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[Music] okay so you have an equation in two variables

and you’re either not willing not able or it’s impossible to solve for

one of the variables as the function of the other but you know that the equation

represents a smooth curve at an interesting point

so there must be a tangent line but how do we find this tangent line

enter implicit differentiation [Music] hi everyone welcome back this is episode

five i’m so excited to be here so at this point now we have covered limits

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continuity differentiation and last time we talked about the chain rule

so now we’re going to talk about implicit differentiation

in fact we have four different parts to today’s episode

we’re going to talk about implicit differentiation we’re going to talk about

exponential and logarithmic different derivatives

and then we’re going to talk about logarithmic differentiation

i love logarithmic differentiation it just seems so powerful

when i’m using it and then we’ll talk about some exercises at the end so i

hope you stick to the end and uh well let’s go ahead and get [Music] started

okay so we’re going to talk about um implicit differentiation first and so we

need to know what is implicit differentiation now the idea

around implicit differentiation is that you may not always have a function of

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x in other words your your solve for y you have y

equals and you can take the derivative so just a quick example if you have

something like y equals x squared you solve for y

and it’s very easy to find d y d x it’s just 2x

and you can either use power rule or you can use the definition of derivative

however you want to find that derivative it’s straightforward because of all the

work that we’ve done when we have a function of x this is a function of x

and it’s very easy to find find the derivative

but what on the other hand if you have an equation something like y to the

sixth minus two x y squared plus y to the third x squared

minus two x equals five so now how do we find d y d x

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in fact how do we even know that there is anything on this graph

if if we just write down a random equation how do we know anything about

it right so that could be a very difficult problem

so we’re going to use implicit differentiation to help us with this

so we’re going to start off by assuming that we have an equation

and this is different from what we did so far because we’re not assuming we

have a function we have an equation and x and y and that

we’re also going to assume that d y over dx exists

in other words the slope of the tangent line exists at some point

the whole point of implicit differentiation though is to be able to find

that value we can find d dydx evaluated at a point

using the following procedure so the first step

is to differentiate both sides of the equation

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with respect to x and this is where we’re going to use the chain rule

and we’ll see lots of examples here in a minute but basically we take derivative

of both sides and then the second step is to solve for d y d x

and then if you need to substitute in a value into the derivative and then

we have our slope of our tangent line so this is the main idea behind implicit

differentiation is this process right here and so look it’s let’s look at some

examples so um again i just want to point out that we have an equation

and that’s how you know that you’re going to use implicit we have an

equation involving x and y and so let’s look at these a couple of examples here

i want to work out here this first example for so we have

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x squared plus y squared equals two so what happens so if we look at this

what we have is a circle radius two [Music]

that’s radius two we have a point on the circle x y

and the equation is x squared plus y squared equals two [Music]

so how do we find the slope of the tangent line through here

and so what implicit differentiation is going to tell us

is the slope of this tangent line we can find using implicit differentiation um

so the process goes like this just take the derivative of both sides

so when i take the derivative of the left side here um i’m going to say 2x

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and here i have 2y and i say y prime and then derivative of the right-hand side

- now why do i have a y prime right here

okay so here’s the thinking behind this i could go and take this equation right

here and solve this for y so let’s see here we have y squared is 2 minus

x squared and then we really have square root and this will be for the top

branch of the circle and then we have minus and this will be

for the lower branch or the lower half of the circle down here so i can

take this as my function right here and i can say look i’m looking for the

slope of the tangent line that’s a local concept that’s a local thing right

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so i don’t need to even think about this right here

this uh function down here i just need to worry about the top part

because this is where i’m interested in that is this point right here

and so around this point right here i can just think of it

as this function right here so in other words

when we’re looking at this problem right here we’re assuming

in the background that y is a function of x

you don’t necessarily need to go find it

for this for this problem right here you uh we were able to find it there’s what

it is that’s the function that’s going right

through there but in general you may not even be able to solve for it

so we have to know somehow some way that there is a function and i’m going

to use implicit differentiation to find the slope of the tangent line so when i

say the derivative is 2x okay that’s just

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a function of x right there we’re taking the derivative

but here y itself is a function of x so i have to use the chain rule here

because what we have is this function y is inside of another function

x a square function so this is chain rule right here so i bring the 2 down

and i say y and now to the first power and now times the derivative of the

inside function and the derivative of y which is just y prime now we need to

solve for y prime now of course you could always use leibniz notation

if you wish d y over dx just depends on how many problems you’re

solving that that notation is very friendly

y prime right in any case now if we look at step two over here

um it says go solve algebraically for d y d x right so here’s d y d x

so to solve for d y d x we’re just simply going to

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move the 2 x to the other side it becomes a negative

and then this 2 y right here will just divide and so we get minus x over y

so if you give me a point for example 1 1 i can plug in here 1 1

and then i can find the slope of the tangent line so one

one is on this circle right plug in one here and a one here so at one one

we have minus one over one which is minus one

so the slope of that tangent line is one is one there

so here we cancel the two so we get minus x over y

and then now i plug in one one and we get minus one

in fact we can find the rest of the tangent line equation because we’re at

one one so we’ll just say y minus one equals the slope

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and then x minus one and then we can go simplify that so this is just minus

x and then minus two plus one so that’s minus one and there’s the

is positive one plus one is plus two right [Music] all right so there we go

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something is not right to me this is minus

x and that’s positive one if i move the one over i get two but it doesn’t go

through that point radius is square root of two okay

so yeah the radius is square root of two and there’s the tangent line equation

right there so it hits right here at 2. [Music] all right good so

there’s an example of implicit differentiation but on this example

the function right here we could go find it if we needed to

so i could have found d y d x straight away

using the chain rule and everything else that we’ve done so far

so we could have just looked right here we could say oh the derivative

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is one half and then two minus x squared to the minus one half times the

derivative of the inside part here which is minus two x

and so then the twos will cancel will get minus x

over square root and then we’ll get 2 minus x squared here so here we’re

getting minus x over this is the y right here

for this little branch of it and so this looks like minus x over y also here

so this right here is sweet because we never need

the exact equation for the y right here all right so let’s do another example

let’s do x squared plus y squared equals minus two

what happens if we just change that to a minus two

so what do you guys think about this if we go and solve this right here we’ll

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do 2x just like we did before 2y y prime equals 0. it’s actually the same

equation now of course the problem with this

right here if we go graph this equation right here

you’ll be actually plotting no points there’s no points on this

so actually there is no tangent line there’s no smooth curve

there’s nothing it’s just a ghost there’s no points

that satisfy this equation but if you go do implicit differentiation to this

you get the exact same thing we got before minus x over y

just move the 2x over divide by 2y cancel the two so you get the same

y prime so the point is is that this right here makes no sense this is not the

derivative because there is no smooth curve and so you know if you use implicit

differentiation without bothering to look at the assumption

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that there is some why um at least locally um that is a function of

x here we don’t have such a thing all right so there is nothing to do on

example two there there’s just no there’s no points that satisfy that

so if someone asks you to use implicit differentiation

you would just say they’re mistaken you can’t do that

all right let’s look at this one right here so we have y to the third plus y

squared and minus five y minus x squared equals minus four

all right so assuming there is some smooth curve how can we find the derivative

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given a an x y value so here i’m just going to assume that y is equal to f of x

so there’s just some curve sitting over here whatever it looks like

but around this point here some point here x y that’s the point here x y

and around x y we we have y equals f of x

so we’re looking at this point here for any point here on here

here or here here we don’t know anything about

where we’re looking here it just gives us this equation right here

so let’s do implicit differentiation so we have three y squared

times y prime remember we take the derivative of the outside

function first so three y squared times the derivative of the inside part the

derivative of y is y prime and then we have two y y prime

and then we have minus five y prime and we have minus 2x and we have zero

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[Music] okay so there’s our first step in implicit differentiation we take the

