

and then we can substitute in at x equals one

um that should be f prime at one equals three we found it to be three

okay so this right here should be a three okay so let’s go on next one

find the derivative of the function using the definition

and this time we’re going to use the cube root

here cube root when you substitute in an 8 you get out of 3 the cube root of 8

01:14

is a 2. so let’s find the derivative and there it is

oh don’t look let’s go do this so our function is cube root

and we’re looking at x equals eight all right so what’s our derivative

it’s the limit as we approach um i’m going to use the h so we’re looking at uh

x plus h minus f of x all over h and so this is our definition so i’m

going to plug in x plus h into our function so we’re going to get the limit

of let’s see here cube root of x plus h minus and now i’m going to substitute in

h into our function so cube root of x and then all over

01:15

h so we’re getting the limit as h goes to zero of cube root of x plus h

minus cube root of x now to find this limit here we’re going

to need to use some type of conjugate so let’s get the same thing going x plus

h cube root minus cube root and h and i need a conjugate here

and the whole idea is to make the h’s cancel

i mean because right now we cannot find this limit if h is going to zero

we’re gonna get cube root of x minus cube root of x that’s zero

and on bottom we’re also gonna get zero so to get away from that we’re going to

need to have cancelling the h’s out now to cancel the h’s out

obviously we need to do something dramatic to the numerator here we need

to rationalize it we need to use the conjugate

01:16

so we’re going to use cube root of x plus h squared plus cube root of x plus h

times h and then plus cube root of x squared x plus h squared plus

cube root of x plus h plus cube root of x squared

so i’m not going to mess with the denominator at all

so i’m going to just rewrite the denominator

so it’s just h times all of that h times all of that

01:17

okay so what happens in the numerator so we’re multiplying these two here

we’re going to get cube root of x plus h here we get cube root of x plus h

squared so we multiply those we’re going to get x plus h out

we’re really getting cube root of x plus h to the third so that’s x plus h

now when i multiply these two right here i’m going to get cube root of x plus h

squared times the h and then here i’m going to multiply here

and get these two right here cube root plus cube root of x plus

h times cube root of x x squared and so we multiply

01:18

this one times those three we’re getting those three right there

now we’re going to multiply this one times those three

and so we have h here we need an h here and we need an h here

i forget to write those down obviously we have to multiply top and bottom by

the same thing now i’m going to multiply cube root of x

times each of these three and i don’t have enough space here so

to me it’s worthwhile of backing up here and making this look better

so again this is an h here and this is an h here we’re going to see why here in

a second but i’m going to write the same

denominator down so this is cube root of x plus h squared

plus cube root of x plus h times h plus cube root of x squared h for all that

01:19

that’s the denominator now i need to do cube root of x plus h

times each of these three so the first one we’re going to get x plus h

and then now we’re going to get plus cube root of x plus h but with a squared

times the h plus cube root of x plus h times the x squared

plus now cube root of this one cube root of the x times x plus h squared

so you see that these two are going to cancel with each other um

we’re going to get one two three and we’re going to get this one

and that’s supposed to be a minus right there yeah those two cancel with each

other and then these two right here when we

01:20

multiply those that’s going to cancel with um that’s going to be the x

and then the h here so that will give us minus cube root of x plus h h with an x

all right that’s really kind of messy let’s go over here and

type this up here uh it looks so much better

all right so there we go let’s look at that very carefully

we’re gonna have x plus delta x so i’m looking at the second line there

and you know we have the conjugate there which would be the cube root of

x plus h squared plus the cube root of x plus delta h or delta x times x

plus the cube root so i use that numerator and denominator

01:21

then when we expand all that out the x’s are going to

cancel out and the delta x’s are going to cancel out so that looks

written up a lot better and it’s very easy to see there

that this is going to work because when we multiply those three out

and then we multiply the cube root of x times those three

then we can see that they’re going to cancel out

then the last step is where the delta x is going to zero

and so now we can substitute into zero there and we’re going to get a cube root

of x squared and then in the denominator we’re also going to get a cube root of

an x squared and then the last one is the cube root of x squared

so we’re going to end up with three of them three cube roots of x squares

and so then that’s how we get the three there in the denominator

01:22

and then the last step would be to substitute in the eight

so if we substitute in the eight there into the derivative

we’re going to get a one twelfth out so we’re going to square that 8 get 64

and then take the cube root of the 64 and get the 4 3 times 4.

