00:00

[Music] okay so you studied inverse functions

in precalculus you know these types of functions are useful

but can be difficult you also know implicit differentiation by now

in this video i teach you a fantastic relationship

between the derivative of an inverse of a function

and its derivative what did i just say let’s find out

okay welcome to episode six derivatives of inverse functions so in this video

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i’m going to give you a brief uh review of what inverse functions are

i’m going to start off with the basics but i’m going to go pretty brisk

but i’m going to talk about what a function is the vertical line test

what a one-to-one function is in the horizontal line test

i’m going to talk about how do you verify if two functions are

inverses of each other and then i’m going to go through and explain

how do you find the inverse of a function then after i do all that

a brief review um i’m going to talk about the inverse theorem

and this will talk uh this theorem will tell us when

a function will have a differentiable inverse function and

then i’m going to talk about the inverse trigonometric functions

um all six of them and so then at the end we’ll have some exercises to get to

and talk about and so let’s go ahead and get started [Music]

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okay we’re gonna start off with a brief review and so let’s try to recall what a

function is um first and let’s you know put it in terms of relations so

you know a function is a relation that has a very special property for

each input there exists a unique output so if i were to give you some examples

right um what would be some examples of some functions so for example if i

gave you this set here s set of ordered pairs say one two and three four

and square root of 2 3. all right so is this set right here a function

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is this relationship a function right so it’s a set of ordered pairs and

you know we can ask the inputs 1 3 and square root of 2

and the outputs 2 4 and 3 and we can ask the question

does each of the inputs have a unique output

so you know to be a function this this relation here

would have to have for each input it would have have only one output

and that’s the key is to only have one output so if i

input 1 into s s returns a value of 2 but it’s not the value that it returns

that’s important to be a function it needs to return only one so for example

what if we are what if we alter s and say s is also 1 3. [Music]

now this relationship right here s if i input 1 now s has two

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different outputs so this right here is as this s right here is not a function

let’s go back to this one right here this is our s right here

when i input one i only have one output when i input three

i only have one output when i input square root of two

i only have one output of three so we would say this right here s is a

function s is a function now if we were to try to

look at a set of ordered pairs where there’s infinitely many ordered pairs

then we would want an alternate approach besides just

looking by inspection at each and every one

and and looking to see how many outputs so you know if we go graph this

relationship here s this function we can say oh we have one and two

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um we go up 2 here so here’s the first point right there

and then we have the 0.34 so we can say it’s about right there and

then we have the point square root of 2 which just say about right here and here

is a three so we can look at those three points and

we can ask does it pass the vertical line test

so recall the vertical line test says if a vertical line crosses the graph

of any of the relationship if it crosses twice or more

it’s not a function and so you can see here from these three

points here no matter where you draw a vertical line it will not cross

twice or more if i draw a vertical line here it doesn’t

cross at all if i draw a vertical line right here at a three then

it will cross once if i draw a vertical line that’s square root of two it will

cross once if i draw a vertical line at one it will cross once

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everywhere else it crosses zero times so nowhere does this cross twice or more

so if you look something like x squared plus y squared equals three

and you look at a circle and now you can say

well yeah that relationship right there that set of ordered pairs

that’s on the circle there’s infinitely many ordered pairs on the circle

but there’s a vertical line and it crosses twice

so this fails the horizontal line test fails the vertical line test

and this one passes the vertical line test

so you can ask uh is is the relation a function

you know if you have a finite set of ordered pairs

and the number is small you can go through and check each one out

just by inspection i just look but a better way

would be to have a understanding of the graph and then you can try to look at

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something like the vertical line test of course there’s a lot of ways to see

if something’s a function but there’s just kind of a basic beginner’s

refresher but what i want to talk about now is the inverse

what is the inverse so you can take any set of ordered pairs here

any set i’ll look at this s we said s is a function

now we can look at the inverse relation now the inverse right here

this is not an exponent it’s just notation that means inverse and it basically

means i switch the inputs and outputs where i switch the x’s and y’s

so that is a set of ordered pairs also and i can go back through and check the

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definition of function again so now i’m asking is

s inverse a function so if i input two i only get one output if i input four

into s inverse my input four i only get one output

if i input three i only get one output so we know that s

inverse is also a function so actually they’re inverse functions of each other

s inverse is the inverse function of s and s is the inverse function of

s inverse and they’re both functions so if we stay if we if we have something

like um let’s say we have something like t and let’s say t is one two

three four and let’s say we have seven four so the question is

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is t a function and the answer is yes if i input one i only out i only have