derivative here 3y squared times y prime here’s the y prime and here’s the y

prime so out of these terms here we can factor our y

prime so i’m going to do that y prime is three y squared plus two y

minus five and then this part right here

doesn’t have a y prime i’m going to move it to the other side

and now the last step solve for y prime so it’s 2x all divided by this

and there’s our y prime so now we can go plug in

a point x and y here and we’ll have our derivative so what happens when the y

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is one we get x is two right so two two ones on the graph right we put a one uh

no it’s not well whatever points on this we can plug it in here and find the val

the slope of the tangent line so let’s look at another example

this one is more serious so we’re given sine of x plus y

equals y squared cosine x so let’s differentiate with respect to both sides

so derivative of sine is cosine leave this inside alone times the derivative

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of this part right here x plus y okay so derivative of sine

leave the inside alone times the derivative of it

and now on the right hand side we have product rule so think of that this is

product of two functions so derivative of the first

is two y y prime remember that’s the chain rule times the second one plus

y prime or sorry y squared times the derivative of the second one

okay we still have to find this derivative here this is cosine of

x plus y times so the derivative of x is one the derivative of y is y prime

and then we have all this stuff over here 2 y y prime cosine x

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plus let’s say minus y squared sine x all right now to solve for the y prime

i need to expand this out because this cosine x plus y

is for each of these so cosine x plus y times one

and then y prime times cosine x plus y and then we have this one right here and

you know what i’m going to move this y this one right here this this one

right here has a y prime in it i’m going to move it to the other side

so that becomes minus and this one i’m just going to leave over here on this

side so i’m going to say minus y squared sine x

so both of these terms here have a y prime in them so let’s factor out y prime

so i’m going to get cosine x plus y and then minus 2y cosine x

and this one right here i’m going to move to the other side it does not have

00:20

a y prime in it so i’m going to move it to the other

side so i’m going to say minus y squared sine minus cosine x plus y

and the last step would be to solve for y prime

so i’m going to take all of this in fact i’ll just put a minus out front

so i’m going to say y squared sine x plus cosine x plus y

all over so i’m going to say this minus this for the whole top because both of

my minus and then i’m going to take all this right here and divide it underneath

so cosine x plus y minus 2y cosine x and there’s dydx

okay so we can go here and see so here we’re taking the derivative of

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both sides with respect to x and i have to multiply

that cosine x plus y times each of those terms

so the next step is to expand out the left hand side

and then to move over the one that has d y d x up to the same side

then we factor out the dydx and then we divide okay so there’s our implicit

differentiation procedure so the first one two three

now the first step and the second step are derivatives

and then the last three steps are just solving it for the y dx algebraically

alright so next example use [Music] so this time we’re going to look at the

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folium of the cartes so we have x to the third plus y to the third

and let’s see if we can do that so i’m going to take the derivative of

both sides with respect to x so i’m going to get 3

x squared and you can say times the derivative of

x with respect to x if you want that’s just times one

and then plus and then three y squared times the derivative of y with respect

to x and that’s not a one and then now equals now on this one

we have product rule so i’m thinking is 6x and then times another function y

so remember in order to do these problems

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we’re assuming that y is a function of x we’re not actually trying to solve and

find that function of x we don’t need to we have implicit differentiation

but we’re assuming that so we can find the derivative of y it’s just d y d x

but the point is is that when you’re looking at six

x you need to think about this as two different functions

six x is a function and the y is an f of x so you really have two

functions of x here but we don’t need to write any of that

so it’s just going to be derivative of the 6x which is 6

times the second function y plus now leave 6x alone and times the

derivative of y now sometimes i use y prime sometimes i use dydx

either one is fine you know just stay consistent within your problem

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don’t need to you know write y prime and d y d x all in the same problem

mix it up in any case we have d y d x here we have a d y d x here

let’s put those on the same side so i’ll say three y squared d y d x

minus six x with a d y d x and everything without a d y d

x let’s put on the right hand side so i’m going to move that over and make it

negative all right so now we’ll just factor out d y d x

so this will be three y squared minus six x equals 6y minus 3x squared

and then now we’ll take this whole thing right here and divide it underneath

so dydx is 6y minus 3x squared all over 3y squared minus 6x

all right so it looks good now we can simplify this a little bit more

00:25

but let’s go back and see what the problem is asking

so use implicit differentiation to find tangent so using implicit

differentiation in particular on the right hand side we have product rule

but anytime you’re using implicit you’re using the chain rule also

and so then we divided and then looks like we canceled some threes

and it also looks like i played switcheroo with the minus signs

let’s go see that so let’s factor out a minus 3 on top so i’ll say 2y

or i’ll say x squared minus 2y and then i’ll factor out a minus 3 down here

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and i’ll say 2x and then minus y squared and so

that that will cancel the minus three so i factored out a minus three on both of

them okay so we’re going to be interested in a tangent line at 3 3 and so our

tangent line equation is so we’re looking at three three so whenever you

plug in three and three into the equation it’s satisfied so that’s a

point on the graph and now the you know the tangent line is y minus

3 equals the slope of the tangent line we’ll have to come here and plug in 3

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3 and we do that we’re going to get some number so what should we get we

should get so let’s say here d y d x at 3 3 is so we’re going to get 9 minus

6 over 6 minus 9. so what is that three over minus three so we get minus one

all right so the slope is minus one so minus and then x

minus three and so here we’re gonna get minus

x and we’re going to get plus three and then another three so we’re getting

plus six so whatever this curve looks like at 3 3 we have a tangent line there

so actually let’s see what it looks like um so the the slant isotope there

00:28

is filled in um and it’s actually not part of the graph

should be dash but but there you see the shape of

the folium and you see it three three right there it has a nice tangent line of

minus x plus six okay next example okay here’s another um

famous curve we’ll see what it looks like at the end

um but this one right here is two times we have the x squared plus y squared

to the squared looks very interesting very this looks fun

so maybe it would look more fun if you knew the shape of it

00:29

this this curve has been widely studied looks like this

like an infinity symbol you’ve got about three going on here and minus three

and one and minus one somewhere and so we’re asked to find um what

well don’t take a peek at that let’s go do it all right so here we go [Music]

when we look at the left hand side here we have a power

and we need to deal with that power first so i’m going to bring this 2 down

and multiply i’m going to say 4 and then x squared plus y

squared and then reduce the power by 1 now one now times the derivative of this

00:30

inside part here so that would be two x plus two y

times the derivative of y so y prime okay so good so we took the derivative

of the outside function times the derivative of the inside

function and now on the right hand side we have well the 25 goes for

each piece so i’ll say okay i’ll just say 25 and then i’ll say 2x

minus 2y y prime so derivative of x squared is 2x times dx over dx which is 1

times and then minus 2y times dydx okay so here if we’re going to work with this

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some more we need to solve for y prime here so we need to

let’s say we have here um these twos come out and make an eight

and here we have an x square y square and let’s say all that’s times an

x for that term right there and then we have an

eight and then an x squared plus y squared

and then times a y y prime right there okay so that’s taking this right here

and expanding to each of these pieces here or said differently

let’s take this right here along with that 2

out here’s an a so i’m thinking about is 8 x squared plus y

squared all that times the x and then all that times the y

y prime there oops and so now on the uh right hand side um again we have

these two so i’m going to say 50. i should have just canceled the twos but

00:32

who cares so anyways we have the um x and then minus the 50

and then the y y prime all right so canceling two’s everywhere i’ll cheat

and say this is a four and this is a four and these are back to 25’s [Music]

