00:00

[Music] okay so you know how to differentiate a function using a definition

and some derivative rules on the other hand lots of simple

basic functions do not fall under these techniques

in this video i will teach you the chain rule and its

proof the chain rule is a powerful method

for finding the derivative of a function whose input is another

differentiable function the result is fantastic and will practice using it

00:01

[Music] welcome back everyone and this is our fourth episode in the cactus one

um explore discover learn series and so i’m really excited really pumped

um so far we’ve talked about limits in episode one

in episode two we talked about continuity in episode three we talked about

differentiation um and now in episode four we’re going to talk about the chain

rule the chain rule really takes differentiation in my opinion to a new level

um it’s one thing understanding the definition to find a derivative

and it’s another thing using some of the basic uh rules

such as the product rule and the quotient rule

but the chain rule the chain rule is very powerful

and um i’m so excited to show it to you we’re also going to cover the proof and

so we’re going to make sure we give a really good rigorous proof

and so but we’re going to go slowly through the proof so that you understand it

00:02

it will test to see if you’re understanding all the concepts

that we’ve discussed so far in terms of continuity

and differentiability and so we’ll put all that together in a really good proof

and then we’ll do lots of exercises we’ll do some easy exercises we’ll do

some challenging exercises and so yeah i’m just really excited to

go let’s get started [Music] okay so here we go we have the chain rule

so we’re going to start off with what is the chain rule let’s look at the

statement of it um and read it and see if we can

understand it and get started and let’s see how

let’s see how it looks so the chain rule here is starting off with a

differentiable function so far that’s pretty pretty simple we know what that

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means from our last episode f is a differentiable function ah but

here’s the interesting part f is the differentiable function u

which is another differentiable function so the chain rule’s working with the

composition of two functions both of which are differentiable

so in other words u is a differentiable function

which is inside of a differentiable function

f and so we want to take the derivative of f

with respect to x so how do you do that well we can take the derivative of

f with respect to x by taking the derivative of f with respect to u

first and then take the derivative of u with respect to x so this is the chain

rule in its simplest form and basically it allows us to take the derivative

00:04

of a composition of functions using this product here dfd u times d u dx

and so we’re going to see how this works uh through some simple examples first

so here’s number one let’s look at number one here so number one says

y is equal to all of x to the third plus two x squared minus 12x plus five

and we have all that to the fourth power now if we don’t know the chain rule

then how we take the derivative of this right so if we don’t know the chain rule

this is a fourth power so we can simply take this expression right here

and multiply it out four times or just in other words just expand this out

and then after we expand it out we can take the derivative

term by term right so this is x to the third

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and we got a fourth power so this will start off with x to the twelfth power

and then it’ll have a lot more terms and then it’ll end with five to the fourth

power right and so that will be very long that’ll be the technique that we don’t

want to use instead this is a good example of when to use the chain rule

now to use the chain rule we need to talk about what is f what is u

right so i’m going to call this function here f and

i’m going to come up with a function u so u is actually i’m going to call that

f sorry let’s leave that as a y i’m going to say that

f is a function of u and i’m going to use

all of this is my u and i’m going to say u to the fourth

and then for u of x i’m going to say i have

x to the third plus two x squared minus twelve x plus five

00:06

so when we look at this function uh this chain rule over here

says f is a differentiable function of u right so f is a differentiable function

of u and that is a differentiable function of u we can find its derivative

the derivative of u now when i put a prime here

that’s a little bit misleading because you may think i’m taking the derivative

with respect to x no no so let’s put the full leibniz notation

so df du is going to be 4u to the third right and now we have u right here

and so what is d u dx now i’m taking the derivative of u with respect to x

so we have 3x squared plus 4x minus 12. so

what does the chain rule say about all this well it says df dx

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the derivative of f with respect to x is the product df du [Music] times du dx

and so what do we get for this product right here

so dftu we found right here 4u to the third du dx which is all this

now you may not like your answer with used in x’s right

so let’s just substitute back in what is the u

here’s the u right here so it’s four u to the third

plus five so i’m doing four u and then to the third

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and then times all this right here so we have three x squared plus 4x minus 12.

so there’s our derivative right there this is dfdu which is right here

times d u d x which we found right here [Music] so this is d f d u

and this is d u d x so this is the chain rule right here

let’s put it over here this is d u d x now this may seem like kind of long and

cumbersome but actually after you practice it it’s very short and sweet [Music]

and so i just wanted to show you how to do it

by looking at what the u is and what the f is

and taking those derivatives separately and then we’ll put them all down here

00:09

together right so let’s do another example [Music]

so that was number one for number two we’re going to change that power to a 40.

now if you don’t know the chain rule working that out for the 40 by hand is

just simply something you’re not going to want to do to the fourth power

well maybe you have a lot of patience but to the 40th power

you know that’s just not something you want to do that

problem is much longer the power of x will start off with 120

and the last term the constant will be 5 to the 4

to the 40th power so that’s just not something you want to mess with

but we can take the derivative of this function very quickly

so let’s see that again let’s write down our function y is x to the third

00:10

plus 2x squared minus 12x plus 5 to the 40th power

so how do we take this derivative so for just one more example i’m going

to show you what the f and what the u are just so

that you can see the chain rule right here so the f is a function of u

that’s what the chain rule right here says f is a function of u

so the u in this case is the inside function right here

this is what i call the inside function so f of u will be u to the 40th

and then you use the inside function here x to the third plus 2x squared

minus 12x plus so we’re just saying that this is the u and this is the f of u

so f is a differentiable function of u and use a differentiable function of x

00:11

and we mean they’re differentiable because we mean we can find their

derivatives right so what is the derivative of f

with respect to x so to find the derivative with respect to x

i find the derivative of f with respect to u so what’s the derivative of f with

respect to u that’ll be 40 u to the 39 times so the chain rule over here says

times d u dx so i need d u dx so what’s the derivative of u

with respect to x so this will be three x squared three x squared

plus four 4x minus 12. okay and then the last step is to say

you know i don’t want to use n x’s so let’s just come back here and

put in the u right here so this will be 40 and then for that out

00:12

now for this u i’ll just write this back in here

so this will be the derivative of f with respect to x

derivative of f with respect to x so all this 40 u to the 39

and then times the du dx minus 12. so there’s the derivative so

we did this pretty quickly um you know compared to

expanding this function out and taking the derivative term by term

so let’s look at number three so this time we’re going to

be taking the derivative of sine of x to the third so let’s go here and say

00:13

let’s look at it this way so y equals sine of x to the third

we learned in the last episode that the derivative of sine is cosine

so i got to remember that so i’m going to say that f of u is and u of x is

so u of x is the inside function here which is x to the third and this is sine

of u so if i put u inside of the f of u then i get back the original

function right here and so this is the derivatives now this

will be the derivative of f with respect to u derivative of sine u is cosine u

and the derivative of u with respect to x is 3x squared

okay so once we identify what f is and what u is

00:14

we can take those derivatives now we can put it all together

so the derivative of f with respect to x is we have the chain rule here

so it’s the derivative of f with respect to u

times the derivative of u with respect to x

right so this is a product here so the derivative of f with respect to u we

found is cosine u times and then now 3x squared

and then the last step would be let’s put in what this u is

this u is x to the third so this will be cosine

x to the third times three x squared and perhaps it’d be better if we put

parentheses here in fact it would probably even just look

nicer if we wrote the 3x squared in front so here’s our function right here

00:15

and here’s our derivative right here using the chain rule now in these first

six examples i’m going to keep doing this the long

way but as we’ll see as we do more and more examples

we’ll shorten this out a lot in other words when you become

skilled at finding the derivative you want to be able to go from here

to here and a lot less work in fact you want to be able to go from

here to here just doing it mentally if you do enough exercises next one

alright so now let’s compare this with sine to the third

so let’s go and look at that so let’s do that over here if we can

y equals sine x to the third and now what is our

00:16

u going to be what is our f of u sine to the third uh x right now this

one’s different than this one because this one

the third only goes for the is this part of the angle here

this third just means this is sine x times sine x times sine x right

sine to the third so here it’s the third function that’s on the outside so i’m

going to say u to the third and then u of x is sine

so in other words if i take this u right here and put it in there

then i get sine to the third there [Music]