one output two if i input three i only have one output of a four

if i input seven i only have one output of four

now i can go make the inverse this is just a

inverse relation and then i can ask the question

is the inverse relation also a function so we said t is a function [Music]

t is a function um what about t inverse though it is certainly an

inverse relation it’s a set of ordered pairs and i found

inverse just by switching all the inputs and outputs but now

if i input 4 how many outputs do i have i have two different outputs so t

inverse is not a function so when i’m using t inverse is not a function when i’m

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using this notation right here of t inverse i’m referring to

inverse relation not an exponent t inverse just simply means the formal

of switching the inputs and outputs switching the x’s and y’s

now if we were to go graph t here we would have 3 4 would be about right there

and then 7 4 would be about right there so we have a we have a four three oh so

oh yeah sorry we have a three four and we have a seven four so we have

on t we have those two points right there now you know t is a function

it passes the vertical line test no matter where you draw a vertical line

so i’ll just go ahead and put one two here [Music]

so one two is right here and then three four seven four so you can see this

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passes the vertical line test passes vertical line test that of course

means that it is a function and if we look at t inverse it just

means switch the x’s and y’s so now instead of one two

now i have the point two one and four three and four seven but what i want to

think about here is what about the ver the horizontal line test

so the horizontal line test is very similar to the vertical line test

vertical line test it has to pass if every vertical line

crosses once or less it fails the vertical line test

if it crosses twice or more if you can find a vertical line

that crosses twice or more it will fail the vertical line test

so we have similarly for the horizontal line test if you can find a horizontal

line that crosses twice or more the graph twice or more

and as you can see right here we have a horizontal line right there y equals

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four that passes through both of those so it fails

so t fails horizontal line test so fails with t

and so what that means is that t does not have an

inverse t inverse is not a function so you want to be able to look at it in

terms of ordered pairs but you also want to be

able to think about the vertical line test and horizontal line test

when we looked at the circle before as you can see for any circle

it fails the vertical line test it also fails the horizontal line test so

when you write an equation like this something like that you get a circle and

it’s not going to pass either one but here’s an example where both s

and s inverse are functions so s passes the vertical line test and

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it passes the horizontal line test but here this t it’s a function it

passes the vertical line test but if you look at the graph of t like

right here you see that t does not pass the horizontal line test and that means

it fails and that means it’s not a function and so

you know when we talk about domain and range

we talk about the inputs right so we have the

domain of s is 1 3 and square root of 2 and the range of s is 2 4 3

and the because s n versus the function we can talk about the domain of the

function 2 4 3 and the range of s inverse would be 1 3 and square root of 2.

and so we can so t is a function so we can talk about its domain

one three and seven and the range of t would be two four would be it would just

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be two and four okay so there’s some stuff there that i hope rejogs your

memory this is obviously very brief review but the next thing i

want to talk about is what is an inverse function and how do

you really know if you have an inverse function

um especially when you have more than just say

a set of a finite set of ordered pairs right

so there’s two equations that we want to look at

so using the composition of functions we have this function right here and we

have this this is f inverse [Music] and so when we look at these two here

we can ask this f and f inverse inverses of each other and we have to check that

00:15

these are satisfied so let’s look at an example [Music]

how we know if two functions are inverses of each other

suppose i’m giving this function here is 2x minus 3 with parentheses and g is

the function g is let’s go with one half x plus three so the question is

are these inverse functions of each other

so we’ll have to check these two things check one and check two

so we’ll check these two things here so here we go

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first we’re going to put g into f here’s the way some people denote this this is

f of g of x so g of x here is one half x plus three

so i just plugged in what g of x is and now i take all of this and plug it

into f so f is two and then it’s input minus three so one half x

plus three and then minus three [Music] and now the threes add up to zero

and then the twos cancel and we get one x so we get an x

so number one works for this f and g number one works because we

started with an x and we did their composition we got out

x so that’s what this is saying so if you input an x and you go to f

inverse and then you take that and put it through f and then you get

back with where you started let’s check this one also holds

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so now let’s go with g of f so now the x is going to go into f first

here the x went into g first so now the x is going to go into f first

so this will be g of f of x and what is f of x i’ll take it right here

so this will be g of and then f of x is the you know oops 2 times x minus three

so i have g of f of x g of f of x and i take all of this right here and

plug it into g so what is g it’s one half times the input that’s whatever the

input is now the input is all of this so one half times the input i’ll put

square brackets there 2x minus 3 close square brackets

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and then so g is one half the input one half the input plus three so plus three