all right so there’s that example good um now to refresh our memory on the

tangent line equation and this theorem the way i wrote it up instead of an a

i used x naught as a fixed number and so there’s our tangent line equation

at a given point or given number x naught so by definition the derivative is the

slope of the tangent line and so we’re gonna have

01:23

line passes through the x naught f of x naught

and that’s the equation of the tangent line and so this is the proof here

the proof is very simple simply because of what the derivative

is the derivative is the slope of the tangent line

all right so let’s do another example here so we have three times

the square root of minus x so let’s see if we can find the derivative here

and we’re looking at x equals -2 now there’s two ways to do this we can

find the derivative function and then we can plug in minus two but i

would like to show us also another way so first let’s stick to

01:24

the way that we know which is the limit as h approaches zero

of f of x plus h minus f of x all over h and so here we go limit h goes to zero

this is going to be three times the negative x plus h minus three times

square root of minus x all over h and so here we’re going to

use the conjugate again i’m going to say 3 times x x

plus h and then we’ll use a plus here and then 3 times minus x here

there’s the conjugate so finding the derivative here

01:25

i’m going to use the conjugate of this so in this denominator here um i’m just

going to say it’s h times all of that so this will be 3 negative x plus h plus

3 times square root of minus x and then close parenthesis

all right so that’s just the denominator h times all of that

but in the numerator we’re going to get 3 times this so this will be 9 times

the negative x plus h and the mixed terms will add up to zero

and so we’re then we’re going to get minus nine times minus x

01:26

so here we’re going to get a minus 9x a minus 9h and a plus 9x

all over h times all of this again so we can see the minus nine x’s or the

x’s add up to zero the h’s are going to cancel and we’re

going to get a minus 9 out of that so the we’re going to get -9

over oops still have limit so the x’s add up to zero and the h is

cancelled so we’re going to end up with a minus nine

and then we have the threes come out so we still have a square root of minus

x plus h and then plus the square root of x

01:27

so we’re gonna get minus three over h’s are going to zero and we’re gonna

get two of these so there’s our derivative function so we’re gonna get here the

h’s cancel so we get minus 9 and then so we cancel the h’s so we’re

going to get 3s out of here we can factor the 3 out

and that’s how we get minus 9 over 3 so we’re going to get a minus 3

and then we just get that and that now when h is going to 0

we’re going to get square root of minus x and we’re going to get another square

root of minus x so we get 2 square roots of minus x’s

and so here’s the derivative function here minus three over two square roots of

minus x’s all right so let’s see if we can go on with our problem

01:28

oh wait i wanted this to show us one more thing though first so

perhaps this isn’t a good example to show it but sometimes you want to find the

derivative at the number two at the same time

instead of instead of plugging in minus two at the end

so in that case we would just have a minus two here

and we would carry that minus two for the x wherever we went

and we could put minus twos wherever the x’s are

or we can plug in the minus two here at the end

in this example i think it’s probably just easier to leave an x everywhere

like we did rather than plug in minus two that would have just been

more writing perhaps but in any case now it’s time to use the minus two

so there’s our derivative that we found as you can see typing it up i show less

01:29

work but that work is just fine because we’re just

factoring out the three everywhere and using the conjugate

in fact looking at the work typing it up

i simplified it because when we did this right here i said the conjugate was

3 and a 3 and so then i used three and a three into three and three and i didn’t

really need to do all that so typing it up i realized oh let’s just

factor out the three first and then let’s find the limit of the

rest of that so the conjugate was simpler

so the derivative for any function right less x less than zero

so we substitute in minus two into the derivative

and now we get our slope so it says find the slope of the tangent line

so the hard part is finding the derivative once you find the derivative

you plug in your number you get out the number

01:30

you have the slope of the tangent line so now we can find the equation of the

tangent line by simply using that slope and using our point

our point is x is negative two and so from that we can substitute in right

if we substitute in minus two we’re gonna get three times the square root of

negative times negative so we get three square roots of two

all right so let’s do another one find the equation of the tangent line

we’re looking at the graph or we’re looking at the function here g of x is

one minus x two x plus one so here we go the derivative g prime

01:31

is the limit as h approaches 0 of g of x plus h minus g of x all over h

and let’s see if we can at least do the first step right here

so we’re going to get the limit as h approaches 0.