all right so now one of our derivatives derivative of

f with respect to u is u to the third uh sorry three u squared and our

derivative of u is just cosine x and so you can see the difference

between these two examples over here i wanted to do them side by side

00:17

so you can see the difference between them there’s

a huge difference where that three is here it’s x to the third

and here this is sine to the third all right so

what is our derivative here derivative of f with respect to x chain rule

derivative of f with respect to u so that’s right here three u squared

times this one right here cosine x and then that last step is to go to only

x’s right here so what is the u the u is sign

so this will be sine squared so we have three sine squared times cosine x

so there we go derivative of sine to the third

is three sine squared times cosine all right very good let’s do another example

00:18

[Music] now let’s look at square root of two x plus five here we go

so y is square root of two x plus five and what’s the inside function and what’s

the outside function u is the inside function it’s two x plus five

and what are we doing to that two x plus 5 we’re taking the square root of it

so the inside function is the u and then we ask

well if we put in the u what are we doing to the u we’re taking the square

root of it all right so here’s the derivative of f with respect to u

so this is u to the one half right what’s the derivative of u to the one

half it’s one half u and then minus one half

right so what is this right here this is 1 over

00:19

2 square roots of u because that’s a negative power so that’s

1 over 2 square roots of u so we’ll put that here

and then what’s the derivative of u the derivative of u with respect to x is

just two all right so let’s find our derivative df dx

with respect to x is derivative of f with respect to u so 1 over 2

square root of u times derivative of u with respect to x which is 2.

so we see the twos are actually going to cancel

and we’re going to get 1 over square root of u but

what is this u the u is 2x plus 5. so there’s the derivative

here’s our function and here’s our derivative

00:20

so it goes without saying that this is what my f of x is

so this is f of x this is d f d x using the chain rule

all right very good now let’s look at uh number six [Music]

let’s go over here and do number six now and so we’re looking at y is equal to

the third root [Music] the third root of x squared minus five x plus one

so y is equal to this this is our function f of x

all right so what is the inside function and what is the outside function

so the inside one is x squared minus five x plus one and the outside

function is well if i put in that u what are we gonna do to it we’re gonna

00:21

take the cube root of that u all right so the fdu

right how do you take the derivative of cube root of u

so cube root is one third so i’m going to bring that power down say

one third u and then one third minus one right so that’ll be minus two thirds

and then if we write that better that’ll be like a one over three

and then cube root of u squared okay so let’s put that over here so it’s

1 over 3 cube root of u squared so that’s the cube root of u squared

all right and now the derivative of u with respect to x that’ll be 2x minus 5.

there we go so now we’re ready to find the derivative

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of this y here without using a definition so here we go d y d x

so this time i put d y d x or you could put f of x equals and say d

f d x i’ll say d y d x is d y d u times d u dx [Music]

so this will be one over three cube roots of u squared but the u is right here

[Music] so one over three times the cube root of u and then squared

and then times d u dx which is 2x minus five

and i’ll say times two x minus five or i’ll just put it over here

so there’s our derivative let’s double check this right so this will be

00:23

d y to u so we got it right here 1 over 3

cube roots of u and i’m putting this in for the u

so it’s u squared cube root times this right here times d u dx so i put times

and i put it over here instead there we go so there’s six examples

using the chain rule we’re going to work some more more

difficult examples in a minute so there was number six we did

um so hope you’re liking these so far all right it’s time for the proof of the

chain rule so how do we prove it well first of all

00:24

i want to say that if you go online no surprise here but if you go online

and look at youtube there’s some good proofs out there but

there are also some proofs that take the shortcut and the proof isn’t

valid um it’s just skipping some cases and it just kind of simplifies it too

far the proof i’m going to give you is rigorous it’s complete there’s no

flaws in it um so let’s get started with it let’s look at the proof

so let’s um state what the chain rule is again suppose f is a differentiable

function of u and that u is also a differentiable function of x

and so we’re going to put u into the f and we’re going to get this product

right here so the proof is going to start off with

hey this is what we want to show i want to show df over dx is equal to df

over du times du dx that’s that’s what we want

00:25

to show that’s that’s what our goal is so in order to do that

we’re going to use the definition of the derivative

again that’s something that we went over last episode and so we want to use that

limit definition and so here here’s what the the definition is the

derivative of f with respect to x here i’m using a delta x

sometimes i’ll use an h but here i’m using a delta x

so delta x is approaching zero and in the denominator we have delta x and then

in the input i have an x plus delta x and then i have minus f of u

so that’s the limit that we’re trying to find and we need to show that it’s the

product of those two derivatives on the left

so here we go to do this i’m going to come up with an auxiliary function

00:26

and when i say i’m going to come up with it well i don’t mean actually i came up

with it this proof has been um looked at by a lot of people over a lot of years

and the best way to prove it is to actually define

this function here so i’m going to call it g of t

so this function g is a function of a variable t

t is not in our original statement if we look at the theorem there’s no t

in there so this is g of t and we’re making up this function g of t

in terms of all the other things that we’re already given

right we’re given a function f we’re given a function u

and we’re using this variable t and so g of t is broken into two pieces

if t is zero then g of t is zero if t is not zero then i have to go

calculate all these things up and i get my output for g of t so what

00:27

do we got to calculate we have to calculate uh f of x plus t

and put that into f minus and then we have to put u into f and that’s

going to be all over t and then we have to subtract off

the derivative of f with respect to u now this g of t the variable

is t that’s important to remember um and so it may have other um things in the

in the function but it’s variable is t so we’re going to describe

three properties of g and use those three properties in our proof

so that’s the strategy of this proof here so i’m going to show you three

things that hold about g and once i show those three things then

00:28

we’ll be able to go through the steps of proving this product dfd u

times d u dx all right so the first thing we’re going to prove

about uh this statement here is that um well first we want to look at what um

delta u is so first i want to talk about delta u so delta u is that right there

so delta u is just u of x plus delta x minus u of x now that’s not anything

unusual itself because we use this delta notation all the time

and we’re just giving a change of input here into u

and then subtracting this difference here but the thing i want to point out

00:29

to you is that i’ll be using u of x plus delta x here

i’ll be using this expression right here and so what is it equal to

right so if we move this minus u of x to the other side so we have delta u

plus u of x so i’m going to be using this expression right here

and what follows and and the reason why is because if we look at what g

is we look at what g is g is going to use this right here and so

we’ll see that in a minute [Music] okay so nothing exciting about

that statement there but property number one so let’s write

let’s write g of t right here real quick g of t is equal to f of u of x plus t

00:30

minus f of x u of x all over t and then this is minus d

f d u so this is what g is when t is not zero that’s what g is there [Music]