so now we can see the twos cancel and we have x

minus three plus three which is just an x so 1 and 2 are both satisfied

and so these two functions right here are inverses of each other

so now we can conclude thus g is equal to f inverse so at the beginning

i was unsure if they were inverses of each other so

i just named them f and g now that i checked both of these to two equations here

i or i evaluated the two functions at x and got back x

so now i’m i can say g is actually the inverse of f so now i don’t have to

00:19

use a g i can just say f inverse if i want okay so this is verifying

two functions are inverses of each other

um and we’re going to do another example in a minute but

um so the next thing to talk about is how do you find an inverse in other

words i i guess that this f and g and in in in fact how did i guess

well i said okay if i input something first i’m going to subtract 3

and then i’m going to multiply by 2. so when you want to undo that how would you

undo that well someone had just multiplied by two

so this input right here is gonna take away that two take away

so what’s the opposite of multiplying by two right it’s

dividing by two and then because they took something away from three well i’m

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going to go ahead and add it back right and as you can see my guess worked out

so let’s do um another example now number three here

how do you find the inverse of a function so let’s try to do that [Music]

okay so i’m going to look at this function here f of x is going to be

2x minus 3 over 7 minus 5x so that’s the function i’m looking at

now if i go graph this function in fact let’s do that real quick so

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yeah if i go graph this here i will have something like this

so this is a rational function and it has a vertical isotope at what

seven-fifths so we’ll just dash that in about right here

and this is the line x equals seven fifths this is where the denominator is zero

and the numerator is not zero we’re going to talk about vertical and

horizontal isotopes in a great deal in an upcoming episode but for right now

that’s a vertical asymptote and it has a horizontal isotope at what

um y equals negative two-fifths so i’ll put that about right here

so let’s see that’s y equals negative two fifths

now what does this function look like it’s a rational function

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so it looks something like this [Music] this is a horizontal isotope this is a

vertical isotope and then it looks something like this

now it crosses right here at what um three over two right so anyways

looks something like this but um how do you find the inverse so so

the reason i wanted to look at the graph real quick

is just that to see just so that you can see

because there’s infinitely many points on here you’re not going to be able to

go through a whole list of ordered pairs and check each one has unique output

check check every input has unique output now we know this is a function for

different reasons it’s a rational function it’s a polynomial function

divided by another polynomial function this is a rational function

so we know it’s a function that’s not really in question the ques the

questionable part is does it pass the horizontal line test

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well you know as you can see it passes the vertical line test of course we know

that because it’s a function but it also passes the horizontal line

test no matter where you draw a horizontal line

it will not cross twice or more now maybe it looks like it levels out here

but it doesn’t it’s just going to keep getting closer

and closer to this horizontal isotope same thing here it’s just going to keep

getting closer and closer so there is no horizontal that

crosses twice or more this passes the vertical line test

passes the vertical line test now another way of saying that is

the function is one to one and what that means is

not only does every input have a unique output but every output has

unique input so that’s something you should also review

if you’re not familiar with that but we’re going to stick to this item number

three here that is how do you find the inverse of a

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function so how do i find the inverse of this

so just like when you want to find the inverse of relation

you switch the x’s and y’s so i’m going to set this equal to y

2x minus 3 over 7 minus 5x and now i’m going to switch the x’s and y’s

wherever i see an x i’m going to put a y wherever i see a y

i’m going to put an x okay so i switch the x’s and y’s and now

i’m going to solve for y now to solve this for y i’m going to multiply here

and then i’m going to expand this out so 7x minus 5xy equals 2y minus 3

so we’re getting closer to solving for y so i’m going to put this

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minus 2y minus 5xy equals minus 7x minus 3.

let’s get rid of all the negatives 2y plus 5xy 7x plus 3. and now let’s

factor out a y i’m going to get 2 plus 5x 7x plus 3 and then the last step would

be to divide and so that right there is the inverse function [Music]

so we started with the original function right here this was f of x with the y i

just used a y but this was our original function and we ended up down here with

our inverse function and we knew it had an inverse function

because it passes the vert sorry because it passes the horizontal line test

sorry hopefully i said that right a minute ago it passes the horizontal line

00:26

test no matter where you draw a horizontal line it crosses no more than once

and here’s the inverse function here so we can go and check we can check

that these two equations hold if i start with f of x first and apply

that to the f or if i start with f first and apply that to the x either way

if we input an x we should get out an x so we can go do that we can verify

[Music] i want to keep track of that though so let’s write them over here

there’s our function and here’s our inverse function that we’re going to check

00:27

7x plus 3 over 2 plus 5x so this was our function here [Music]