so we’re going to plug in x plus h everywhere it’s going to have 1 minus

x plus h over two plus x plus h and then minus g of x which is one minus x over

two plus x and then all over h so there’s our first step there’s our

limit limit as h approaches 0 and we’re getting the difference of the g’s over

the h okay so here we go next step well we’re going to need to get a common

denominator here so here we go limit as h approaches 0

01:32

and we’re going to have 1 minus x minus h 2 plus x plus h times

2 plus x over 2 plus x minus 1 minus x over 2 plus x times so 2 plus x plus

h over 2 plus x plus h and then all that times

h is in the denominator so 1 over h so there we go there’s our next step there

all right so now we have a common denominator here and here

so we’re going to get 2 plus x times 2 plus x plus

h that’s our common denominator and let’s see what we’re going to get in

this numerator here so we’re going to need to expand this out so that we

can see what adds up to zero so because you know when we substitute in

01:33

h is zero we’re gonna get the same thing

oops minus the same thing it’s gonna end up being zero of zero

so you know we need to cancel the h’s out let’s see if we can do that so two

times the one two times the negative x two times the negative h

x times each of those all right now here we go for these so we

have a negative here so we got to keep that in mind

so we have 1 times 2 so we have negative 2 negative x

negative h and then we’re going to be multiplying by a positive

x multiply these three by a positive x so we’re going to get 2x h squared hx

xh and then all divided by the h so we’ll put the h down here there we go

01:34

now let’s see if we can get this everything without an

h should cancel this 2 should add up with this

minus 2 this minus 2x should add up with this 2x so let’s factor an

h out and see what we end up with and the bottom let’s just write that

down again i just wrote the h first and the top

everything without an h should add up to zero so everything should have an h

left so i’m going to factor out that h at the same time

so cancel cancel i’m going to have a minus 2 left

cancel cancel i’m going to have a minus x left cancel cancel minus 1 left here

cancel cancel and then i’m gonna have an h and x left

so actually the x’s add up to zero also here

01:35

and so let’s see here what do we get on top here

so on top we’re going to get a minus three aren’t we

so we’re going to get a minus three right there those add up to zero the h’s

cancel we end up with a minus three over the two x plus h and the two x plus h

and so this limit here is um let’s find the limit now so

we’re going to get minus 3 over x plus x sorry let’s just say it’s x plus 2

and then we’re going to get times each other so it’s squared so this is g prime

all right so very good and now we need to look at

01:36

minus -1 so we can plug in the minus 1 and we’ll get the slope of our tangent

line let’s do that over here so there’s the limit that we just calculated

i’m using a delta x instead of an h so i plug in the x plus delta h

minus i plug in the x and so we get difference quotient there

we have a lot of simplification but in the end we get the same

minus 3 over 2 plus x squared now to find the equation of the tangent

line we plug in -1 into our derivative we get out minus 3 and

the equation of the tangent line there is given to us by our formula

01:37

so for our for our derivative our slope we’re gonna use the minus three

all right looks good all right so now let’s talk about when

the limit doesn’t exist so far in all the examples for example this one

right here the the limit existed the limit of this right here

oops i forgot an equal sign the limit of this right here

um exists it’s equal to this but sometimes the derivative limit the

limit that we’re using to find the derivative sometimes that limit doesn’t exist

so in this example we’re going to give three

functions where f is not differentiable but it is continuous

so we have the absolute value function let’s recall what the absolute value

function looks like and let’s see why it’s not differentiable

01:38

so if we look at the absolute value function here

you know another way of writing the absolute value function is x

and then minus x x if x is greater than equals zero

minus x if it’s less than zero this is another way of writing the

absolute value function but anyways the absolute value function looks like this

right recall it looks like this and so when we’re looking at zero we’re asking

what is the slope of the tangent line right

so how can we say there is none right if we look at this part right here the

slope of the line that goes through there is just one

we look right here the slope of the line that goes through that point is just

minus one what about if we look at zero right so i’m going to look um

actually let’s not look at that let’s look at

01:39

trying to find it on our own so what happens if we try to find the derivative

function for this so we’re going to say the limit as h approaches 0

of x plus h minus f of x all over h i mean just like we did

before right it’s just the limit of the difference quotient what

could possibly go wrong well when i’m looking here at h is going to zero

so there’s you know there’s two ways we could go to zero we could go to zero

from the right or from the left but you know when i use this f of x right here

which piece do i use do i use the x or the minus x right here

so you know we’re because we’re looking at f prime at zero so

that’s gonna be f of zero right there which will be the zero so this would be

01:40

zero right here right in fact let’s just go ahead and try to do that

this is the definition right but what’s f prime at zero

right because that’s what the question was asking the question was asking

is a differentiable at x equals zero so at zero the definition just becomes

zero plus h so that’s just f of 0 now all over h so the derivative at 0

does this exist question mark so this is zero right because if we plug in zero

we get out zero but how do we find this one you see because zero plus

01:41

excuse my zero zero plus h is well depends on what h is if h is point one

then i’ll use the top piece here i’ll use point one if h is negative point two

then i’ll have to use the bottom piece here

so we really don’t have a way to do that but what we can do is we can take the

limit from the left and the limit from the right at zero

and then we’ll be able to use this right here

so let’s do that right here let’s say here’s the limit from the left

and here’s the limit from the right the reason why this matters is because

if h is approaching 0 from the left then this is negative right here and so then

i’m going to whenever i put an x in i’m going to say this is minus x

01:42

so this will be putting in an h i’m going to use a minus h

so this will be the limit as we approach h from the left of h

over h which is one um but this will be a minus h sorry and

so that’ll be a minus one and so this right here will be the limit

as we approach h from the right and this will be h and then minus zero

so this will be h over h and this is one and that makes

sense because as we’re approaching from the left

the slope of the tangent line is minus one if we’re approaching zero from the

right the slope of the tangent line is one so these make sense

but what these tell us is that the limit from the left of the difference

quotient and the limit from the right of the difference quotient

the limits don’t agree the one-sided limits don’t agree

so that means this does not exist now when you say the limit of the

01:43

difference quotient doesn’t exist or the limit

in the derivative definition doesn’t exist you just say

the limit or the function is not differentiable

so when i look at the limit from the left here i get minus one

and when i look at the limit from the right here i get one

and so the two-sided limit doesn’t exist so the derivative doesn’t exist this is

the limit of the difference quotient right so the two-sided limit does not exist

so when that happens you say the function is not differentiable

this is an example of a corner we look right here

this is a corner right here it’s a sharp corner

and whenever you have a sharp corner like that the limits on both sides

have problems they don’t equal and so then the function is not differentiable

so this is one type of the example of a function that’s not differentiable

01:44

it’s because of a corner point so we can summarize this our

argument right here we summarize this whole arg argument right here

by saying we have a corner point now saying that i have a corner point

is very easy to memorize oh there’s a corner point there

but what you really have to do is to know the calculus

why is it a corner point what do we mean by corner point

so we have to know the definition of the derivative

and when we look at those one-sided limits we realize they’re not the same

for some reason and so the two-sided limit doesn’t exist

so it’s not differentiable all right so let’s look at another type of example

so this time let’s look at the cube root of x squared

it’s also not differentiable at zero let’s see why

so let’s look at this cube root of x squared

01:45

at x equals zero so that’s cube root so let’s just go find our derivative

like we normally would so cube root of x squared at x equals zero let’s

go find our derivative so it’s the limit as we approach h of cube root of

x plus h squared minus cube root of x squared and then all over h

okay now again if we substitute in zero we’re going to get cube root of x

squared minus cube root of x squared we’re going to get zero over zero