all right anyways getting on to solving for y prime

this one has a y prime here and this one has a y prime here

so let’s say this is four and then we have an x squared plus y squared times y

y prime minus 25 y y prime equals we have the 25

x minus and we still have this right here we need to move over to the side

so we have four and then x squared plus y squared and then an x

00:33

okay now solving for the y prime we have 25 x minus let’s just say this is 4x

and then x squared plus y squared and then so i’m factoring out a y and

i’m going to divide it underneath and so here we’re going to get [Music] the

[Music] this part right here and this part right here so let’s say

minus 25 y and then plus 4 x squared plus y squared times y okay so

00:34

looks good how are we getting this right here

that is negative here when i move it over it becomes

positive so that’s positive there there we go perfect there’s the dydx okay so

we have this right here with the y prime and this right here with the y prime so

i’m going to move this one to the other side

it becomes positive and i’m going to move this term right here over with the

negative and so now these two have y primes in it

got a y prime and this one got a y prime and this one

so y prime equals this whole side right here

divided by because we have to factor out the y prime

and the part that’s left over we divide by it so we have 25y

00:35

plus 4x squared plus y squared right there times the y all right very good

so we’re going to look at this at 3 1 so we can find dydx at 3 1. [Music]

and what we’re going to get here is an equation of the tangent line

this will be minus 9 over 13 after you plug in 3 1 everywhere here

and so equation will be y minus 1 equals minus 9 over 13

and then the x minus the 3. right so we have here y minus the one equals the

slope and then x minus the three so that that simplified will just be

00:36

y equals minus 9 over 13 and then x and then plus 40 over 13. [Music]

all right so good and we can see the shape of it right here

and you can see there at 3 1. three one here we’re about right there

and we have this tangent line equation right there right there at three one

[Music] all right so next one now in this one we’re asked to find

00:37

horizontal changes so the shape is similar if you look at

the equation it’s very similar to what we had a minute ago

some constants are different um but we’re asked to find where is the horizontal

line uh where’s the tangent line horizontal so let’s go and

look for that we need to to find a horizontal tangent right

we need the derivative to be equal to zero so let’s go here and find our d y d x

so let’s see we’re going to bring the two down

and then we’re gonna have the x square plus y squared

to the first power now times the derivative of the inside part here

derivative of x squared is 2x and remember y is the function of x so

this is the chain rule here so this is 2y times the inside function which is y

so that’s times y prime that’s the chain rule right there

00:38

now on the right hand side we have 4x squared

or we can think about this as just a constant four times

two x minus two y times y prime so there’s step one of implicit differentiation

find the derivative on both sides of the equation

all right so we can cancel some twos here but what we really need to do is take

this x squared plus y squared times each of these and then try to get to

the y prime actually this has a 2 in it so we can

say a 4 so i can cancel that 4 right there

we’re still going to have a 2 over here though so i’ll just write it one more

step here x squared plus y squared factoring out those two and putting it here

so we get x plus y y prime and we cancel 4 from both sides

00:39

so now i have a 2x minus 2y y prime all right but we still need to multiply

the x squared plus y squared times the x [Music]

and the x squared plus y squared times the y and we still get a y prime

and then we get um i’m going to say plus 2 y y

prime and that’s equal to 2x and now i’m going to move this one with no y prime

to the other side and out of these two i’m going to factor the y prime

so y prime is well wait a factor first so this will be x squared plus y

squared times y all right let me write it better so factoring out a y prime

00:40

and here we’re getting x squared plus y squared times y

and here we’re getting a plus 2y now you might say oh why not factor out

the y also well it’s not really needed you could but

i’m going to get a 2x over here and this right here has no y prime in it so i’m

going to move it to the other side so i’m going to get minus x squared plus

y squared times x and so then the last step

is to take all of this and divide it by this so the right hand side is 2x minus

and then x squared plus y squared times the x

all right so that’s the numerator there and then all divided by

this right here so we get x squared plus y squared

00:41

x squared plus y squared times the y and then plus the 2y all right so good

so like i said we could factor out an x and a y

and we can look at that some more but what do we really need to do we need to

find where the derivative is zero don’t we so this needs to be set equal to zero

and we need to find those x’s and y’s and so that is [Music]

so yeah maybe they’ll factor out the x’s and the y’s

so what happens if we factor out an x here we’re going to get what 2

minus the x squared minus the y squared and in the denominator we’re getting y

00:42

and we’re going to have a 2 and then plus an x squared

plus a y squared right x squared plus y squared we’re factoring out that

factoring out that so there we go so this is equal to zero when so let’s take a

look at the sketch well it looks something like this

same shape as before this goes to two right here and minus two

and you know we can see there’s some places where

00:43

there’s a horizontal tangent line now it’s not going to have a horizontal

tangent line at zero zero it’s just not not going to have a

horizontal tangent line zero zero so we can rule out here zero zero

so you know we need to have um x squared plus y squared needs to be

equal to what um the top here well let’s just put it like that

minus and that’s equal to zero right so that’s

x squared plus y squared equals two so we’re looking along that circle there

um now with the original we look on the uh left hand side we have

the x squared plus y squared so we actually have a four so using the original

we have four because the left-hand side is x squared plus y squared that’s two

00:44

so we have two squares that’s four equals four times x squared minus y squared

so we have this equal to one now x squared minus y squared is equal to one

and so what we’re going to get now we’re gonna get you know [Music]

this right here is going to lead us to 2x squared equals 3.

so just simply taking this one and this one right here

00:45

and adding them up to get rid of the y squares

so we’re going to get two two x squares equals 3 that will give us the

x’s there so we’re going to have two x values so we’re going to have the

or minus square root of three over two and then now coming back and finding the

y’s we end up with plus or minus square root of one half

and so those are the x and y values and we get four of them where we have a

horizontal tangent line [Music] so we can see that in our sketch here

so we use the implicit differentiation to find dydx

00:46

we’re setting dydx equal to zero and we’re going to use the numerator

equals zero right i mean i don’t lose anyone there so

you know if you have zero over ten it’s zero zero over a hundred

is zero right so i need the derivative to be zero

so i need the numerator to be zero and actually the denominator will not be zero

we’re rolling out zero zero there’s no horizontal tangent line there

so i’m not worried about the y being zero there

this will not be zero that’s two plus x squared plus y squared

so we need this right here to be zero that’s what everything hinges on

right because of this property to get zero i just need zero up here so we got

that worked out [Music] so but of course it’s not enough to have

x squared plus y squared equals two and we can use the original because

00:47

the point that was looking for is not only on the tangent line where the

derivative is 0 but it’s also a point on the original graph right so

using those together we get the 2x squared

equals 3 we get the x’s and we get the y’s

we can see a nice graph and you can kind of visualize

where those points are on that graph that gives us a horizontal tangent line

[Music] okay one more example and this time

we’re going to ask for vertical tangent so let’s see if we can do this here

okay so we’re going to start off with implicit differentiation

although we could perhaps try to solve this for y

we would get two branches we don’t need to do any of that

00:48

let’s take the derivative with respect to x on both sides so we end up with 2x

and then okay so here we have product rule

so actually let me write this down so we have x squared minus 3xy plus 2y

squared equals minus 2. so here we have product rule so i’m thinking about it as

minus 3x is one of my functions and then remember we’re assuming that y is a

function of x so i have i have two functions there a product of

two functions so we have a 2x minus now derivative of the 3x is 3

times the second function and then another minus because this minus goes

for the whole product rule so minus and then leave the 3x alone

and now times the derivative of y and that’s y prime

plus now we bring the 2 down so 4y and then times the inside function y prime

00:49

and the derivative of the other side is 0. so we have these two right here that

have y prime in it so let’s say we have 4y minus 3x

so we have a 4y with the y prime and we have a minus 3x with the y prime

so y prime and the ones that don’t have y prime let’s move them to the other

side so that minus 3y now becomes positive

the 2x now becomes negative and now we can solve for y prime

it’s the 3y minus 2x all over 4y minus 3x so there’s our y prime right there we

just simply take this right here and divide it underneath

and we have the y prime [Music] okay now in this example we’re looking for

00:50

a vertical tangent line so horizontal tangent line

is where the numerator is zero and the denominator is non-zero

and now we’re looking for where the denominator is zero and the numerator’s not

so we can solve for that we need 4y minus 3x to be zero [Music]