all right good there’s our function and there’s our inverse function

that we just found now to be inverse like i said has to

satisfy both of these equations here um i want us to do that let’s just practice

so let’s do that over here if we’re so brave f f inverse of x

that should be an x let’s see what happens so this is f of f inverse of x

f inverse of x is right here i’m going to take all of that for f

00:28

inverse it’s 7x plus 3 over 2 plus 5x so all that right there goes into the f

i think if this is some big x and it goes into f so what

happens when you plug something into f well we have to put wherever we put in

we have to put it here and here so here we go two times all of this

minus three all over seven minus five times the input

the inputs all of this so seven x plus three

over two plus five x and you might say that doesn’t look like x

well hold on we’re gonna need to get a common denominator everywhere

so it looks like our common denominator is 2 plus 5x

so i’m going to multiply this 2 here and get 14x plus 6 minus 3 times

00:29

2 plus 5x all over so each one of these has a two plus five x here

and a two plus five x here and then we’re gonna get seven times two plus five x

and then minus and then here we’re gonna get 35x

and then minus 15 and again all that is over two plus five

x so i cancel the two plus five x’s everywhere so last step 14x plus six

minus 6 minus 15x all over 14 plus 35x minus 35x

minus 15. and let’s see here the sixes add up to zero

we end up with minus x over those add up to zero when we get a minus

one so we get an x so we get minus x minus one so we get x

00:30

all right so we’ve checked this one right here

we input an x and we trace it all the way through

we get out an x so the next thing would be to try the other one so if we input

x into f first then we’ll get out an x okay so i recommend you try that there

these two functions are inverse functions so we found this inverse

using switching x’s and y’s and then solving for y

we did that in fact you don’t even have to check that that procedure works

i just wanted to check one just for fun but i’ll leave the second check for you

00:31

okay it will work all right so there’s a brief review

one two and three how do you find an inverse of a function

and how do you validate the the inverse you check

both equations we check the first one and then i’m asking you to put in the

dot dot for the second one there okay so let’s go on to the next thing

okay so now we’re going to talk about the inverse function theorem

and the inverse function theorem begins with this question what happens if you

apply implicit differentiation to this equation or to either one

00:32

especially the one on the left so let’s look at that let’s go right

here and look at that go right here and so we have f of f inverse

of x equals x so in the last episode we covered implicit differentiation and

so let’s apply implicit differentiation so this is the same

thing as f of f inverse of x [Music] and so we’re going to take the

derivative of the left side and the derivative of the right side

now i’m going to just go ahead and assume that f is differentiable

so the derivative of f is f prime and we’re going to leave the inside alone

now we’re going to do times the derivative of the inside part here

00:33

so the derivative of f inverse and so there’s the left side and the

right hand side is just one so derivative of the outside leave the

inside alone and then times the derivative of the inside here so now

i’m just wondering what is this thing right here the derivative of the inverse

[Music] is one over the derivative of f at inverse f inverse at x

this is going to lead us to the inverse theorem to find the derivative of the

inverse we can look at the derivative of f at some value here [Music]

00:34

so let’s look at that [Music] so if of course this works

under certain assumptions if f has an interval um you know

as domain or you can restrict the function that you’re interested in to

look at some interval and the function is differentiable the

derivative exists and is never zero you want to worry about division by zero

here then then then the inverse itself without having to go look at what it is

so then the f inverse is differentiable at every point in its domain

moreover you can find the value of the derivative of the inverse at a point

um by using this formula here so that’s the

you know formula that we just got there and so this says the derivative the

00:35

derivative of the inverse at some number v b where a is equal to f inverse of b

you can find it just doing one over the derivative evaluated

at that a which is f inverse of b so it’s a very nice beautiful formula

there that we’re going to go verify here and so

i wanted to look at that in more detail here there’s our inverse theorem there

and i want to look at this function again

so there it is with the vertical isotope and

and you know we drew a sketch of that pretty good last a couple minutes ago

but i want to look at that around x equals 0

and i want to verify this formula right here so the function that we’re looking

00:36

at here is the same one we looked at before which is 2x minus 3 over 7 minus 5x

so there’s a couple things i want to do to this function here

first thing is what is its inverse right so do you remember what the

inverse was well it was pretty easy to find it was just

set this equal to five switch x’s and y’s [Music]