so we’re going to need to use the conjugate again so here we go

cube root of x plus h squared now we could go mess with that conjugate again

01:46

but probably it’d be easier to take advantage of

we’re looking at zero so if we’re looking at zero

where’s the derivative at zero so rather than going to

that mess again let’s put 0 here and that means we have 0 here so 0 plus

h and that means we have a 0 here and so let’s try to find this limit and

now we see we don’t need the conjugate because this is a zero right here so

this will be cube root of h squared over h and we can simplify that

to one over cube root of h and this is what is this limit

01:47

so this is a problem here and in fact i want to look at both sides

here because i’m not sure what’s happening here

so i’m going to look at from the left and from the right and these are h’s

h is going from the left and from the right cube root of h

so i’m not sure what’s happening here with this so i’m going to look at these

instead so i have a fixed number and this h here is going to zero

the denominator is getting really small which means the whole fraction is

growing really large but if i look at one sides then i can

find out what these are if h is negative right here like

negative point one or negative point zero zero one

then that’s growing larger and larger but they’re negatives

so this is going to be minus infinity and this one’s going to be positive

infinity now the h’s are positive so this is going to be positive numbers

01:48

here but these are growing really really really small right here so one over

that’s going to go really large this is unbounded

so this one right here does not exist this was the derivative right here

so we just started trying to find the the derivative at zero

and we you know use zero plus h and then zero and then over h so this

is the derivative at zero here but it doesn’t exist

because the one-sided limits are not the

same eleven from the left and limit from the right

now if we were to go graph this function we could see what’s happening better

so let’s look at the graph down here so this is y equals the cube root of x

squared and so if we look at this function right

here what’s happening is it’s coming in right here

and it’s coming in near vertical but it’s not vertical

and then it’s coming back out like that so if you

01:49

were to find the equation of the tangent line you would have a positive slope

and then you would have a positive slope but the closer you’re getting to zero

right we’re looking at the derivative at zero what’s the tangent line when we’re

really close to zero you see those tangent lines

the slopes are getting larger and larger and larger when

we’re at zero right so this is something called a vertical tangent

the tangent line is just the vertical line

so we would say the tangent line is just x equals zero

but if the function is not differentiable the function is not

differentiable at zero the reason why the function is not differentiable zero

is because of a vertical tangent now that’s important to remember it’s a

catchy phrase vertical tangent but again you have to understand all of

01:50

the calculus we went and tried to find the derivative at zero

and we found a limit that doesn’t exist and the reason why didn’t exist

is the limit from the left and the limit from the right are not equal to each

other all right so very good so we calculate the derivative from the

left or the limit from the left of the difference quotient

which is sometimes called the derivative from the left

and we get these two different values for these limits

so this proves the two-sided limit the derivative does not exist so

this is an example of a function that’s not differentiable

it’s called a vertical tangent okay so here’s another example

i’m looking at this piecewise function here two lines are pieced together

and we’re looking where it’s pieced together now remember in the last video

01:51

we looked at piecewise functions a lot and we asked the question where is the

function being pieced together and is it continuous there is there a jump

is there a hole or something like that so now we’re asking

the similar question this function is pieced together at zero

is the differentiable there is it smooth there or is there a sharp corner

right so we can look at the derivative from the left

at zero and the derivative from the right so this is the derivative from the

left and why is it from the left well h is approaching zero from the left and

we have the difference quotient i’m using for my x equals zero so i’m

using zero for my x and we get one and you can kind of see

that if you look at the piecewise function there the derivative

of the top piece the x minus one is just one the slope of that line is one

01:52

and if we look at the derivative from the right we get two

so the two-sided limit doesn’t exist so this is how you show that something

is not differentiable by using the definition

okay so this is going to be an example of a corner point

all right time for the next break okay so now we’re going to be looking at

differentiation rules we have um looked at these um

limits of these difference quotients enough and now we’re going to be looking

at some differentiation rules um but we do have one more thing to do

before we look at the differentiation rules and this is a very nice

01:53

theorem right here it says the function is differentiable

than it is continuous and then we’re going to be looking at the

differentiation rules here so now there’s two ways we can write this down

if f is differentiable in other words the derivative exists if

f is differentiable then f is continuous so how about if we try to write it down

like this if f is continuous then f is differentiable

so this is the one that’s true and this is the one that’s not true

so if f is continuous then f is differentiable why is this not true

well the quintessential example is the square root function i mean the

01:54

absolute value function so this function right here is continuous at zero

but as we saw earlier it’s not differentiable at zero

so you cannot play around with these words in such a way

we’re going to show in a moment that if a function is differentiable

then it must be continuous but you cannot just switch the two and say the

same thing it’s not true now here is another way that you can say it though

if f is not continuous then f is not differentiable

so this is called the contrapositive so both of these are correct

um but the other way that i wrote down here a second ago is not correct

all right so this is just the contrapositive of this statement in other words

if we can prove this statement is true we’ll automatically know this one’s true

01:55

because the contrapositive so how do we prove that the function is continuous

if a function is differentiable then it must be continuous

so we’re assuming the function is differentiable now

and so what that means is that limit right there must exist

the two-sided limit right there exists that’s what it means to be differentiable

and so we can use the product rule for limits and so what we’re going to do is

we’re going to look at this limit right here and we’re differentiable at c

so we’re going to put in here a delta x and then a delta x so

that’s just multiplying by one but we’re going to use the product rule here so

we’re going to take the limit of the first one and the limit of the second one

now the limit of the first one is something in fact it’s the

derivative but the limit of the second one is zero

01:56

right so we have the derivative times zero so that tells us that the

limit of this expression right here the limit as delta x goes to zero

of x plus delta x minus f of c is zero so the first limit if you trace all the

equal signs is equal to zero so that’s what this means

the limit of c plus delta x minus f of c is zero

so just move the f of c to the other side and so

there we have it we have that the function is continuous that’s all we need

okay so now let’s look at uh differentiation rules

we have a whole bunch of them the first one is the constant rule

the derivative of a constant is zero now these are theorems which means they

01:57

can be all proven using the definition of the derivative and

we have a power rule here so the derivative of x to the n is n times x to the n

minus 1. we have the sum rule for derivatives if your function f can be

written out as a sum of two functions g and h then to take the derivative of

the function we just take the derivative of each of those g and h

and add those up derivatives up we have the same thing for difference

and we have the same thing for linearity putting three and four and one together

um well yeah we have linearity rule so if your function can

be written as a linear combination of g and h for any constants a and b

then you can take the derivative and then we have the product rule so if

01:58

your function can be written as a product g times h then we have this

product rule here derivative of the first times the second plus the first

times the derivative of the second and then we have the quotient rule here um

that assumes that the denominator h squared

is defined but actually the function is written as a quotient

so h is already in the denominator there all right so

let’s find the derivative of this function right here f of x

so you might think that the fun the derivative is um bring the three down

and say four thirds and then pi is constant and then r squared

but that’s not true derivative is just zero

01:59

this one is false so the reason why it’s false is because the

r is not the variable the x is the x is the variable

so all of this right here is a constant well all of this right here is a