all this divided by t now what i want to do is substitute in here

delta u for the t what happens if we substitute it in delta u for the t

so now we’re going to get f of u of x plus the t which is the delta u

minus f of u of x there’s no t in that part

all over t which i’m using is my delta u now and then minus d f d u [Music]

so this is what g of delta u is [Music] so we’re going to use this here

00:31

coming up that g of delta u is equal to all this right here um

provided that delta u is not zero right because if delta u is zero

then when i plug in zero i just get out zero so as long as the input’s not zero

right if the input’s not zero this is what g of t is so if delta u is not zero

this is what g of delta u is right here so that’s property number one right

there all right now property number two is going to work with this a little bit

and simplify it so what i’m going to do so this is property number one

now for property number two we’re going to take this

derivative right here and we’re going to move it to the other side

00:32

so this will be plus df du and it’ll be equal to all of this so f of u of

x plus delta of u minus f of u of x all over delta u

and now the last step i want to work on is i want to multiply

both sides by delta u and we can multiply both sides by delta u because

we’re assuming delta u is not zero still so here we go we’re going to have g of

delta u plus the this derivative right here all that times delta u

and that’s going to be good to this numerator f of u of x

plus delta u minus f of u of x and this is property number two that

we’re going to use [Music] so this is f of u of x plus delta u

minus f of u of x here this is just the numerator

00:33

so in other words i’m just playing around with this number one

and i’m just adding this to the other side then i’m taking that right there i’m

assuming it’s not zero so i’m taking that right there and i’m multiplying

both sides and i get this expression right here and so we we have isolated this

difference right here because we’re going to use that right

there it’s very important that we have an expression for that

okay so this property number two now for property number three

we want to show that g is continuous at zero so here’s what g is so we have

this and then we have zero that’s t is not zero

and t is equal to zero and that’s what g of t is

00:34

so to prove that g is continuous at zero we need the limit as t approaches zero

of g of t to be g of zero which is zero so we need

this limit to be zero right here now when we’re approaching zero we’re

going to be looking at this top portion here for this limit and so

what’s the limit as we approach zero of this f of u of x

plus t minus f of u of x all over t and this right here is my uh minus all this

right here we’re looking at this limit right here [Music]

so if we look at this limit right here though um we have u of x plus t

00:35

and then we’re over t right here and so this is the derivative of f with respect

to u um [Music] you know if we used a delta u

here and a delta u and a delta u you might recognize it

but remember you don’t have to use the delta u you can use an h or a t or

anything else you want but this limit of this part right here is d f

d u and now we’re subtracting d f d u off of this so in other words i’m taking

the limit of this one i’m taking the limit of this one

and so in the end they’re equal to each other because that’s the derivative of f

with respect to u and i take that limit of that so we’re

going to get zero for this right here and that’s equal to g of zero

and so that says that g is continuous so g is therefore g is continuous

at t equals zero so those three properties that’s

00:36

property number three that we need those three properties are going to give

us the proof that we need okay so so far at this point we have

defined this auxiliary function g and we’ve done three properties with

this function g g of t the first one is that we plugged in delta u

into it provided delta u is non-zero and we got this output and then step two

was we played with it a little bit we moved the dfd u over

then we multiplied both sides by delta u and then the third property is we

noticed that g was continuous at zero and the observation that made that

happen was this limit right here is equal to dfdu

okay so now we have those three properties established

for g now we’re ready to write out our formal proof here

00:37

so we’re going to go through these steps so we can write df

df dx as follows so it’s the limit remember we’re going to use the limit

definition here the dr the derivative of f with respect to x

so i’m using a delta x here now what i’m going to do is i’m going to

take that u of i said earlier we’re going to use this property right here

[Music] if i have u of x plus delta x right here

i just move that over and i have this now so u of x plus delta x

is delta u plus u of x and so that’s what we’re using right here

so on this inside part here where we have u of x plus delta x now we have u of x

plus delta u so we just simply made that small substitution right there

but that’s going to pay off huge because we’re going to make the connection with

00:38

g here so this numerator right here and this is property number two that

we’re using right now that numerator can be rewritten in terms of

g it’s g of delta u plus the derivative of f with respect to x

now we can use the product rule for limits right here

the limit of the first times the limit of the u with your delta u over delta x

okay so let’s look at the limit of each one of these

so when i look at the limit of the first expression

i have a sum so now i’m going to take the limit of each

term in the sum the limit of g of delta u plus the limit of

d f d u delta x goes to zero and we still have the limit of delta u over delta x

00:39

so now here’s a big step here it seems kind of minor but it’s a big

step we know that g is continuous at zero so we’re going to pull the limit

inside of the function g and we can do that because of the

limit composition theorem that was the theorem we covered in episode one

and we know g is continuous there at that point so

we can pull g inside so now we’ll take the limit

limit of delta u as delta x goes to zero right so we can look at that right here

real quick [Music] so if delta x is going to zero then we’re going to get u of x

minus u of x right so delta u goes to zero as

delta x goes to zero so that’s we have right here so now we have g of zero now

just leave everything else the same the limit of delta u as delta x goes to zero

00:40

that gives us the zero so now we have g of zero

when we have g of zero well that’s just zero

and then finally we can find our last two limits and if you

if you’re paying attention those are dftu

because when delta x goes to zero there well there is no delta x’s in that

expression so it’s just dftu and the same thing with the last one here

delta u over delta x as delta x goes to zero

that’s the derivative of u with respect to x and so there’s the chain rule so

i see seven or eight steps here eight or nine steps there and each step

is justified by um a limit definition or one of the three properties of g

00:41

so there’s each step very clearly laid out and you know we can say exactly

what is holding or or the reason why each equality holds

and the justification for each step so there’s the proof of the chain rule

and believe me that takes some time to digest

but once you understand all the limit definitions and you understand the

theorems that we applied then it should be then it should be able

to you should be able to make sense of it all right

now it’s time to look at some more examples now these examples we’re going to do

we’re not going to take the long approach that we’re taking

before before we were taking very close attention to what was

f of u and what was u of x and now we’re just going to start going a lot faster

00:42

but we’re also going to work some interesting problems

for example this right here this is probably not the first

exercise or example you want to see with the chain rule but we’re going to

see it but it won’t be first let’s do another one let’s do

let’s do something else real quick let’s do something simpler

because i want us to get used to um taking the derivative using the chain rule

without identifying the inside function the u of x

without writing it out so let’s say we have 2x plus 3

squared just something very easy and basic

so the way that we do this is we bring the two down

we’ll leave the inside function alone and by leaving it alone i just mean

write it down again now i reduce this power down by one so two minus one

00:43

so one and then now times the derivative of the inside part here

okay so that’s the chain rule you know we said times 2 right here that’s the

chain rule so i take the derivative of the outside function so 2

leave this alone to the one power and then now times the derivative of the

inside function there and so in the end this will just be four

times two x plus three there’s the derivative and

when we write it out quicker like this now we don’t necessarily have to use

leibniz notation because there’s only one variable that

we’re we’re writing down here so let’s try another one so this time

we’ll say x squared minus 4x to the fifth power how do we find the derivative

okay so we’ll use the derivative of the outside function

which is 5. so we’re going to say 5 bring the 5 down

00:44

x squared minus 4x and then reduce the power by 1

times the inside function right here so the derivative here is 2x minus 4.

and so we could write this a little simpler by taking out a 2 and saying

10 x squared minus 4x to the fourth and then x minus 2.