so how can we find those x’s and y’s that make this zero there’s

there’s infinitely many of them that make this zero

um well we can solve this for y right so you know move the 3x over

divide by 4 so y is 3 4 x there’s infinitely many of them

but we can use this so move the 3x over and divide

by 4. but we can use this with our original

00:51

so our original here has the x’s and y’s in it

and i’m going to go and substitute in the y is 3 4 x

so we’re going to get x squared minus 3x and for that y i’m going to use the 3 4

x plus 2 times y squared equals minus two so just plug in the y

and this will give us an x value so if we calculate this up right here

we’re going to get x is plus or minus four well how do we get that

so we get x squared and then here we’re getting what minus nine fourths x square

and then we’re getting get 9 times 2 or let’s just say 9 over 16 cancel

00:52

so 8 equals minus 2 oops so uh obviously we need an x squared here

and then here we’re getting 9 and we’re getting 16

and then 2 over 16 is 8 and then still x squared

all right so if we add up all the x squares how many of them do we get

we get x squares so think of this one as eight over eight so here we’re getting

eight of them and then minus twelve of them so

eight minus 12 and then we already have plus nine of them

so what does that give us um five of them five over eight minus two

00:53

so x squared equals well this is certainly impossible here

perfect square on the left and we get minus two there [Music]

so we’re putting in three fourths x here and three fourths x here

and it has to be equal to minus two so [Music] nine fourths x squared

and here we’re getting 9 over 16 there’s a 2 there

okay so this is all over 8 again this obviously has to be a negative

number up here but we’re going to get 8 minus 18. plus nine

00:54

and so what is that um minus one over eight so x squared is sixteen

x is plus or minus four all right so just adding those fractions

together we get minus one over eight and then so you know we get x is plus or

minus four [Music] okay so once we get the x’s plus or minus 4 [Music]

now we can you know go find the y’s from right here so we got the point here

four and minus four so this one will give us the three from right here

and when we use the minus four we get a minus three

00:55

so there’s the two points there all right so good

so we use implicit differentiation we take the

derivative with respect to both sides don’t forget those d

y’s in there that’s the chain rule we solve for d y d x now we’re looking for

a vertical tangent line here so i’m going to go with the denominator is 0.

4y minus 3x is 0. i solve that for y i put that y back in the original function

to find my x’s once we get the x’s then we go and get the y’s

now for each of those points how do we know we have a vertical tangent line

well for example 4 3 well 4 3 makes the denominator 0

and what does it do to the numerator right so it’s

nine minus eight so it’s one over zero so that’s definitely going to give us a

00:56

vertical tangent line there if we sketch the graph of this equation

you can see that at minus four minus three there’s a vertical tangent line there

all right so that looks really nice [Music]

okay so now we’re going to talk about derivatives of exponentials and logarithms

and we’re going to start off with the theorem first for derivative derivatives

of exponentials and in order to find derivatives of exponentials

we need this theorem first this theorem is going to give us the value of

some limits so when i’m looking at this limit right here as h approaches 0

00:57

of x to the h minus 1 over h so that limit is 1 and

there’s only one number that satisfies that limit x when that limit is 1

right so there’s only one number x it’s unique

and there is a number x where that limit is zero

where that limit is one sorry and that’s called euler’s number

moreover this limit is always equal to natural log of x

whatever x is as long as x is greater than zero

so this is a nice theorem here which will help us immensely proving the

derivative rules for exponentials so if we look at the derivative of

exponential function now when i’m talking about exponential function what

am i talking about here just to make sure we’re on the same page

00:58

so here’s a exponential function now what do i mean by exponential function

well b is the base it’s either between zero and one or it’s greater than one

that’s what i mean by exponential function it looks like it has this form

right here whatever b is it has to satisfy one of these two right here

so that’s what i mean by exponential function

so you can see the derivative right there the derivative of an exponential

function is the derivative of an exponential function

is the exponential function times natural log of b put parenthesis around that

so why is that true well typically when we have a function like

00:59

this we’re going to go apply the definition of derivative because we

don’t have a derivative rule for this yet so this would be the limit as h

approaches 0 of b to the x plus h minus b to the x

all over h and if we look at the numerator here

we can say okay b to the x plus h that’s the same thing as b to the x

times b to the h and so because this right here has a b to the x

in it we can actually factor b to the x out so we have b to the x times

b to the h minus one all over h and in fact this b to the x here does

not have any h’s in it so b to the x is the constant

01:00

with respect to the h here so i can pull the b to the x out

and so this is b to the h minus one over h

and now we can use that theorem that we just stated

that this limit right here is natural log of the b

this would be natural log of b [Music] all right so long story short the

derivative of an exponential function is the exponential function times

log of the natural log of the base and it just comes from the definition of

the derivative and knowing that this limit right here

knowing the value of this limit right here [Music] now in the special case where

b is actually equal to euler’s number right so what happens if we have euler’s

01:01

number what changes well euler’s number actually is

greater than one it’s approximately 2.718 something something

it’s an irrational number but anyways the derivative will be e to the x

times the natural log of e natural log of e is just one

so e the derivative of e to the x is e to the x

and you can see that if you follow this proof here this is e to the

x plus h this is e to the x the e to the x’s can be broken up this is an

exponent this is exponent law [Music] so we can factor out an e to the x

everywhere we still have e right here so this would be e to the h

minus one all over h we can bring the e to the x outside of the limit

01:02

because the limit as h is approaching zero so this will be e to the x

and we still have h’s approaches to zero and everything with the h in it

except we have an e right here and from our theorem that the previous theorem

this limit right here is one so we have a natural log of e this would be e

and so we have e to the x okay so it’s the same proof whether your base

is arbitrary or it’s e it’s just that in the case of e the

natural log of e is one so things simplify great deal

01:03

okay so let’s go look at some examples now [Music] and let’s do this example

so on this example here we should be able to do this example here because

this is the product rule so our function is e to the x secant of x

we’re looking at x is pi over three so to find the um

bind and equation of the tangent line that should say an equation of the

tangent line and find any equation of the tangent line in any case

when we find this derivative right here we’re going to be using the product rule

so i think of this as a product of two functions

so here we go derivative of the first derivative of e to the x is e to the

01:04

x times the second function plus the first function times the

derivative of the second derivative of secant is secant tangent [Music]

secant times tangent so there’s the derivative now of course we

can factor out an e to the x and in fact the secant if we wish

secant x 1 plus tangent but in order to find an equation

of the tangent line we’re going to need to plug in pi over 3

to get the slope of that line so this will be e to the power three

and then well what’s secant um at pi over three well in case you don’t

remember that’s just one over cosine pi over three

that’s one over one half or you know you probably just remember it’s two

01:05

so this would be two and then one plus and then what’s tangent of pi over three

what’s tangent to pi over three right so that’s just square root of three

tangent pi over three is just sine of pi over three cosine pi over three [Music]

right so you get some twos that cancel there in any case

this is the slope of the tangent line uh

perhaps i’ll write it with the two first [Music]

and in fact maybe i’ll just call that number m

just writing the two first makes it look slightly nicer

but so i don’t have to keep writing all that down i’ll just call it m

all right so what’s the equation of our tangent line well

they tell us pi over three is where we’re at for the x

so what’s the output so what’s the function value at pi over three

01:06

so the function value at pi over three so the function is e to the x

so e to the pi over three times secant pi over three

um oops secant of pi over three right so that’s just two e to the power 3.