constant four thirds by r to the third x is the variable so all of this is a

constant what’s the derivative of a constant is just zero so this one is

nothing so the derivative is zero so since this is a constant the

derivative is zero all right next example now this one

right here is a sum and difference of a bunch of functions right here

so i’m going to write this function out as

three x to the fourth minus seven x to the third

and then plus and what’s my power on this x

02:00

x to the two thirds and then minus nine and so what’s our derivative here so

the the differentiation rules say that i can look at each one of these

terms by themselves and take the derivative so that’s what

one differentiation rule said and then another one said you know you can do the

same thing for subtraction so i just need to look at

the diff at the derivative of each of these terms separately

now to look at the first the first one was the power rule

the power rule said to bring the four down and multiply

and then reduce the power by one now take the derivative of the next term

so bring the power down and multiply and i reduce the power by one now the

next one i bring the power down and multiply

and i reduce the power by one so what’s two thirds minus one

02:01

that’s minus one third and now what’s the derivative of a constant

now actually you might think of it like this x to the zero

right x to the zero is just one bring the zero down and multiply

right so the derivative of a constant is just zero so i would say plus zero

so there’s our derivative now obviously that’s much faster than if we had

taken the derivative using the definition

so the derivat so the definition of the derivative of course is very important

if you’re trying to find something for example is not differentiable

or if you’re trying to prove theorems for example

so those are two cases where the definition of the derivative is

extremely important but if you’re looking at functions that look like this

then the derivative theorems which of course rely upon the definition of the

derivative in any case there’s the derivative right there

all right so using the power rule linearity rule and sum rule

02:02

we just take the derivative piece by piece there all right so here’s another one

so let’s go to our function let’s make it look like something that’s

easier maybe so we have what’s the power on that x so it’s three halves

and then this is what’s the power on that x right there

right so we have the two plus a half so right five halves and it’s in the

denominator so i would say minus five halves

so what’s the derivative three x squared

and then three three halves minus one so that’s one half

plus let’s bring that power down so minus five halves x to the

02:03

and then what’s five halves minus one so three halves well minus so we have

minus five halves minus one so minus seven halves

so there’s the derivative right there and then obviously we can rewrite that

in terms of radicals all right so there’s the derivative right there

and then if someone wanted you to rewrite it in terms of radicals for some

reason or another all right so there’s the derivative

again all right now next example now this time we have the product rule

we have the product of two functions so we’re gonna use the product rule

so here we go oops this one all right so here we go the derivative is

so i’m going to take the derivative of the first one and the first one is going

02:04

to be the 6x squared plus 3 and then times the second function x squared minus 3

plus now i’m going to leave the first function alone

times the derivative of the second function which is just 2x

there we go there’s the product rule right there derivative of the first

function times the second plus the first function

times the derivative of the second now could we simplify that sure

we could expand it all out and collect like terms

and do all that but that’s not calculus that’s pre-calculus now the question is

should i do that or not so here’s a rule of thumb i like to to think about

is that ready for the second derivative oops what’s the second derivative well

what happens if you want to take the derivative

02:05

of the derivative that’s called the second derivative

is this something you’d want to take the second the derivative of

i’d have to use the product rule here so you could expand all this out

and simplify it and that’s probably easier to take the second derivative of that

however having said that if we look back at the original function right here

if i knew that i wanted to take the second derivative

i probably would have already expanded that out at the beginning

so if we do that for example our original function here is

you know 2x to the third times the x squared that’ll be 2x to the fifth

and then minus 6x to the third and then plus 3x to the third

and then minus 9x right so it didn’t take us long to expand that out

02:06

it’s just 2x to the fifth minus six x to the third and then plus three x to

the third and then minus nine x so what’s the derivative

so the derivative is 10x to the fourth minus 18x squared plus

9x squared minus 9. so the derivative was actually faster this way wasn’t it

and in fact if i wanted to find the second derivative

that would be very easy also now you might say wait these derivatives don’t

match well if you were to expand all this out

and simplify it you should get this you should get this answer right here

also so in this example although i used the product rule to find the derivative

02:07

and we found the derivative very quickly here using the product rule

you have two choices you know either either approach

all right so what did i do here so i said here’s the

product rule so g is the first function and h is the second function and i have

a product so there’s the product rule derivative

of the first times the second plus the first times the derivative of the second

so we get that right there which simplifies to all that

um seems like that’s missing a 19 minus 18 x squared right there

which we found right here okay so um what if we have three terms

find the derivative of this now we can use a alternate version of the product

02:08

rule a product rule with three products so i could say this is f of x right here

is equal to g of x and then h of x and then what would be the third

function here called gh let’s call it d of x so we have a

product of three functions here and so we can find the derivative

of three products here we could say derivative of the first

and then times those three and then now the derivative of the h so

i’ll just say g of x and then h and then d of x

and then plus and now we have g and h and the derivative of the b

02:09

function there so we could go find that derivative there using that

rule right there so i used a product rule here

and i took the derivative of the first function

the first function g i said was two functions there

so g was the first two and h was the third one

so i’m going to have to use the product rule again

because to find the derivative of g g is already a product

i have to use the product rule there so it’s a little bit quicker and easier

to use that derivative rule there which is a shortcut

and so i’m showing you in this example of how

whichever approach you take you’re going to get the same answer

so i found the derivative of g right there and then we can come back and put all

02:10

that together so there’s the derivative of g times the other one

plus this one times those other two there so we can use the product rule

with three products or more you can either do that using a generalized rule

or you could just use the product rule over and over and over again

and then in either case we get this simplified version by expanding it and

simplifying it which is certainly something much nicer

to take the second derivative of all right so let’s look at the quotient

rule now so we’re going to use the quotient rule and

we’re going to use g to be our numerator and h to be our denominator

02:11

but first we need to calculate the derivative of g

and we need to calculate the derivative of h

so those that’s what that’s what i get right there for the derivative of g

um in fact we calculated the derivative of g on a previous slide i believe and

there’s the derivative of h let’s look at the derivative of h maybe so h of x is

x minus 4 and this is x squared and so what would be the derivative of 8 here

so this is x to the third minus 4x squared so the derivative right so i just

expanded it out so the derivative will be 3x squared minus 8x

02:12

so that’s how we get the derivative of h all right so i looked at what my

numerator is i looked at what my denominator is i found those derivatives

now we’re ready to find the derivative of f the derivative

of a quotient so we have low or denominator times the derivative of high

which is the derivative of g i plug in there minus high

or the numerator so the numerator product

times the derivative of the low and all that’s for the numerator

and then we have the denominator squared so there’s the quotient rule there

02:13

so let’s do that here from scratch so here we go f x derivative it’s going to be

low times derivative high so here we’re going to have x to the third minus 3x

times x squared minus three so i have low times derivative high minus high

times derivative low so then we have derivative of the x minus 4 x squared

and then we have derivative times all that all right so i ran out of room there

all right so anyways there it is there so we have low

02:14

derivative high minus high times derivative low

and then don’t forget to square the bottom all over the denominator squared