we could write it something like that okay so in these examples that we work

through it’s important to remember the chain rule we can’t just take

uh the derivative like this this is x to the fifth

what’s the derivative 5x to the fourth and then you might say times the

derivative of x what’s the derivative of x is just one right

so we don’t need to write times one derivatives by

5x to the fourth but if you do have something inside here you can’t just say

00:45

times one now you have to say times the derivative of the inside part here

okay so let’s try to go to this example first let’s take a peek at it

um perhaps let’s do one more real quick that involves the quotient rule right so

let’s do a quotient rule real quick what if we have something like x minus two

over x squared plus three and then well that doesn’t use the chain

rule does it that’s just quotient rule so let’s put a power on there let’s put

a square let’s put a third on there now that’s going to use the chain rule

because they got this function inside of the cube function

so now we’re going to use the chain rule so first thing i’m going to do is i’m

going to say ah just use take the derivative of x to the third

00:46

so what is that it’s three times all of this

squared now times the derivative of all this

and so i could take that derivative or i could just say

derivative and then i’ll take it on the next step

so we’re doing three all of this squared times the derivative of this so let’s

find the derivative x minus two x squared plus 3

squared now what is the derivative this is quotient rule here so it’s low

derivative high minus high derivative of low which is 2x

all over denominator squared [Music] okay so now we’re done taking all of our

derivatives and you can simplify this further

and you definitely want to if you had to go find a second derivative

00:47

so i would not think this is simplified at all

we would want to see what cancels there and maybe something will cancel here or

there we have this to the squared we have this to the squared

so we’re going to get the denominator to the fourth power so we can continue

simplifying that but there’s the chain rule right there i take the derivative

use the 3 x squared but this is the x then i take the derivative of it

all right so let’s go now and see if we can solve our problem

so let’s see here our function is our function is x

over square root of x to the fourth plus four

and let’s write it like this x times um x to the fourth plus four

to the minus one half now do you like to write it like that or

00:48

do you like to just keep the quotient rule in here

so if you write it like this you could try to use the product rule

i have the product of two functions here but maybe you just want to keep it as

the quotient rule so let’s see if we can do that let me

write this a little better square root of x to the fourth plus four

all right so here’s the derivative quotient rule low derivative high minus high

times derivative low ah what is derivative of low what’s the derivative of this

right all of this has a one half power so we’re really asking what’s the

derivative of this what’s the derivative of this

00:49

right so we have one half coming down x to the fourth plus four now one half

minus one and there’s the derivative right nope chain rule

so times the derivative of the inside part here four x to the third

and so what does this come out to be we have a four

we’re dividing by two so we have a two and then an x to the third and that’s

all in the numerator then we have square root of this right

here in the denominator so the derivative so here we go again

low derivative high minus high times derivative

derivative of low is this right here so we had to go do that scratch work

00:50

right there so here we go low derivative high minus high

times derivative low and derivative of low we did over here all divided by

denominator squared now the reason why i picked this example is because

simplifying this is a good problem is a good exercise

we’re gonna need to find the common denominator the common denominator is

square root of x to the fourth plus four so

let’s think about it like this i have x to the fourth plus four square root

um times one so i’m going to put it again here and divide by it

minus and we have this x here so that we can say

00:51

2x to the fourth over this denominator here and all that is times

our denominator which is has a square square root so we’ll just say one over

x to the fourth plus four okay so what power are we going to get in the

denominator in the denominator we’re going to get

x to the fourth plus four to the three halves power we have a full one over here

and then the common denominator is a half power so one plus a half

and so we get x to the fourth plus four to the three halves

and then what about on top well we’re getting x to the fourth plus four

minus two x squared i’m sorry minus two x to the fourth

00:52

so it looks like we’re getting four and then a minus x to the fourth

okay so let’s put all that together so we’re going to use the chain rule and

the quotient rule and here’s the quotient rule so we have

low times derivative high minus high times derivative low

all over the denominator squared so notice i’m waiting to take the derivative of

the square root of x to the fourth plus four

i just simply wrote i’m going to take the derivative of it

now i’m taking the derivative of it and i plug it in there

and then now we’re going to simplify and what we

what we found in our in our work below is that we simplify to four minus

x to the fourth right that’s we got right down here right

00:53

four and then we got a minus right here so there’s the write up right there

all right so let’s look at another example um here we have a quotient

to a power this is not something you would have to take the derivative of

using a definition hopefully not um we’ll use the chain rule

and of course the quotient rule so let’s see here

are we gonna need to do any side work let’s see first thing is we’re going to

do the chain rule so i’m looking at that third power that’s my outside function

my inside function is the quotient so i take the derivative of the

third power so i bring the three down and i leave the inside alone

and then i reduce it by one so i get a squared

and then now i take the derivative of the inside part

and so that’s a quotient rule and so we have low

00:54

times derivative high minus high times derivative low all over

denominator squared okay so does that simplify to anything nice well we have 3

plus 2x all to the fourth in the denominator the the middle factor

there has a 2x plus 3 squared all of that squared so the denominator squared

and then the last factor also has a 2x plus 3 squared

so we end up with 2x plus 3 to the fourth and then in the numerator well we

have a 6 because a two will factor out from the last part there

um maybe we’ll look at that here let’s let’s look at that simplification just a

tiny bit so how do we get that numerator there so we end up with a 3

00:55

and then a 3x squared minus 2 squared we end up with a 2x plus 3

times the 6x minus 3x squared minus 2 the squared that’s we end up with in the

numerator so how does that simplify so let’s not worry about this part right

here right now what do we get here so this would be a

12x squared plus 18x and then we’re getting a minus 6x squared and a

right so minus 6 right because we have a minus here

and then we have a minus minus so plus 4.

and so let’s see what we get here we’re going to get a 6 x squared here

00:56

and then plus 18 plus 4. it looks like we get a 6x squared plus 18x plus 4.

so out of these right here we’re going to take a 2. and put it with the three

and we get three x squared plus nine x plus two

and then now for the last step here for some reason or another

we can do a minus one times minus one because this is all squared i’m just

putting in a minus one minus one so this one right here

this minus one i’m gonna pull inside of here and this minus one i’m gonna leave

00:57

out here so this minus one will go to the two and

then i’ll say a minus three x squared and it’s still all squared

and that’s a nine so this minus one gets a squared so it goes away

so i’ll just say equals six and this is what we have here

on the notes right here so we get a 2 minus 3x squared here

even though the original has a minus 2 with a positive 3x squared so because

this is squared we can do that and i showed you how right here how that works

now the reason why that may happen to you um is because of someone rewriting it

00:58

this way um ascending order constant and then an x and then x squared and so on

and so someone may prefer to write it in that manner [Music]