and so equation of the tangent line will be y minus 2 e to the power 3

and then m our slope that we found right here and then x minus the pi over three

and so you know there’s an equation of our tangent line okay good so

there’s all the steps typed up there um gotta find the

01:07

so there’s another form of the equation of the tangent line

and this time i didn’t use an m to shorten it up but um

yeah derivative of e to the x is e to the x

work out a lot of fun examples with that all right so what about the derivative

of the logarithmic function so let’s make clear what i mean by a

logarithmic function again so when i’m talking about

a logarithmic function i have this base b and b is between 0 and one

strictly and or it could be greater than one so this here is the base some real

number satisfies one of those two all right so

01:08

you know just the usual logarithmic function that you’re used to

but um how do we find the derivative of this

um now in the next episode we’re going to talk a great deal about

inverse functions and how to find derivative of inverse functions

and so you might have seen that exponential functions logarithmic

functions they’re inverses of each other right now we’re going to use the

property that you may remember from precalculus that e to the natural log of x

is x and the reason why that looks fun is because we can use

implicit differentiation to this right i mean we know how to do implicit

01:09

differentiation and we know what the derivative of e to the x is

but this isn’t just e to the x this is a function in here instead of an x so

we’re going to have to use chain rule so let’s do this what is the derivative

of the left hand side well the derivative of e to the x

is just e to the x but if we’re going to use chain rule

if we have a function in here say u of x

now the derivative will be e to the u of x times the derivative of u of x

times d u dx and you know that’s just the chain rule

right there right if you have a composition of functions

so here we go so derivative of e to the natural log of x will be e to the

natural log of x times the derivative of natural log of x

01:10

which will we’ll just write it as derivative of natural log of x

all right times the derivative of x but that’s just one

and now on the right hand side times the

derivative equals to the derivative of x which is one

so we have the derivative of the natural log of x

is one over so i’m just saying this is derivative of the natural log

and then i want to divide this underneath so this is e to the natural log of x

but what is e to the natural log of x that’s

x so this is 1 over x so the derivative so the so e to the natural log of x

that’s just the x so this is just 1 over x

so the derivative of natural log of x is 1 over x

01:11

so that’s sweet derivative natural log of x is 1 over x so let’s do some

exercises [Music] or we could do this for any base base b um

using similar property we don’t have to use base e we can use any base we want

so let’s see here uh b to the log of b of x is x also

right so we can do this for any base right between zero and one and greater

than one so again the derivative will be you know we found the derivative of the

natural log or sorry the derivative of the exponential which was the exponential

[Music] and then times natural log of the base

01:12

and then now times the derivative of the numerator of the

uh of the inside function so that’s the derivative of log derivative

of log base b of x and then on the right hand side we get

one right so this is the derivative of an exponential it’s the exponential

times log of the base so to find the derivative of log base b of x [Music]

it’s just 1 divided by this right here so we’ll put log base b first and then b

to the log of base b of x but what is b to the log base b of x

that’s the same thing as x so this is 1 over log of b times the x

so derivative of log base b of x is 1 over natural log of the base times the x

01:13

all right so that’s that theorem there all right so let’s look at another

tangent line problem make sure you get lots of good practice in [Music]

so here we go we need to find the derivative here y prime [Music]

think of this as the product of two functions x squared

and then times another function natural log of x

so let’s see here derivative of the first one is 2x times the second plus

the first times the derivative of the second and we’re looking at e squared

01:14

so there’s the derivative and so we can um plug in the e squared whatever we

want and get the slope of the tangent line there okay so let’s just go here and

[Music] all right so there’s our derivative that we just found

simplify the x squared over x and we’re looking at the point e squared

now if i substitute e squared in then we get 2 e to the 4th right because

y of e squared is e squared squared times natural log of e squared

and so that’s 2 e to the 4th yeah [Music]

right so if you substitute in e squared into y

then it’s e squared squared and natural log of e squared

right 2 comes down natural log of e is one so that’s a two

01:15

and that’s just e to the fourth [Music] all right so that gives us the slope of

the tangent line by substituting in e squared into our

derivative so we get 2 times e squared natural log of e squared

remember natural log of e squared is just 2.

all right so therefore an equation of the tangent line

so we’re looking at the point e squared comma two e to the fourth so there’s our

tangent line equation [Music] and that simplifies a little bit

all right very good let’s do let’s do something a little bit more different

besides just taking a derivative so here we’re giving a y

and we’re given a differential equation this is a really good time to talk about

what a differential equation is in case you’ve heard of course called

differential equations right if you’re a calculus student

01:16

then one day you may take a course of differential equations

but the basic idea is this you’re given a differential

equation and you’re asked to find the solutions of the differential equation

is this one of them is this y for some a and b does it satisfy this

differential equation so the way to do that is to construct the differential

equation using the y so here we have y equals ax natural log of x

plus b e to the x and our equation over here wants us to

uh look at the second derivative so we need the first derivative

all right so here we go thinking about this is a product of two functions

a of a times x and the other function is natural log of x so

01:17

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

the second plus and now the derivative of b times e to the x

is just b is a constant and derivative of e to the x is e to the x [Music]

so we can simplify this a little bit a natural log of x plus a

plus b e to the x and now because our differential equation has a

second derivative in it we need to find our second derivative so this will be

so a is a constant derivative of natural log is

one over x derivative of constant zero and derivative of b e to the x is

constant b times the derivative of e to the x

and so we need to know when the second derivative this is the second derivative

[Music] we need to know when the second derivative minus y

01:18

and this is our y right here a x natural log of x

and then this is the plus but the minus is going for the whole y

so minus b e to the x so this is y prime minus the y

and then that has to be equal to zero so can this ever be equal to zero

ah the b’s actually add up to zero don’t they all right so let me do that again

so this is y prime y double prime minus y equals zero

so this is y double prime is a a x plus b to the e to the x

that’s y double prime minus the y and so the original y was a x natural log of x

plus b e to the x and that has to equal zero the original

01:19

says has to be equal to zero so this b e to the x minus

this b e to the x they add up to zero which means b can be anything it’s free

and we get a to the a over x minus a x natural log of x has to equal

zero so a has to be equal to zero and so that’s the solution that we’re

looking for there is in order to solve this differential equation

or in order to check if we have a solution to the differential equation we

simply build the differential equation so the differential equation has a

second derivative in it so i find the first derivative i find the second

derivative and then i plug all that in and i simplify

and i set it equal to zero because that’s what the second derivative that’s

what the differential equation said in any case a is zero

01:20

and b is any real number all right so one more derivative this is an easy one

this problem should not take long at all right y of x is x squared

minus 3 e to the x so the derivative will be 2x minus 3 e to the x so

derivative of e to the x is e to the x alright next one how about this one

so this time we have one minus two and then we have e to the minus x squared

so derivative of the one is zero now we have a constant so minus two

01:21

now derivative of e to the x is e to the x

but we don’t have an x we have a minus x squared

so we have to use the chain rule times the derivative

of minus x squared so what do we get for all this

minus 2 e to the minus x squared times the derivative of minus x squared

so that is what minus 2x and so we could just simply write it all as minus 4x

so derivative is so why do we do times the derivative of minus x squared

because that’s the chain rule we have a composition of functions

and then we just simply bring those minus twos together

01:22

and now let’s do this one all right so this one has an exponential and

a natural log so here we go y prime is well

first let’s make sure we can visualize it 5 2x squared

and then natural log of 4x so i need to think about this as a product of two

functions so i’m thinking about this is the this is the function here

and this is the function here so that 4x is

function inside of the natural log so i’ll definitely need to use the chain

rule there and this is the 2x squared that function

is inside of an exponential function so i’ll definitely need to use the chain

rule there so lots of chain rule in this problem

but the first rule that you should be looking at is of course product rule

this is a product of two functions all right here we go

01:23

derivative of the first what is derivative of this

sorry what is the derivative of this well the derivative of

any exponential is the exponential times natural log of the base

but we have the chain rule also times the derivative of two x squared

so that’s four x so that’s how we find the derivative of this exponential

derivative of exponential is the exponential

times natural log of the base that’s not the chain rule right there that’s just

the derivative of the exponential [Music] now times the derivative of the inside

function here the derivative of 2x squared is 4x

all right let’s go back to our problem derivative of the first

so that’s an exponential so exponential times natural log of the base

01:24

times derivative of the function inside here and then times the second one plus

now leave the first one alone [Music] times the derivative of natural log of 4x

derivative of natural log is 1 over and now times the derivative of the

inside function 4. [Music] so this is simplified a little bit let’s see

how we can write this up okay so that’s exactly what we had right

there on our scratch work and yeah i mean okay let’s see if we can simplify this

i don’t know that simplifying this will be useful or not [Music]