so that’s the quotient rule there and this simplifies to that

is that really a simplification well yeah i think that one looks better

so in other words we’re just expanding out the

numerator there and simplifying it so let’s look at something smaller f

of x is let’s say we have two over x plus two do we use quotient rule

02:15

or product rule well we have a quotient so what’s the derivative

so we’re going to find it’s going to be low times derivative high minus high

times derivative low all over denominator squared

so low derivative height what’s the derivative of two

so that’s a zero there that’s zero so we’re going to get minus two

and what’s the derivative of x plus two it’s just one

so we have our derivative there all right so let’s do another one that’s

02:16

perhaps um more interesting how about if we have like a

2x squared minus 3 all over x minus 5. 2x squared

so what’s the derivative of this one right here we got the quotient rule again

so let’s say here we have low derivative high so we have 4x

so low times derivative high minus high times derivative of low all over

denominator squared and so we can simplify this a little bit

so we’re going to get four x squared minus twenty x minus two x squared

02:17

plus three all right and so i think we can simplify it one more step

we’re going to get two x squares and then minus 20 x and then plus three

and this is something that i would want to take the second derivative of

right so we don’t want to take the second derivative of that we don’t want

to take the second derivative of that if i had to take the derivative of the

derivative i would want to do this in other words this is simplified

all right how about this one um ax plus b c x

plus d we don’t know what the a b c and d are what’s the derivative of this one

02:18

so we have low derivative of the numerator a minus high times derivative low see

all over cx plus d and then squared okay so we have this is a

so we have a acx plus ad minus acx minus bc all over cx plus d squared

so the x’s cancel out so we get ad minus bc over cx plus d squared

as our derivative all right so whatever the a b c and d

are we can plug that in here into our derivative and we’ll have it already

02:19

okay so next one ah for some reason i wanted to use the product rule

oh that just looks messier let’s use the quotient rule

so we have g is the as the high and h is the low and so there it is cx plus d

squared on the denominator this the quotient rule is easier all right so if

f is a differentiable function then its derivative is also a function

for example we saw this one right here what was the original function the

original function was ax plus b c x plus d

there’s our original function and here’s our derivative function

so if we have an original function and it’s differentiable

then the derivative is the function also we can input an x in here

02:20

and you can get out a number so if f is differentiable then its

derivative is also a function so the derivative may have its own derivative

which we’re going to call the second derivative so we use the notation

f double prime or f prime of f prime so this is called the second derivative

the second derivative may also have its own derivative so for example here

we could go find this derivative right here and if we were able to find this

derivative right here that would be the second derivative

and that would also be a function so the second derivative may be differentiable

in which case you would find the derivative of it and that would be the

third derivative or denoted by three primes

and the fourth and so on for fourth and higher we usually use a 4 there with

02:21

parentheses around it so it’s to distinguish it from being an actual exponent

so you may be able to do this 4 times 5 times 10 times

so the derivative of the nth derivative is the nth plus one derivative

so in leibniz notation we use the d y over d x’s so y prime is just

a substitute for d y over dx and y double prime

is the derivative with respect to x of the derivative

and so you can see the special notation for that

we have the third derivative notice where the threes go

on the derivative on the third derivative is d

to the third y over dx over dx to the third

all right so find the first second and third derivatives of this

so what i would want to do if i had to find that many derivatives

02:22

is i would just rather expand it out first

so i expand all this out it doesn’t take long

you’re just multiplying out a couple terms

you’re just doing minus three x squared plus x plus eighty two all that times

the two x all that times the minus twelve or

perhaps you’re right to bring in the x to the third

times the two x first and the x to the third times -12 first and then expand it

from the right either case you simplify it to the same

then we can find the derivatives very quickly

so we bring the power down and multiply and we subtract one and we do that for

each and every term these are the differentiation rules

all right so we have the first second and third derivatives there

all right so now we want to find the first second and third derivatives of

02:23

this one right here so we have a quotient rule where we have g over x and the

h is the denominator and the g is the upper

the denom uh derivative of the low is one and the derivative of high

is six x plus seventeen so here we go derivative of f

is low times derivative high minus high times derivative of low which is one

all over denominator squared and then we expand that numerator out and simplify

it the reason why is because we want to take this derivative of that

we want to take the second derivative so here we go

we have low x minus 2 squared times derivative high which we get 6x minus 12

minus so so low derivative high minus pi that’s the numerator times derivative

02:24

of the denominator how do we get the derivative of the

denominator it’s x minus 2 squared i wrote 2x minus 4. let’s check that out

is that right what’s the derivative of x minus 2 squared so the derivative of

x minus 2 squared so i’m just going to say that’s just

x minus two times x minus two which of course is

x squared minus four x plus four so what’s the derivative

is two x minus four okay so we have that 2x minus 4 there right and

then all over the denominator squared so this is the

second derivative right so we squared a squared all right so then now

02:25

we have to simplify that whole numerator there now

perhaps i should show you that how to simplify that yeah let’s do that

so in fact let’s start from the first derivative

the first derivative here we see is 3x squared minus 12x minus 116

over x minus 2 squared let’s start there all right so let’s look at the second

derivative so it’s low derivative high which is 6x minus 12x

or minus 12. so low derivative high minus high

times derivative of low which we said was 2x minus 4.

02:26

we check that all over denominator squared so i already have a square so if i

square that square i have a fourth all right good now i want to show you

how to simplify this because when i’m looking at this all right here

i don’t want to expand all that out right away i’m looking at an x minus 2 here

and if you factor the two out we’re also looking at x minus

2 so when you look at 2x minus 4 think of that as

2 times x minus 2 there in other words this subtraction right here

is separating this and that subtraction is separating this

and both of these have x minus twos in them

so i’m gonna factor out an x minus two now i’m still going to have an x minus

two left and then i have a six and then an x minus 2 left

and then here we have this x minus 2 came out but we still have a 2 left

02:27

so i have a 2 and then i have a 3 x to the squared minus 12x minus 116

and then all that over x minus 2 to the fourth

so the point is is that these x minus 2s cancel

and i’m going to get an x minus 2 to the third in the denominator

and then we still have to expand all that out there

all right so good so that’s how we get this

when we look at the second derivative there the middle equation

that’s how we get x minus two to the third now to get to 256 you just have to

expand all that out apparently all the x’s cancel out and we

get to just the 256. now let’s take the derivative of the 256 over x minus 2.

02:28

so there’s our function there and let’s take the derivative of this

let’s write it over here 256 over x minus 2 to the third

this one shouldn’t be so hard this one’s the third derivative

because there’s a constant in the numerator so so we’re gonna go low derivative

high which is zero minus high times derivative of low what’s the derivative of

x minus two to the third right so it’s low derivative high minus high times

derivative of low we need derivative of x minus 2 to the third right in there

and then all over x minus 2 to the third but squared so that’ll be to the sixth

so we’re missing the derivative of x minus two

02:29

to the third here what’s the derivative of x minus two to the third well

this is um x to the third minus four x squared minus four x

plus minus eight is that right is that right so

this is x minus two times x squared minus four x

plus four so that’s x to the third minus four x squared plus four x

minus two x squared plus eight x minus eight so that and that gives us

minus six yeah minus six and that and that gives us the twelve

plus twelve and then minus the 8. so let’s just check that real quick

02:30

so x to the third minus 4x squared plus 4x

and then minus 2x squared and then plus 8x and then minus 8. all right so there

we go so now what’s the derivative that we’re missing in here what’s the

derivative of all this 3x squared minus 12x plus 12.