so you know that’s just the way that the world works some people like to

see some things one way or another [Music] okay so there’s our problem there

there’s our derivative of g with respect to x uh next example

all right some trig so this one right here we have to remember that the

derivative of cotangent which we uh worked out last time the derivative

is remember what the derivative is of cotangent

so you know i get students saying i don’t know what the derivative of

00:59

cotangent is right so okay i believe you the derivative of cotangent is

what so you certainly remember the quotient

rule right derivative of cotangent is what cosine over sine so what will this be

so low derivative high minus high times derivative low

derivative of sine is cosine all over denominator squared

and so what do we get here minus sine squared plus cosine squared all over sine

squared and so let’s see here this is minus one over sine squared

and so this is minus cosecant squared so the derivative of cotangent is minus

cosecant squared so you know maybe you remember maybe you’d like to

01:00

work it out but certainly you need to remember it at some point

all right so now for our problem we have this other stuff here and let’s do some

work here with some trig functions so let’s kind of ignore the problem for a

moment and let’s say here what if we have cotangent of let’s say pi t

what would be the derivative of this so we have a function

sitting inside of a function and we’re going to use the chain rule so i’m

going to say the derivative of cotangent is minus cosecant squared and i’m going

to leave this function alone leave the inside alone now times the

derivative of the inside function which is pi

so that’s times pi and that’s because of the chain rule

and now we’ll probably write the pi out front so we’ll say minus pi

01:01

and then cosecant squared of pi times t so there would be the derivative there

what if we added 2 to our function so cotangent pi t plus 2

what would be the derivative of that so again i’m taking the derivative of the

outside function first which is cotangent

so the derivative of cotangent is minus cosecant squared

and i leave the inside function alone times the derivative of the inside

function derivative will be pi that’s it derivative of the two is

zero so still times pi difference is of course we have a plus 2 here

and so again we’ll write the pi in front so nothing terribly exciting if we add

01:02

the 2 there okay what if we add a 2 out front what if we added 2 out here

well that’s a constant so i’m going to say 2 minus 2 because i

need the derivative of cotangent is minus cosecant squared so i’m going

to put that minus out front so i’m going to say minus 2 cosecant squared

and leave this part alone now times the derivative of the inside

part which is pi so now i’m going to get minus two pi

cosecant squared of pi t plus two so there’s our derivative okay

but our original problem has a cotangent squared in it so let’s start

thinking about what the cotangent squared does

so what happens if we have a cotangent squared let’s back up here

let’s forget the rest let’s just say oh what if we have a cotangent squared

01:03

what’s the derivative okay so now the derivative

the outside function is the squared in fact let’s put a t

here just to match our problem let’s say our variable is t

cotangent squared of t so the cotangent so the t is inside of the cotangent

and the cotangent t is inside the squared so another way to think about it just

like this cotangent of t and then squared right let’s cotangent of t

times cotangent of t now writing it like this is not very

useful because you know i’m not going to use the product rule

writing it like this you have more intuition perhaps that you

can see that cotangent of t is the inside function but we usually don’t write

our trig functions with parentheses around them and then the powers out here

we usually write them with the power sitting right over the

01:04

trig function so they mean the same thing though right so how would we take the

derivative for example if you had 3x minus 5 squared

how would we take the derivative of that

we would bring the 2 down and then leave the inside alone

and then times the derivative so in other words we’re taking the derivative

of the outside function the square function first

and then we’re taking the derivative of the inside part here

so it’s the same thing here we’ll take the derivative of the outside function

first so we’re going to bring the 2 down and then leave the cotangent alone

and then cotangent of t and then times the and then to the first power

and then now times the derivative of cotangent which is minus cosecant squared

t and then now times the derivative of t but the derivative of t is just one so

just that’s it so the derivative of cotangent squared

01:05

we don’t need to write that the derivative of cotangent squared is two

cotangent of t to the first power times the derivative of cotangent t so

long story short there that’s the derivative there

all right so let’s change it a little bit what about if we have cotangent

squared and now let’s say we have a t minus five something like that

so what would the derivative be so now the two comes down to cotangent

of the t minus five to the first power now times the derivative of cotangent t

so minus cosecant of t minus five so i took the derivative of the outside

function the two so two comes down cotangent of t to the first power now

times the derivative of cotangent which is minus cosecant

01:06

leave that alone now times the derivative of the t minus five which is one

so that would be the derivative there well what if we have a pi t pi t

so here we go bring the two down two cotangent of t times pi of t

cotangent to the first power now now times the derivative of the cotangent

leave the inside alone times the derivative of the inside so times pi

so this would be the derivative here bring the two down cotangent to the

first power leave the inside alone all of it now times the derivative of

cotangent minus cosecant leave the inside alone

now times the derivative of the inside so this is the chain rule we’re actually

chaining it together we used two times there so let’s solve our problem now

01:07

i think we’ve done enough work now where we can just solve our problem

so we have h of t is two cotangent squared of pi t plus two [Music]