01:25

so this addition is separating terms i know the fours cancel we can also

factor out a 5 to the 2x squared and what are we going to get here

so a good way to write this would be a 4x times let’s say 4

with a natural log of 5 and then all that with an x and natural log of 4x

and then plus we factor that out plus one over x [Music]

um and that plus may may look like something [Music] so we could write this as

yeah i don’t see any advantage of simplifying it anymore

all right let’s go on to the next one [Music]

all right so now we have exponentials chain rule

quotient rule and we need to find the slope of the tangent line

01:26

so here we go i’m thinking quotient rule now so we’re going to have low

derivative high so what’s the derivative of e to the minus

x derivative of e to the minus x is the exponential

times the derivative of the function on the inside

so that’s the chain rule right there [Music] minus so that’s low derivative high

minus high times derivative of low derivative of the denominator well the

derivative of one is zero and the derivative of e to the minus x

we just found that that’s minus e to the minus x

and then all over the denominator squared so let’s see if we can simplify this a

01:27

little further [Music] this will be 1 plus e to the minus x

times a minus e to the minus x plus an e to the minus two x and then all over

one plus e to the minus x squared and so will this simplify anymore

so i’m going to have minus e to the minus x times the 1 and then a minus e

to the minus 2x again and then i have a plus one of them

and so sorry so i did minus e to the minus x times the one good

01:28

and then the minus and then positive so that gives me a minus

and then e to the now i’m going to add the exponents so i get minus two

and i keep that one the same these add up to zero here

so it looks like we’re going to get a minus and then e to the x over

one plus e to the minus x squared [Music] and somehow okay [Music]

so let’s see here what we interested in oh yeah find the slope of

the tangent line at zero so i’m gonna plug in zero

into my derivative here and we’re gonna get minus and then we’re gonna get a one

01:29

over and that’s gonna be a four so we’re gonna get minus one fourth and so

that’s gonna be the slope of the tangent line at zero minus one fourth there

okay good so you know we find the derivative and we simplify it um

you know actually derivative was done on the first step and so you know

you can substitute in zero there much quicker if you’d like

either way you get minus one fourth and so let’s just quickly recap here

um the derivative of the natural log is one over x

and if we have a differentiable function inside of that natural log then we’re

going to use the chain rule so it’ll be 1 over the u times du dx

so we just saw some examples doing that and actually if you remember the change

of base formula natural log of b of x is equal to

01:30

natural log of x over natural log of b now 1 over natural log of b is a

constant so that’s another way to think of the derivative rule there

is the derivative of a natural log uh sorry the derivat

the derivative of log base b can be found using the change of base formula

in any case if use a natural uh if use another differentiable function

then we have to remember to use the chain rule so the derivative

of log base b is one over the natural log of b and keep the u the same

times the derivative of the inside function u [Music]

okay so here’s an example let’s do this example here

01:31

what if we have natural log of [Music] four x plus nine [Music] so

the derivative will be [Music] one over and keep the u the same the

inside function times the derivative of the inside function here four

so this is just four over four x plus nine that’s it [Music]

okay so how about this one so we have here

y of x is natural log of x to the third over x plus one

so what i’m going to do before i take a derivative

is apply some basic log properties so the log of a quotient is

01:32

the difference of the logs so log of the numerator minus log of the denominator

and the reason why i do that is because now when i take derivatives

it’s a little bit easier in fact there’s another property that says i can bring

this three down so there’s two properties of logs that

hopefully you remember now there is no property for logs for addition like that

so don’t think that that’s natural log of x plus natural log of one

that’s certainly not a property of logs but the point is is that now i can take

the derivative much faster what’s the derivative of

natural log of x that’s three over x and then this will be just one over x

plus one and then times the derivative of x plus

one but the derivative of x plus one is just

one so in fact there’s the derivative so because the derivative of the natural

log of x plus one is just one over x plus 1.

the derivative is very quickly found so in other words

01:33

i used the property of logs i used another property of logs

and then i took a derivative so if you use those log properties first

the derivative is much easier because each one of these is a lot easier

now we didn’t have to do that we could have just taken the longer way

which has been what [Music] one over all of that

times the derivative of all of that and when you find the derivative of all

that you’re then going to go and use the quotient rule right so i don’t want to

do that this way is a lot nicer use the log property use another log property

now take the derivative of things that are a lot easier so that is the main idea

01:34

behind logarithmic differentiation that’s a very nice observation

let’s do this one more time [Music] so let’s look at

so there’s the derivative right there um so [Music]

putting those together and getting a common denominator

that last step is unnecessary um but if you had found the derivative

the long hard way the way we didn’t want to do it using the quotient rule

perhaps you would have found that in any case

let’s look at a even you know something more fun

this one’s so much more fun right because we’ve got all these properties in here

of these base of these logs we can use so we can write y of x

01:35

i don’t want to use the quotient rule anywhere here

so i’m going to bring that 1 4 down this is 1 4 log base 2

and then we have 3x plus 2 over x squared minus 5. so there’s our

function i just bring the 1 4 down that’s a log property now there’s

another log property that says the log of a quotient is the difference

of the logs so i can take log of the numerator [Music]

and then minus and then log of the denominator

now this isn’t correct because this one-fourth goes for the whole log

so if i break this log up into two pieces this one-fourth has to go to each

piece so you can either fix it by putting parentheses in or you could just go

01:36

ahead and say 1 4 to each piece so maybe you’ll like parentheses instead

so there we go 1 4 times the log of this and the log of this is that those two

the difference of those two logs now there’s no more logs to apply so now

i’m going to take derivative and it should be easier that’s the point

so this will be 1 4. now what’s the derivative of log base 2 of that

well it’s 1 over that times the derivative of this which is 3 and then times

1 over the natural log of the base and then minus

so derivative of a log is one over times the derivative of the inside part

01:37

here which is 2x and then don’t forget 1 over log of the base

so then we can try to simplify this if we’d like this is the three

we have a natural log of two we have a one-fourth

so we could write it something like three over and then this

four here we could say four natural log of two and then three x plus two

minus and then what will all that be we go to two x

over natural log of two of x squared minus five

so three and then we have a four and a natural log of two

and then a three x plus two and here we have a two x we have natural log of two

so there’s the derivative [Music] the derivative step right there is not

hard but you have to remember chain rule derivative of natural log is a sorry

01:38

derivative log base 2 is natural log of 2 and then 1 over that

and then times the derivative of that that’s the chain rule

okay so different ways of writing the answer

but the first step is to break it down into

simpler components so that’s something that precactus students often practice

is to take a complex looking log expression

and break it down into simpler pieces now we can apply derivative rules [Music]