all right there we go but this term right here is zero

so that greatly simplifies that and then now we need to do minus 26 or

256 times each of those and then simplify that all right so here we go

in the end we’re going to get minus 768 over x minus two to the fourth

and the reason why is the x minus twos are going to

02:31

cancel out um that’s going to take some work there to expand all this out though

and see that um that we can factor some of this stuff in here

and cancel some x minus twos and that’s the goal

of this problem here is to is to do that that’s what’s left to be done

all right so there we go there’s the function

and we took the first second and third derivatives and as you can see this takes

a lot of practice to get really skilled at it all right next one

trigonometric derivative sine cosine tangent cotangent

we have these derivatives here so you know

something like what is the derivative of the tangent

you prob you might have heard the derivative of sine is cosine and the

derivative of cosine is minus sine if you stare at this long

02:32

enough you start to see patterns for example the derivative of a cofunction

is always negative derivative of cosine derivative of cotangent derivative of

cosecant they’re all negatives how do we get these though well you can

get some of these using our previous theorems so for

example what’s the derivative of tangent well that’s the derivative of sine over

cosine and what’s the derivative of sine over cosine this is the quotient rule

so low derivative high derivative of sine is cosine

minus high times derivative of low cosine squared and then all over cosine

squared so let’s see here low times derivative high

minus high times derivative low what’s derivative of cosine

02:33

actually it’s a minus sign ah so this comes out is positive sine

squared plus cosine squared and that’s all one

so what’s the reciprocal of cosine right so that’s secant squared

so derivative of tangent theory is the secant squared

and similarly we can prove the derivative of cotangent just use the

quotient rule um in fact the same thing with all of them um

let’s see where let’s do the last one let’s just do one more

what’s the derivative of the cosecant what’s the derivative of the cosecant so

that’s the derivative of one over sine so let’s use the quotient rule so low

02:34

times derivative high minus high times derivative low all over

denominator squared right so low derivative high minus high times derivative low

so this will be minus cosine over sine squared

now that doesn’t look catchy enough derivative of cosecant is minus cosine

over sine squared so let’s write it as cosine over sine and then we have one

more sine and so by looking at what cosine over sine is and what over

sine is that’s how we get this right here the cotangent and the minus cosecant

there i think it’s pretty catchy to say the derivative of cosecant

02:35

is a minus cosecant times cotangent all right so very good so what is the

derivative of sine x plus cosine x and then we have sine x minus cosine x

in the denominator what’s derivative something like this do that so we have

sine x plus cosine x over sine x minus cosine x now of course

you know you cannot just start cancelling these things here

right the signs don’t cancel the cosines don’t cancel okay well anyways

so we have low times derivative of the numerator derivative of sine is cosine

derivative of cosine is minus sine the low derivative high minus high

02:36

times derivative low derivative sine is cosine derivative of cosine

is minus sine so i’m going to have a minus minus so plus sine and then all over

sine x minus cosine x squared so low times derivative of high

minus high times derivative of low derivative of sine is cosine derivative

of cosine is minus sine so i have a minus minus okay so can we simplify this

so we have sine x minus cosine x squared all right so we’re going to get sine

times cosine minus cosine squared and then plus another sine times cosine

02:37

minus sine times cosine minus sine squared and then cosine times cosine

and then cosine times time oops sine times cosine

all right it’s a little hard to read but that says sine times cosine

all right so let’s see what we’re getting here we’re getting this one and

this one and they cancel um actually that was a minus there that

cancels with this one here also that should have been minus sine times cosine

minus here in fact let’s just double check that minus sine cosine

minus sine sine squared and then minus cosine this should be a minus cosine

squared also all right so good so all those mixed terms

02:38

go to zero and what are we left with here we’re left with cosine squared

cosine squared the sine squared okay so let’s see what we get

all right there’s our quotient rule there and now to simplify that

we’re supposed to get a minus 2 there and then how do we get the 1 minus sine

2x there so let’s go back here and see that

so we’re going to get sine times cosine we’re going to get sine times sine so

it’s a minus sine squared we’re going to get cosine squared here

02:39

and then plus cosine times sine and then minus sine times cosine

and then minus sine squared and then minus cosine squared

and then the last one is minus sine times cosine

and then all that over sine x minus cosine x squared

okay so these go to zero these add up to zero

and then we’re going to get minus sign squared minus sign squared so we’re

gonna get minus two sine squared and we’re going to get the same thing

for the cosines and then now here let’s multiply this out so cosine squared x

02:40

okay so this gives us a minus 2 up here because we can factor out a minus 2

and then a sine squared plus cosine squared so it’s

2 up here and now now this gives us a 1 and then we have minus 2 sine x

cosine x and so then the last thing is just to remember a trig identity

this is 1 minus sine 2x all right now look at all that

we’re just taking the derivative of sine plus cosine over sine minus cosine

we end up with you know the quotient rule and then a lot of simplification into

something very sweet down here and i call this sweet because this is

certainly something that we could take the second derivative of if we wanted to

it’s much nicer than that much nicer than that this is nice

all right so that’s how we get that derivative there so

02:41

this example shows us that sometimes when you’re working with trig functions

your derivative can greatly simplify using trig identities or whatever

so on this one right here there’s really nothing special about this one right

here nothing’s going to simplify on this one right here

so if we just take the quotient rule we have low times derivative high

so the derivative of the x squared that’s where we get the 2x

derivative of tangent is secant squared so good so we have low times derivative

high minus high times derivative of low so derivative of 3x is 3

and the derivative of tangent is secant squared

and then we have the denominator squared so the denominator is supposed to be

02:42

squared so we have low times derivative high minus high times derivative of low

so that looks like this all right that should be definitely be

oh sorry you didn’t see it so that should be definitely be square down there

okay next example oh time for a break all right see you see you after the break

okay so now we’re going to talk about the um

derivatives role in the real world so for example

how do we interpret the derivative okay so we know it’s the slope of the tangent

line that’s not the real world that’s geometry we know the slope

is uh so we know the derivative is instantaneous velocity we talked about

that before but now let’s talk about a concrete real world problem

we’re going to talk about the free falling body problem

02:43

so we’re going to talk about rectilinear motion

motion of an object that can be modeled along a straight line

so and the type of example we’re going to use is

something called the free-falling body problem

and we can model this problem if we neglect air resistance

and we can use this equation right here which gives us our position

s is the position and g is the acceleration due to gravity so on earth

it’s going to be minus 32 minus 32 feet per second squared and the t

is going to be in time and the v naught is the initial velocity and the s naught

is the initial position so if you remember back earlier we worked with

this ability billiard that was dropped from a height of 500 feet

so the s naught was 500 feet and the the gravity was well we’re

02:44

assuming we were on earth so it was minus 32 over 2 or minus 16

so let’s see another example of the free falling

body problem right so we have initial height and initial velocity

okay so this time instead of just dropping something we’re going to throw

something up from the ground and we’re going to have an initial

velocity so that’s going to be v naught it’s going to be 160 feet per second

and we’re going to ask the question when will it hit the ground

so we’re just throwing it up we’re going to ask when is it going to hit the

ground so the thing is is that we’re asking this ball to go

up and then come down so we can determine when the ball hit the ground

by looking at our equation s of t and plugging in our given values

so we’re using g is minus 32 we have initial velocity given to us

and we’re throwing it from the ground so we’re going to assume s naught is zero

02:45

so now we have our function minus 16 t squared plus 160 t

plus zero so we have factor of 16 t out we can see that when t is zero

when t equals 10 well of course it’s at the ground when t equals zero

but now we know ten seconds later it’s also going to hit the ground

so the next question is with what velocity will it hit the ground

how fast will it be going um you know it’s a related question

so the velocity of the ball at time t is given by the derivative

so we’re using the velocity is instantaneous rate of change