so here when i say h prime i’m saying derivative of

h with respect to t we don’t need to write it it’s the only variable there so

it’s understood so here we go i’m looking at the cotangent squared

so i have this function is sitting inside of the cotangent function

and that whole function is sitting inside the square function

so i’m going to use the chain rule twice here because i got

one function sitting inside another and then that function

01:08

sitting inside the square function so think of it like this 2 cotangent of t

of pi t plus 2 and all of that is squared so i have this function sitting inside

the cotangent function and i have all of this sitting inside the square function

so this is just this more streamlined way of writing it

but this is what maybe you’re visualizing sorry you couldn’t see that

for a second so i’m thinking about these well let me

move it up so you can actually see it so i’m thinking about this is cotangent

of pi t plus 2 and then all of that to the square function here this is squared

so as you can see i have this function right here of t

inside of cotangent and then i have cotangent of all of this

that function is sitting inside the square function

so we have two compositions here this one inside this one

and all of this inside of the square function so i’m going to use the chain

01:09

rule twice on this problem so i’m going to bring the 2 down so i have a 4

and then i have cotangent of pi t plus 2 and then all of that to the first power

which we write here which we don’t write now times the derivative of all of this

which is minus cosecant of pi t plus two and then now times the

derivative of pi t plus two which is pi and so then we would probably write that

much simpler writing the pi the minus and the four

all out front and then we can write the cotangent of pi

t plus two times cosecant of pi t plus two

now would it help to write it in terms of sines and cosines

well this is cosine over sine and this is one over sine

01:10

so i don’t really see a simplification in that

so here’s the derivative right here let’s write it up and so

we’re remembering the derivative of cotangent and

basically i pull the minus 4 pi out oh um cosecant squared right

yep so i forgot my squared there derivative

of tangent is cosecant squared cosecant squared there all right

all right so next problem let’s um look at the difference between

when we’re cube rooting our angle and versus when we’re cube rooting our sign

01:11

here so let’s look at this real quick so what happens when we take the

derivative of cube root of sine the cube root of x

and when we take the cube root of sine so how are these different than each

other so we could rewrite these as sine of x to the one third

and we can rewrite this right here as sine of x to the one-third

so when we take this derivative here on this one

i’m looking at the sine function first and i have to take the derivative of the

outside function so derivative of sine is cosine and i

leave the inside part alone now times the derivative of x to the one-third

what is the derivative of x to the one-third what is that derivative well

it’s x to the sorry it’s one third comes down and then x to the

01:12

one third minus one so minus two thirds right and so we can rewrite this as

one over 3 and then cube root of x squared so this will be 1 over 3

cube roots of x squared so on this function on this function

right here i take the derivative of sine first and i leave the inside alone

and then i take the derivative of the inside part right here

that’s different on this one this one i need to take the derivative of the

one-third power first so here i have derivative

and i’m going to bring the 1 3 down i’m going to leave this alone

and i’m going to say 1 3 minus 1 so minus 2 thirds

now times the derivative of the inside part which is cosine

01:13

and so we can simplify this this will be cosine over one-third

so three’s on the denominator cosine and then we have a

cube root of sine squared so as you can see the derivatives are

very different so for the cube root of sine this is our derivative

the cube root of sine this is our derivative for sine

of the cube root of x this is our derivative right here

and as you can see the derivatives are very different

all right so let’s go to our problem now and solve our problem

so we have y is equal to sine of cube root of x plus cube root of sine x

so we’re just putting them all together in one problem right

so when i take the derivative of sine to the

01:14

x to the one third i’m going to take the derivative of sine

first sine of cube root of x times the derivative of cube root of x which is

3 over cube root of x to the third i took the derivative of the outside

leave the inside alone now times the derivative of the inside

derivative of cube root of x plus and now this one right here

so this is sine x to the one third power

so i’m going to bring the one third down and then sine x to the minus

two thirds power times the derivative of sine

so y times derivative of sine that’s the chain rule

so we use the chain rule right here and we use the chain rule right here

and then the last step would be to write that fraction down

and so that’s what we got here using the chain rule

01:15

and we just write that as cosine of three over sine x to the two thirds

just like we did before all right so i hope that seeing those right next to each

other you get a good feel for how they’re different

all right so now let’s work on some more let’s chain these

chain rules together here let’s see if we can get this here

all right so let’s see here what our function is

so y is equal to sine to the fourth power of x squared minus three

minus tangent squared of x squared minus three

and let’s do the derivative so i’m looking at sine to the fourth

fourth is the outside function right here so i’m gonna bring the four down

and i’m gonna leave everything else alone

01:16

and i’m gonna reduce the power by one so instead of all of this to the fourth

power now it’s all to the third power now times the derivative of sine of all

of this which is cosine of all of this and then now times the derivative of the

inside part right here times 2x so as you can see we use the chain rule

twice here depending upon how many functions you have inside of other functions

will determine how many times you use the chain rule i have this function

inside the sine function and then i have that function inside of

the fourth power so i got two compositions there all right so then now minus

so now we bring this two down and then tangent of

all of this now times the derivative of tangent and the derivative of tangent is

secant squared x squared minus three and then now times

01:17

the derivative of x squared minus three and so there’s our derivative there

and let’s see here the only thing we can simplify out of all this is the 4x

so that’s not going to be really all that exciting what’s important is

that you see the chain rule used twice here so you might write it like this if

you’re still trying to get it sine of x squared minus three to the fourth power

and to find this derivative you might build it up first first what

is the derivative of sine what is the derivative of x squared

minus 3 and then what is the derivative if you put that inside that

and then what’s the derivative if you put all that inside the fourth power

01:18

so you could try to build up those derivatives and piece it together

but in the end you should be able to get this

y right here is equal to this derivative okay so i think we got those type of

examples um good and now i want to look at a different type

of example now now in this example here we’re not given rules for our function

in fact the functions are kept secret from us

so we’re just saying f is a function but we’re given the derivative and we’re

giving another function which is a composition

g g is a composition of functions the function three x minus one

is put into f so g is a composition and we’re asked to find the derivative of

01:19

the composition and then we’re asked the second question another composition

1 over x is a function and we’re putting that function inside of

f so h is a composition of functions also

so we’re asked to find two derivatives g and h and both of them are composition

functions and the reason why i point that out is because

that’s exactly when you use the chain rule when you have a composition of

functions is when you use the chain rule so here we go let’s see if we can

we can get this here so g prime of x is [Music] so first let’s just write g

g is f of three x minus one and so g prime is derivative of the outside function

leave the inside alone times the derivative of the inside

01:20

and we don’t know what f is but we know what the derivative of f is

and so i need to go put 3x minus 1 into that derivative so the derivative

is 1 over x but this is what the x is we’re putting 3x minus one

in so the input squared plus one and then times three

and so this is the derivative of g it’s just three over

and then we could just say three x minus one squared plus one

and we can expand that out if if we want to or need to

but there’s the derivative of g so it’s asking for that derivative g

prime is sitting right there so what about h so h is f of one over x

so again this is a composition of functions i have this function one over x

01:21

sitting inside of f so i’m going to use the chain rule

so derivative of the outside function leave the inside alone

times the derivative of the inside what is the derivative of 1 over x

right that’s just x to the minus one so minus one comes down

and then x to the minus one minus one in

other words that’s just minus one over x squared

all right so that’s what the derivative of one over x is

that comes up a lot that should be probably memorized

after doing enough exercises but any case back to h

the derivative of h derivative of outside function

leave the inside alone times the derivative of the inside function

and so then probably we’ll just put this in the front oh

01:22

actually we need to put one over x into f prime so f prime is

one over the input squared and then minus one over x squared

okay and so this probably simplifies a lot because this is one over x squared

and so let’s write it like this this will be one over x squared

and this one will be say x squared over x squared

and this is minus 1 over x squared actually it’s probably easier just to do that

there’s an x square sitting right here just multiply it through

so we’re going to get minus 1 x squared over this one that just gives us the 1

and then x squared times the 1 is x squared

and so that certainly looks a lot nicer for h prime

it’s just minus 1 over 1 plus x squared okay and so that’s what we get here

01:23

let’s see by the chain rule there’s the derivative derivative of g

is derivative of f leave the inside alone times the derivative of the inside

and then we plug in 3x minus 1 into f prime

they tell us what f prime is they don’t tell us what f is but they tell us what

f prime is so i put 3x minus 1 into f prime for h we apply the chain rule so for

h h prime we find the derivative of f we leave the inside alone times the

derivative of the inside and the derivative of 1 over x is minus

1 over x squared i put the minus out front and then we

put the 1 over x squared in and then we multiply through by the x

squared to just make it look nicer so there’s an example of hey find the

derivative even though you don’t know what the original functions are

but you know what their derivatives are that’s kind of nice

01:24

all right so here’s a similar problem we don’t know what the function f is we

know what f of 2 is and we know what the derivative is

find the derivative of g at that number 2.

so let’s see if we can work this one out here

now what’s the clue that this is a chain rule problem

so we’re asked to find g prime and if we look at g it’s a product of

two functions x squared times another function and that other function is

a composition of functions because it’s f

and we’re putting the function x over x minus one

into f that’s a composition of functions so let’s just see how we can do this

here okay so i’m going to say g prime here is

we have the product rule here x squared times the other function

01:25

so here we go derivative of the first 2x times the second one leave it all alone

plus first one times the derivative of the second so what is the derivative of f

of x over x minus one what is that derivative right there i’m

looking at that derivative there so i’ll just say dx what’s the derivative

so i take the derivative of f and i leave the inside alone

times the derivative of the inside now when i take the derivative of the inside

here i have quotient rule so i have low derivative high minus high

times derivative low all over denominator squared

all right so when we take the derivative of this

01:26

we get all of this uh does it simplify any so this becomes x minus x

it looks like we’re getting a minus one here so i’m gonna say minus one over

x minus one squared now keep in mind this parentheses right

here is not multiplication it’s function function notation f

prime and you put in that into the derivative

and they give us the derivative don’t they

they give us the derivative it is square root of

x squared plus five so i need to put in x over x minus one into that derivative

so the derivative is square root of the input squared so x over x minus one

01:27

squared plus five and then times all of that still

okay so the derivative of f over x minus 1

is all of this right here okay so let me do that again

i’m going to put this into the derivative so the derivative is square

root of the input squared plus five and then still times all of this

okay so that’s the derivative of f of x over x over x minus 1 into f

what’s the derivative so there it is all right so let’s go back to our

problem up here g prime what is g prime so g prime is product so derivative of

01:28

the first times the second plus the first times the derivative of the second and

we just found the derivative of the second it was equal to all this down here

which was square root of x over x minus one squared plus five times

minus one over x minus one squared okay so now we need to plug in two

so g prime of two so i put a two everywhere i can so i get a four

f of two over one so f of two plus four times i put two over one so that’s

two two squared four plus five nine so i get

square root of nine which is three and i put a two here

01:29

so i get one square so i get a minus one so the only thing i need left here is f

of two so f of two is given to us as minus three and so

we put in minus three there we’re going to get minus 12 and another minus 12

and we get minus 24. so our g of 2 is minus 24. now we can try to look for a

simpler way of doing this because this was kind of long

could we have tried to gotten g prime of 2 faster

without maybe finding g prime for all x so we found g prime for all x and then

01:30

we put in two but the problem is just asking us for g prime of two

so we have done it faster so let’s see if we can do that so we know that f 2

is minus 3 and we so we can plug in and find out what f

prime of 2 is and so that’ll be square root of n that’s 3 and we know

g of two we can find g of two g of two is four f of uh two over one so two

we know f of two is minus three so that’s minus twelve

so we know these three numbers here f of two f prime at two and g

01:31

of two so g prime at two will be can we find that

without finding the whole derivative so i’ll just put an x here

and see what we can do so we can say this is product rule again 2x times f of

all of this plus now let’s leave the x squared alone

and say derivative of all of this f of x over x minus one

and so i didn’t actually take that derivative that’s the part that made it

made it longer can we plug in 2 already from right here though [Music]