and so then perhaps simplify it if you want to

to be honest there’s not really much simplification there

all right let’s go do another one here we go so this time we have a fractional

01:39

exponent so we have um s of t and so we have a quotient so i’m going

to break this up as a difference so log base 5 of the numerator

t squared plus three minus log base five of square root of one minus t

okay good so far we broke the log of a ratio into a difference of logs

but we can go one more step before we take derivative

so log base 5 t squared plus 3 minus now this this r this square root here

means we have one half power right so that’s the same thing as

one minus t to the one half whether you like to write it with a

radical or exponent but the point is is that the one half

can come down that’s the property of logs and then 1 minus t there’s no more log

01:40

properties to apply so now i’m ready to take the derivative

so it’s 1 over natural log of the base times t squared plus three

now times the derivative of the inside function which is two t

minus we have a one half here so 1 over 2 for the constant

and now we have natural log of the base and we have 1 over 1 minus t

now times the derivative of one minus t which is minus one

all right so very good so we can write that up

first we break it down using our log properties

then we use derivative rules and then we play with the answer at the end if we

01:41

want to all right [Music] now we’re going to do something fun we’re going to do

logarithmic differentiation and the hot the idea behind logarithmic

differentiation is you know take the log of both sides and then use

simplicity differentiation that’s the basic idea is

this is implicit differentiation all over again but

the first step is take log both sides now when i say take log both sides you

can actually use any base you want however since the natural log

has a very nice derivative i’m going to use natural log

on most of the examples so logarithmic differentiation take natural

01:42

log of both sides will be what i do then use

properties of logarithms like we just did you know in our previous examples

and then now use implicit differentiation once you simplify it use

implicit differentiation take derivative of both sides of the

equation and then go solve for d y d x so logarithmic differentiation says

bring logs into the problem then simplify the logs

and then do implicit differentiation so the question is

when would you use logarithmic differentiation as opposed to

implicit differentiation and so the idea is that

if logs will help you then use them so it takes a little bit practice to

figure out if logs will help you or not but you know we do some examples here

i’ll show you in fact for our first example i want to kind of question

whether or not we should use logarithms or not on one hand this is just a

01:43

function of x so why do we even need to use logs at all so you know

we don’t want to use the limit definition but we have quotient rule we

can use the limit theorem we can use quotient rule um this problem

is asking us to do it two different ways so let’s do this with

and without the quotient rule so first let’s use quotient rule

so the derivative is so low there’s my denominator times derivative

high which is all times zero so i’ll put that in brackets times zero minus high

times derivative low what’s the derivative of the denominator

so this will be three x minus one times the derivative of x minus one

01:44

which is one and then minus three times x plus one

times another one and then all over the denominator squared [Music]