so we’re going to find the derivative we know how to find the derivative of

of s and it’s just you know in general it’s just going to be

gt plus v naught the 2 is going to come down and cancel with one half so we get

gt plus v naught so we’re going to use that when t equals

02:46

10 right because that’s we know when it hits the ground

so we’re trying to find v of 10 the velocity when it hits the ground

so we plug in 10 into our derivative and we get minus 160. the minus sign is

important because it’s hitting the ground it’s going

downward so remember velocity is a directed

you can think of velocity as directed speed if you want

so the velocity of the ball is minus 160 feet per second when it hits the ground

so now the next question is when will the ball reach its maximum height

maximum height is when we have a horizontal tangent line

or so differently that’s when our derivative is zero

so we’re going to be looking where the velocity is zero

the velocity is going up i mean the ball is going up

at a certain rate and then it comes down at a certain rate

when is that rate zero when the velocity is zero that’s when we have our maximum

02:47

height so we look at our v of t again and we set the velocity equal to zero

we solve that equation we get t equals five and so we can see the position

at 5 is the maximum height so we go back to our

original function here to find its position so the s of t was giving us position

so we calculate its initial position so at 400 feet

it’s reached it’s reaching a velocity of zero

so that’s going to be the maximum height so there’s an example

all right so you know just to clarify the s of t is giving us position the

velocity is given by v of t which is the derivative and the acceleration

is the derivative of the derivative the derivative of velocity

the second derivative which we denote by a of t

02:48

of course assuming these derivatives exist and the speed of the object is the

absolute value of the velocity all right so let’s look at this right

here a particle is moving along the x-axis so it’s just linear motion here again

and it has position so the x of t is giving us the x coordinate in other

words you give me a moment in time i plug in that time value and i get out

the x position the x the x value and that’s measured in seconds

so we want to find the velocity of the particle at time t

and we want to find the acceleration at time t

and we want to find the total distance traveled by the particle so

to mention the total distance traveled there

for example let’s say i’m moving along the x-axis here

let’s say let’s say i start right here at 1

02:49

x equals 1 and then i move to the right to three and then i moved back to a two

and then i moved back to a four so what was the total distance traveled

you could you don’t want to think of it oh i moved from one to four

so i moved three that’s not the total distance traveled though

we moved to three right so that was one two

and then we moved another one and then we moved

two more so we moved a total distance of five units

we moved two units here and then we moved to third

and then we move two more units there so we moved a total distance

of five units so i just want to keep i want us to keep that in mind when

we’re uh looking at this problem here find the total distance traveled

02:50

in the first three seconds all right so let’s see what’s happening here

so when we look at that x of t that gives us the position

the derivative of the position is the velocity

so the derivative of x which we know how to do very quickly using our

differentiation rules so we’re looking at three times two so

we’re looking at six t squared and then plus a two times the three so plus six t

and then minus 36 so that’s going to give us the velocity

for any time that we choose so we choose 3 seconds so we’re going to plug in 3

and that will give us the velocity all right but what about the acceleration

acceleration is the derivative of the velocity which is the second derivative

of the position so i’m going to take the second

derivative now so now i’m looking at 2 times 6 i get 12

and then plus 6 so there’s the acceleration okay so the velocity is zero uh

02:51

exactly when let’s go solve that equation

so we can factor that we can factor out a six and factor that

we get t equals two and t equals negative three

we’re going to disregard the negative three we’re going to look at the t

equals two so at t equals two we’re going to find the total distance covered so

we’re going to find out the distance between that it traveled in the first

two seconds and then the distance it traveled in the next second now

why well why well because when the velocity was zero

which is how we found the t equals two the velocity zero it may have changed

direction right so you gotta slow down and stop before you can change direction

and that example i was just showing you uh we traveled

two units to the right one unit to the left and then two more units to the right

02:52

so that was a total total distance traveled to five

and so when you change direction uh sorry when you stop yours when your velocity

becomes zero you may have changed direction okay

so we have to look at two places the distance traveled between two and

zero and two seconds we know we didn’t stop at that in that time interval

and then the time interval between two and three so we’re using absolute value

to find the distance traveled so we plug in those numbers x of two we

get minus four x of zero we get minus forty or we get forty

all right and so then x of three we get thirteen

and x of two we get minus four again so plugging those numbers in

finding the absolute value and adding them up we get 61.

so the total distance traveled to be 61. all right so very good

02:53

okay so now let’s look at some exercises so all together i have 55 exercises for

us um but you know this video is already um almost three hours long so we’re not

going to be doing uh these exercises right now in this video

so i just wanted to go through these exercises very quickly

and i mention them to you and let’s see what we have

right so the first three exercises here takes us back

to the tangent line and the average rate of change

problems here that’s the first four exercises

um then the next three exercises here we’re looking at

um some more problems here finding velocity and acceleration

um then in the next same thing we’re given functions s t position

02:54

breast defined velocity and acceleration so more problems here

and then more problems here uh i really like 15 here so

in the comments below if you could let me know which ones you want to see

in an exercise video on on differentiation

then i’d be happy to solve any of these problems

pick them out and let me know in the comments below all right so here’s some

using the definition of the derivative so this would be can be computing out

the limit of a difference quotient and also finding the equation of the

tangent line problem um and so 18 is asking us for an example

something that’s continuous but doesn’t have a derivative at 2 and

then there’s some more problems there so these

these three problems here want us to understand the definition of the derivative

and looking at when something is differentiable and not differentiable

02:55

um and here’s some more definitions 21 there uses a slightly different

definition of the derivative so i’d be happy to work out any of those

problems for you and then here using some differentiation rules

just practicing practice using the rules there

finding more equations of the tangent line and let’s see what else here

equations of the tangent line sketching some graphs

um making sure you understand the derivative there

let’s get off there and see if we can go up here you can see that problem

alright so um uh 3132 we’re asked to find a

parameter in the function the function isn’t fully determined

you want to know what the slope of the tangent line is going to be equal to

02:56

and so you try to come up with that function by finding a parameter and then

33 is finding some more derivatives all right finding more derivatives 35 is fun

because it tests whether or not you know a differentiation rule

without ever actually knowing a function so you’re

you’re going to just apply the differentiation rule and then plug in

the numbers that you’re given so 35 is kind of fun um so there’s 38

37 um finding the derivative of some trig functions there

finding a function and you want it to be differentiable

if you can if you can find the parameter and then we have a function that’s

continuous first derivative ah look at 42. so

you want to see that that function has a continuous first derivative in other

words you want to find the derivative and then you want to show that

02:57

derivative is continuous all right then sketch some more graphs um

and then we got the last couple last batch of problems here

and then let me move over here so you can see that one

and then the last two slides here um and these are some type of other

problems where we’re talking about marginal cost

and stuff like that so yeah if you want to see a video with

uh these 55 problems in it or some of these 55 problems

um i’d be happy to do that there we go all right so

i want to say thank you for watching and

again if you want any of those exercises leave a comment below and i can get to

them in another video also below in the description are my social media links

so if you’d like to keep up when the videos are coming out

02:58

you can either subscribe on youtube or follow me on twitter or something

the next episode is going to be over the chain rule

which will give us even more powerful methods

to finding derivatives of very complex functions and so that would be exciting

video can’t miss it so thank you for watching i hope you’re

getting the value that you need out of these videos

i look forward to making more of them for you see you next time

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