01:32

so we’re going to get 4 and then f of two and then we’re gonna get another four

and then we need to plug in two here so we’re gonna get f of two we

can’t take the derivative of that f of two

so you know you got to be really careful

when you’re plugging in the number right here like

in other words i can’t say this derivative of x

times f of 2. in other words if i try to go straight in and plug in that two

and say two over one two because then i’ll be taking the

derivative of a constant which is zero and so that that that that right there

is not good so if you try to take shortcuts you could end up making

a big mistake so be very careful i’ve seen um you know a

lot of people make mistake on that all right so here we go using the chain

rule and the product rule so where i got the product rule

01:33

derivative derivative of the first times the second

plus the first times the derivative of the second the derivative of the second

is where we use the chain rule so we have f prime at x over x minus one

times the derivative of x over x minus one and that’s the chain rule right there

and so we can say what is the derivative of x over x minus one

right we found that earlier just minus one over x minus one square

so now we can plug in the two and once we plug in the 2 there

we’re going to get minus 12 and then another minus 12

and we’re going to get -24 so there’s a very nice way of writing it up and

we don’t need to necessarily find the derivative of the whole expression

01:34

like we did here i mean you can do that don’t get me wrong you can find the

derivative for all x right here um so i found the derivative of this

part right here and then we plugged in the two and right here

i just said f prime at x over x minus one and i’ll plug in the x there

and then because we know the derivative of f so we can find the derivative of f

at the two so we’re good all right so very interesting very fun

and you know if you are just solving only one type of problem where

here’s a function take the derivative here’s the function take your derivative

now that is certainly important because you have to become skilled

but at the same time you want to be able to also

answer more sophisticated questions such as this one right here where the

function isn’t even given to you can you find the necessary informat

01:35

information out all right so now let’s just play a little game um we know what

f a function is and we know what the derivative is

and so what if you start putting functions into their derivatives

or derivatives into their functions and what happens so let’s look at that

can i put it over here let’s see well we can for the first one

what happens if we put f into the function

and we try to take the derivative right so we’re given a function

a differentiable function and so we have the derivative

and we’re asking what is the derivative if we put f

inside of the derivative so what is the derivative of f prime of f of x

so that’s what we’re asking right here what is the derivative

01:36

when we’re putting the function inside of the derivative

so we have the chain rule now because we have a composition of functions

and so we take the derivative of the outside function right here

so what’s the derivative of f prime it’s the second derivative

and we leave the inside of one alone and then now

times the derivative of f which is just f prime

and then now times the derivative of x what the derivative of x is just one so

that’s it so this is what happens when we put the derivative into

when we put the function into the derivative and then we take the derivative

so we get the second derivative times the first derivative

and it’s the second derivative of the function

times the first derivative so what happens if we switch

now what if we have the first derivative into the function

right so f has the derivative we go find it and then we put that function into f

01:37

what’s the derivative of that function well let’s see

now we’re going to take the derivative of the outside function first this is

the chain rule leave the inside function alone

times now take the derivative of the inside function

which is the second derivative okay so don’t think this is times here so this is

the first derivative goes into the first derivative

and then times the second derivative that’s the first derivative

so again derivative of the outside one derivative of

f is f prime leave this inside alone and then now times the derivative of the

inside so you can see how they’re different here

so that’s kind of that’s kind of fun all right so i just go and write that up

01:38

alright so what’s next well let’s look at some more position velocity

and acceleration and let’s see if we can look at something new from that

so particle moves along straight line okay so before i

before i go into this so in the last video the last episode we talked about

position velocity and acceleration so i’m assuming that you’ve watched that

video of course all video links are below in the description

so here’s a continuation of that so show that if a particle moves along a

straight line and right we’re so we’re using the s of t4 position

and the velocity v of t which is the derivative of s with respect to t

and the acceleration a of t which is the um second derivative

01:39

but for this example we’re going to show that acceleration

is the velocity times the dv ds and we’re going to use this formula to find

dvds and then we’re going to look at an example

where we’re given an explicit function for s an explicit position function

okay so we’re going to use the chain rule so why are we going to use the

chain rule because the acceleration is the derivative of the velocity with

respect to time that’s the definition that we wrote last time

and so we’re going to use the um dv over ds and ds over dt

and so we’re using this uh these differentials here so we have the velocity

which is the ds over dt i just switched the order there

01:40

and times dvds and so that’s just all given by the chain rule there

so then the last thing to do is just simply divide divide both sides by

v of t and so if if you divide both sides by v of t then you get

a of t over v of t so that tells us that dv ds is just simply that a of t

over v of t so if we’re given an s of t then we can find the

ds dt which is the v of t by taking the derivative and

we can find the second derivative of that which is minus 12

t and so then the last step would be to find dv ds which again is the

a of t divided by the v of t and you can see that from the very first

top equation just divide both sides by v of t and so this is the velocity

01:41

uh sorry this is the rate of change of the velocity

with respect to position instead of you know that’s different than dv dt

dv dt is the acceleration which is the rate of change of the

velocity with respect to time but dvds is the rate of change of the velocity

with respect to position so that answers that question and so

that’s just a little fun a little exercise using the chain rule there

so back to equation of the tangent line we covered that in episode three at

great detail and let’s just refresh our memory

our function is more sophisticated this time so we’ll get to use the chain rule

so let’s see here we can do this over here we have the function um f of x is

01:42

9 minus x squared over two thirds to the two thirds power and

so let’s find the derivative of that we need derivative because remember the

derivative is the slope of the tangent line

so in order to find an equation of the tangent line we need the slope

so let’s find the derivative now we have two thirds

that’s the outside function and the nine minus x squared that’s the inside

function so i’m going to take the derivative of

the outside function first so two-thirds comes down

nine minus x squared leave the inside alone now two thirds minus one so that’s a

minus one third now times the derivative of the inside part which is minus two x

and then we can simplify that if we want but

actually we’re looking at a point here so we don’t need to simplify it

we’re looking at the point here 1 four so we’re looking at the derivative at one

01:43

so two thirds and we’re looking at nine minus one

so that’s eight to the minus one third and then this is minus two

so putting a one everywhere and then this becomes what um so this is one half

so that cancels so we’re going to get minus two thirds

okay so that’s the slope of our tangent line that we’re looking for

so remember the equation of the tangent line we need a point we’re given the

point here 1 4. so our tangent line equation is y minus 4 equals the slope

and then x minus the one there’s the equation of our tangent line

and then we can go simplify that if we want

okay so here we go using the chain rule we find the derivative of the function

01:44

and we can simplify it if we want but what we really want to do is plug in one

so we plug in one and we get out minus two thirds

and then now we have an equation of the tangent line

actually important word there is an equation of the tangent line this is not

the equation of the tangent line we could write that equation in a lot of

different ways here’s another way to write it maybe you

like that way maybe it’s easier to graph it

here’s what a graph would look like if we’re looking at this problem right here

so this looks like the tangent line there i dashed in and we’re