so all of that squared now this is zero right here because that’s times zero

and that’s with the minus one and so i’ll take this one more step

this will be um a minus three or let’s just say a three can be factored out

and we’ll have x minus one actually those should have squares on them

my bad is 3 and then x minus 1 squared and now times the derivative of the

inside part and of course the same thing over there

01:45

and so we have squares here and so we can say here minus three

and then an x plus one squared and then all over

and as you can see here it doesn’t really simplify that much

cubed and then squared all right so we factored out the three so

we factored out the three so this should be minus

and so we’re going to factor out a minus 3 to make that a positive

and then that’s a positive well that doesn’t look too interesting

this is just x minus one squared and factoring out a minus three

factoring out a minus three we get a plus one here now and

so that’s what the derivative is if we use quotient rule we can find the same

01:46

derivative using perhaps logarithmic differentiation let me show you that now so

let’s see if we can do that down here so i’m going to put natural log on both

sides alright good so we have natural log on both sides

now on the right hand side we can simplify this further

that’s a quotient so i can say natural log of 1

minus natural log of the denominator so there we go natural log of 1 minus

natural log of the denominator so in short we get natural log of y equals

that’s a zero and so we can just say minus natural log of

01:47

all of this x minus one to the third minus x plus one to the third the point

is is that we simplified it as much as we can so we

did the two steps of natural logs of logarithmic differentiation so far

we’ve taken the lateral log of both sides we simplified it

and now we can use logarithmic differentiation

so the derivative of natural log of y is one over y

times the derivative of the y and here we have derivative of the right hand side

and so it’s going to be minus one over all of this

times the derivative of all this which is three x minus one squared minus

times one and then minus three and then x

plus one squared times one so that’s the derivative

01:48

and then to solve for this y prime right here we will multiply both sides by y

so in the end we’ll get the same answer right here

that we got for the y prime right here so when we multiply both sides by y

so the y will cancel over here that’s that’s the whole point of multiplying

both sides by y they’ll cancel over here and we’ll get a y over here

but we already know what the y is remember the y is one over

this right here so we’re going to end up with a squared here

so you know we have the same derivative it’s exactly the same

and so using the quotient rule or using logarithmic differentiation you

01:49

know you’re going to get the same derivative

the question is which approach do you take to make things simpler

in this case either way is about the same but now let’s look at the next example

how about this one now it’s very clear at least to me which route you

should take you could still try quotient rule if you’d like

but i’m going to use implicit differentiation

so i’m going to put natural log on both sides so we have e to the 2x over

x squared minus 3 squared natural log of square root of x

so i put natural log on both sides now the reason why

01:50

the natural log here is a good idea is because there’s lots of properties of

logs we can use i can take the log of the numerator minus log of the denominator

so i take the log of the whole denominator here

so natural log of the whole denominator now this is a product here so we can

apply more properties so i have the natural log of y

and this will be well actually what is the natural log of e to the 2x

right what is that that’s natural log of e to the 2x that’s just the 2x

so now we have natural minus natural log so we have a product here so i’m going

01:51

to take natural log of the first piece and then minus natural log of the second

one okay so there’s what we look like so far now make sure you have this minus

natural log of a product this minus natural log goes to each piece

normally when you say product you have a sum here in the middle

but because this is a minus for the whole natural log the minus is going to

go to each piece or said differently perhaps you used parentheses

so this is the log of a product it’s the log of the first factor

plus log of the second factor all right so now we have natural log of y

is 2x minus now i can bring this 2 down in front

so 2 natural log of x squared minus 3. so that piece is done this piece is done

plus now this is a one half to this x so i can bring this one half

01:52

down in front of the natural log so i have natural log of one half

of natural log of x now let’s put parenthesis here this is

the natural log of all of that so it’s the natural log of all of this

and now we have product rule again so i have one half times

natural log of x so i’m actually going to simplify this one more step before i

take a derivative i’m gonna keep all this the same

plus now i have natural log of a product

so it’s the natural log of the first one

plus natural log of the second one which is the natural log of x

okay so there we go now we’re ready for derivative

01:53

so let’s just check it real quick natural log of the quotient so that is the

difference of natural logs and then natural log of a product is the

sum of natural logs then the two comes down the one half comes down

the one half doesn’t come down in front of both logs the one half

is for the x so it comes in front of the first log but then i can break this

product into some the reason why i did that is because the

this is constant there’s no more x’s here so the derivative of this part will be

zero but in either case we’re ready for the derivative now

so this will be one over y times y prime

remember that’s the chain rule there the derivative of the natural log is 1 over

y times the derivative of the inside function now on the right hand side

derivative of 2x is 2 minus 2 times one over this x squared minus three

01:54

times the derivative of x squared minus three so times two x plus

now derivative of this part is zero now the derivative of the natural log of the

natural log right so i take the derivative of the outside function first

natural log and i leave the inside alone times the derivative of the inside is

one over x okay so we’re almost done so y prime

is now going to multiply both sides by y if i multiply over here by y it cancels

the 1 over y and i end up with y prime and over here i’m going to say y

so y times all of this but before i write down the y

let’s ask ourselves what is the y y was given to us y is e to the 2x over

x squared minus 3 squared times natural log of x so i’m

01:55

multiplying both sides by y but instead of y on the right hand side

i’m just replacing it with what they give us for the y and then now times

2 minus that’s a 4x right there so 4x over x squared minus three

and then this right here we have an x over an x so oh sorry

this plus right there so plus one over x natural log of x

there we go there’s the derivative that was that was pretty nice i like that

all right so we’re going to use the natural logarithmic function take

natural log of both sides and simplify our

01:56

properties there so i’ve simplified in that first step there

and then we’re going to get some simplification there all right so

let’s go on to the next one now what about if we have x to the sine x

so here’s another good example of when to use logarithmic differentiation

because if you take the natural log of both sides you’ll be able to apply

logarithmic properties so in other words when do you use logarithmic

differentiation well if you try to use logarithmic differentiation if you put

natural log on both sides then you should be able to use some

logarithmic properties if there are no logarithmic properties to use

then don’t use logarithmic differentiation

01:57

so what about here we have y equals x to the sine x should we use logarithmic

differentiation well if we try all right so i’m trying

do i have any log properties to apply if yes well then it’s probably a good idea

so what log property do we apply well we have natural log

and we have an exponent to a power so we can bring that power down in front

so this is sine x times natural log of x and so now we don’t we no longer have

exponential exponentials so this will be product rule here

sine x is the first function natural log of x is the second

so we’re ready for derivative derivative of natural log of y is one over y

times the derivative of y and now here we have product rule

01:58

so derivative of the first function times the second

plus now leave the first function alone times the derivative of the second

so there we go so we have cosine x times natural log of x plus

sine x over times 1 over x now to find the derivative we’re going

to multiply both sides by y so we have y equals all of this

now of course we already know what y is y is the sine x sine to the x to the

sine and so let’s just substitute that back

in so we can put everything in terms of only x’s if possible that’s always the

01:59

nicest idea so there’s the derivative of x to the sine x

so the function is x to the sine x and our derivative is x to the sine x times

that expression right there cosine times natural log plus sine over x so

there we go we substitute in that y and that’s it well

what if we wanted to do something a little bit more interesting

instead of sine x what if we want to do something like

all right so we did x to the sine x what if instead we don’t have sine what

02:00

if we just have some function g we don’t even know what g is

so don’t think about sine for a minute just think about any function there

so we would take natural log of both sides

we can bring that function down because that’s the property of logarithms

and so we’ll have a product rule g of x times natural log of x a minute

ago we had the sine x there but we could have any function there we

want as long as it’s differentiable so i’ll say 1 over y times y prime

right 1 over y times the derivative of y and now we have product rule again

derivative of the first times the second plus the first

times the derivative of the second and now if i solve this for y prime

i multiply both sides by y but we know what y is y is x to the g of x

02:01

so if i multiply over by y on the left hand side the y’s cancel

if i multiply by y on the right hand side i’m going to go ahead and

substitute what what it is there’s y times the whole right side

and so there’s our function this is like a new derivative rule

x to the g of x the derivative is x to the g of x times the derivative of g

naught times natural log plus g over x pretty sweet i like that um

how about we do how about we do g of x to the h

of x is that something we can even do is that a function we input a real

02:02

number in and get an output well let’s assume we can let’s assume g and h are

differentiable functions natural log of y natural log of the right hand side

bring this power down just like before that’s h of x times natural log of g of x

right so i just bring the power down in front

and now we have the same same thing before one over y times y prime

i have a product rule this function times this function

derivative of the first times the second plus leave the first one alone

times the derivative of the second what’s the derivative of natural log of g

02:03

of x it’s 1 over g of x times the derivative of the inside

well it looks a little complicated no not really now we’ll multiply both

sides by y so i’ll have y on the left hand side

and they cancel and i’ll multiply the whole right side by y

but we know what y is so y is g of x to the h of x power times

all of this which is h prime of x natural log of g of x plus h over g

times g prime and so there’s our derivative formula there [Music]

02:04

how would we do something like um well how would we do something like this

um g of x to the h of x like we just did plus p of x to the s of x

assume that all functions are differentiable that the derivatives

exist wherever we need them to exist what would be the derivative rule for that

g of x to the h of x plus p of x to the s of x

right so p is a function of x s is a function of x

now it might be tempted to take natural log of both sides

all right let’s just call this y natural log of y

natural log of that whole side that would be a bad idea though because

we don’t know a property for the natural log of a sum

02:05

so i wouldn’t do that natural log of both sides there

but on the other hand y prime would be the derivative of this one

plus the derivative of this one well we just found the derivative of this one

it’s that and we could find the derivative of this one easily

just substitute in instead of a g everywhere put

a p and instead of h everywhere put an s right so yeah we could go find this

derivative it would probably fill the space up here too much

but i’ll leave it for you to write that out

yeah so just don’t take the natural log of both sides there

i think that’s the the key thing for example

if you have x plus 2 what you don’t want to do is to take the log of both sides

because there’s no property for the natural log of a sum you know

02:06

and besides you know how to take the derivative of that really fast anyways

the derivative is just one all right so logarithmic differentiation is really

powerful um but you have to kind of really know

the logarithmic properties if you know them

then you can use them to your benefit to simplify things first

before taking the derivative so that’s the big idea behind that [Music] okay so

now we have now we have some exercises to look at and let’s do that now

so let’s look at some of these exercises here

um so we talked about that on the right hand side

02:07

when f is different functions like that but let’s look at some exercises now

so exercise one is find the derivative d y d x and those are straight forward

we just need to use the chain rule on those and then exercise two we need to use

implicit differentiation on those and we need to know the derivative of

the exponentials and derivative of signs and

stuff like that so those problems are straightforward

use implicit differentiation again on number three [Music]

then on number four we need to find the second derivative so

how do you find the second derivative well the idea is to use implicit

differentiation on the equation and that will give you dydx equals something

and that will have expressions with x and y’s in it

02:08

and so then you’ll want to go take the derivative of that

and you’ll get the second derivative now when you do that you have to be very

careful because you want to make sure that you use the chain rule appropriately

so let me just work out a quick example of that

i know we’re running kind of long here but just

very briefly you know if you have something like

x squared plus y squared equals two as we looked at before so what will be

the first derivative so it’ll be 2x plus 2y times y prime equals

so y prime was 2x over 2y in other words minus x over y

so if i want to and that’s dydx right and so if we want to go find the

second derivative how would we do that here so now we have

quotient rule so we have low times derivative high minus high

02:09

times derivative low all over denominator squared and so this will be the second

derivative here so this will be minus y and then plus x y prime

and then all over y squared okay so that’s just a minus y and that’s

a plus x y prime and that’s all all over y squared now you might say

well wait a minute that’s not x’s and y’s it has a y prime in it

well that’s true but we have x but we have y prime so we can put the y prime

in here now so we can say minus y plus x and for y prime we found that to

be minus x over y and then all over y squared and so then we can

simplify that you know get a common denominator if we wish you know we can

simplify that as much as we want to but as we saw in some examples

maybe you don’t want to simplify that too much if you’re looking to evaluate

02:10

your derivatives you may just want to plug in you know numbers right in there

and get the values that you want out of it so in that exercise there um

yeah and that exercise there it’s asking us to find the second derivative

so you find the first derivative using implicit and then you find your second

derivative just using your regular derivative rules but you may

you’ll you’ll probably you me you might get a y prime and all that

wherever you see a y prime just replace it with what you found and

then you should get nothing but x’s and y’s

all right so here’s some more exercises here five and six and seven and

there’s seven right there and let’s move up so we can see

seven right there seven’s talking about um a vertical tangent line like we

discussed earlier number eight is logarithmic differentiation

02:11

so i picked some examples there where if you take natural log of both sides

you will be able to apply some log properties

in particular number six you may want to work out those

derivatives separately take the derivative of

x to the tangent x power separately and then take the derivative of x to the

cotangent x separately using logarithmic differentiation and

then add those derivatives together to get that last one

and then we have exercise nine there and then we have two more examples here so

you know there’s 10 and 11 and then there’s the last one was which is 12 there

so you know any of those exercises you’d like for me to work out

please leave a comment below and i’d be happy to make another video

with these exercises in it let me know if that’s what you want so

02:12

i want to say thank you for watching i hope you got the value that you needed

out of this video if not if there’s something that you

need to see and it’s not here you know put a comment below

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and our next video is over derivatives of

inverse functions um so we’ll talk more about

the exponential and logarithmic and how they’re inverses of each other

but there’s so much more to inverse functions besides those

um for example there’s also inverse trigonometric functions we’ll talk about

and so um i hope to look forward to you next time thank you for watching

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02:13

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