looking right there at 1 4. okay so just to keep that fresh in our mind find the

equation of the tangent line all right so now let’s find out what points

for this graph we have a horizontal tangent line

01:45

so again we talked about horizontal tangent lines

in the last episode and we’ll do another example here to find out a

horizontal tangent line we need to know where the derivative is zero

right so let’s look at the derivative so we have our function right here

and we solved one similar to this before where we had x divided by a square root

and i believe we used the quotient rule so let’s go find our derivative so we

have low derivative high minus pi times derivative low what’s the derivative

of this let’s work that over here right so that’s 2x minus 1

01:46

square root and so that will be 2x minus 1 to the one-half right

so what’s the derivative of that so we’ll bring the one half down

two x minus one and then we have one half minus one times the derivative of the

inside part times two that’s a two so we see the twos cancel and we end up

with one over square root of two x minus one

less scratch work there you might might want to write

and if you write it in your way then you have to do this over again

all right so here we go derivative of f low derivative high

minus high times derivative low we found the derivative

01:47

1 over square root of 2x minus 1. so there we are low derivative high

minus high times derivative low all divided by denominator squared [Music]

all divided by denominator squared now we want to simplify this um or do we well

do we have a number to plug in no we don’t have a number to plug in

like we did on the previous example so let’s just simplify it real quick

so i’m going to get a common denominator here

so i’m going to say this is going to be square root of 2x minus 1 times square

root of 2x minus 1 all over square root of 2x minus one i’m

just rewriting the square root of two x minus one here

01:48

times one and then this is minus x over square root of two x minus one

all right and all that times or divided by square root of 2x minus 1

with a squared and so what’s happening here so we’re getting 2x minus 1 minus x

so 2x minus 1 and then a minus x and then in the denominator what are we getting

square root of 2x minus 1 but then we’re also getting another 2x minus 1

right there so in the end we’re going to get 2x minus 1

to what power well we have um one power here and then we’re adding

another so we’ll say three halves and so that will give us an x minus one

01:49

over two x minus one to the three halves and that is the derivative

and that looks like a nice derivative there so if we trace back all of our work

we started with the function we did a little scratch work over here

when we needed to but what we’re really using is the quotient rule here

low derivative high minus high times derivative low

all over denominator squared and then we have the fun part

of simplifying that because remember we’re trying to find a horizontal

tangent line so we don’t really want to go set that equal to zero right now

we want to simplify it first to find horizontal tangent line

we need to set our derivative equal to zero so let’s just simplify it

there’s the steps to simplify it now we have this right here and now we’re going

to set the derivative equal to zero in fact if we set the derivative equal

to zero what happens well x is just one i mean that’s all there is to it

when x is one we’re gonna get zero over one

01:50

which is zero and you can kind of see that just by looking at the equation you

don’t really need to do much if x is one that gives us zero

right you can just set the numerator equal to zero if you want and go solve that

all right so let’s write it up so we have the quotient rule there

and then we’re going to simplify and we’re going to set it equal to zero

so the only point where f has a horizontal tangent line is at one one

um and that’s again found by setting the derivative equal to zero

and so then i made a little sketch for us if we go sketch the graph of x over

square root of two x minus one our original graph our original function

if its graph is sketched there then we have this horizontal um

01:51

i’m sorry we have this horizontal tangent at y equals one and

it happens at one one when x is one right we set the derivative equal to

zero we found x is one and if we go back to our original function and plug in

one we get one over square root of one right we get a one so we get a one one

out and so 1 1 that’s where we get our horizontal tangent line

all right very good okay so now it’s time to look at some exercises

and you know the videos already almost uh close to two hours so

what we’re going to do now is look at these videos here

look at these exercises here and see which ones

of these exercises you want me to do in another video

01:52

so i got some good exercises for us so the first couple of exercises there

help us concentrate on how to use the chain rule in terms of

what is the inside function and what is the outside function

so i worked some examples like this at the very very beginning before we did

the proof and even though that’s a little bit longer way to do the chain rule

it really helps you understand what the chain rule is and how to use it properly

before you start trying to go do lots of examples

and and not really understanding what you’re doing but just kind of memorizing

it so you want to have a balance between you understand it

but then also you’re good at it so to give you an example of what i mean

i mean you probably already know what i mean but just to make sure

suppose you’re asked to find seven times eight

so on one hand you have it memorized you should just flat out know what it is

without even thinking why you know it or how you know it

it’s just 56 you know that it’s really fast

01:53

but if someone asks you why why is that 56 well you would tell them i have eight

plus eight plus eight plus eight plus eight plus eight seven times one

two three four five six seven and i add them all up and i get 56. so

you not only know this very quickly without any thinking at all

but you also know how to explain it so of course this comes up in everyday life

right who knows when you’re gonna multiply

seven times eight well if you become a scientist or engineer

who knows when you’re going to use the chain rule you could use it every day

or maybe you don’t use it at all and either way a good course in calculus

will prepare you so you not only need to know how to use the chain rule

like really quick through lots of exercises

but you also need to understand why it’s true what you know

understand what the statement of the theorem is

01:54

so this first couple of examples do that they want you to explicitly write out

the u and the f of u and piece it together and write out the the theorem so now

we get into some more exercises it just says find the derivative

and so now you want to get into lots of exercises where you’re getting faster

and faster at it and the exercises are more and more complicated

and you’re understanding how to chain the chain rule together

and so you know do some of those and number five

well we’re given some information we’re not given the function f and g

explicitly but from that information on that table

can you then find out what all of those are

one through nine can you find out what those derivatives are

at the given value of x so for example number three x is one

so you have to take the derivative of that expression

01:55

using the quotient rule and then plug in one

and then use the table of information on the left hand side

to find your actual value those are fun gotta practice some of those we did some

of those earlier if you look back [Music] all right and now look at those on the

right on number eight those look like really fun exciting problems

so see if you can do those if you can’t leave a comment below

and uh you know i can make a video help you check your work

um six and seven are also important all together there’s 13 problems here

so here’s 9 10 11 and 12. if you look at 12 it is the

chain rule where we chain together more functions so we have y as a

function of u u’s a function of v and v is the function of x

right so this time we got two intermediate uh differentiable functions so we

01:56

multiply them out um and then we have the derivative of a trig function

when it’s given in degrees and so that has something to do with the chain rule

so yeah let me know which one of those you want to solve

and i look forward to hearing from you okay so at this point i want to say

thank you for watching i hope you got a lot of value out of

this video we started with we didn’t know what the chain rule was

and we looked at some beginning examples we

went through the proof very carefully then we did lots of exciting examples we

did some that were really easy and basic we did some that were a little bit more

meat to it and then we worked out some where we didn’t even know what the

function was but we’re able to find the value of the

derivatives given some information in any case my social media links

01:57

below and i hope you follow me on social media you can get

feedback in terms of when the next videos are coming out

and you know you can make your comments on social media also but uh i want to

say thank you for watching this video and i hope you got a lot out of it

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