solve for y so seven minus five y times the x two y minus three

so we have seven x minus five x y two y minus three and then we have a y

over here with a so we’ll just move that 2y over [Music]

we have y is 2 plus five x and then seven x plus three right so our inverse is

00:37

seven x plus three over two plus five x so there’s our

function and there’s our inverse we had that inverse just a minute ago so now

what are the derivatives what is the derivative of f well that we can go find

using the quotient rule right so what is the derivative here

slow derivative high minus high derivative low all over denominator squared

so what do we end up with here 14x minus 10x

and then we end up with a 2 times a 5 negative 2 negative 5 so we end up with

a 10 x and then plus 15 minus 15. so we end up with 7 minus 5x squared

00:38

and then this is what a that adds up to zero so that’s a minus one

so we end up with minus 1 over 7 minus 5x squared [Music]

and what is the derivative of the inverse so the derivative of the inverse

what do you think that would be [Music] let’s go find that now so so far we

found the right that’s minus one plus zero

all right so we found the derivative now let’s go find the derivative of the

inverse [Music] so the derivative of the inverse is so

now let’s use the inverse here so we have low times derivative high

00:39

so derivative pi is seven minus high times derivative low which is 5 all over

2 plus 5 x squared so what are we getting here in the top 14 plus 35x minus 35x

and then minus 15. so we get minus 1 over 2x plus 2 plus 5x squared

all right so there’s our functions there’s our derivative

now what does this over here say it says that if i want to find the

value of the derivative of the inverse which is this one right here

00:40

at some b then i plug that b into the inverse and then i plug that into

the derivative so let’s look at all this right here i wanted to look at

sorry i wanted to look at x equals zero [Music]

so at x equals zero what’s happening [Music] f equals zero of course we’re

just at minus three sevenths um but we’re not really interested in

that we’re interested in zero right here and so you know one way to do this

is to put zero right here or to set this equal to zero

what happens if we set this equal to zero [Music] so x will be

00:41

what uh three over two [Music] and so when x is three over two

we get out zero so that’s on f so for f so for f we have the point here three

over two zero so what happens for f inverse we have zero three over two

so when we look at the value at zero we’re going to

substitute in here f inverse the derivative at 3 over 2

and that will be the same thing as 1 over the derivative evaluated at

00:42

f inverse at the b f inverse at the b three over two [Music] so [Music]

so we have a equals f inverse of b and the b is given to us is zero and

00:43

this is three halves so we’re going to input zero here

and so this right here is the three halves

and i need to put three halves into the first derivative right here

and that will give us our value right here sorry f inverse

into the derivative right here [Music] so i need to put three halves into this

derivative right here we’re gonna get minus one and we’re gonna get um one over

minus one over seven minus five times three halves squared or said differently

00:44

minus seven minus five times three over two halves [Music] squared

all right so let’s look at this written up better so in this example here

we have our function is 5x over 1 minus 2x and on this example we

don’t want to go to the trouble of finding out a rule for the

for the inverse can we find the value of the derivative of the inverse at one

without actually finding the inverse so it seems strange to be able to find

the value of the derivative of the inverse without actually even finding

the inverse and so we’re going to use our inverse

theorem to do that so the first thing is

00:45

we take the derivative of the function f using the quotient rule so we have the

derivative of f now because we want to plug in 1

into the derivative of the inverse i need to find out

what is the input when i output a one and so i solve this one equals to the f

of x what is the input which i get out one and so solving

that equation for x i get 1 7. so when i want to put 1

into the derivative of the inverse i use my inverse function i use my

inverse theorem and i plug in 1 into that but i know f inverse of 1

00:46

and that’s why we solve that equation because we needed to find f inverse of 1

f inverse of 1 right so to plug in 1 into the inverse

means to plug in 1 as the output of the function that’s why we plugged in

1 is the output of the function and we solved

so we get 1 7 and now we plug in 1 7 into our derivative and we plug in 1

7 into that derivative we get out 49 over five

and then we do one over 49 over five and so that will give us the five over 49.

now what if you wanted to take the long way what would have been the longer way

00:47

[Music] so don’t take the short way the long way maybe this will

be interesting to look at so here’s our function so i’ll call this equal to y

[Music] i’ll switch x’s and y’s [Music] now i’m going to solve this for y so i

have 1 minus 2y times x equals 5y now distribute so we have x minus 2xy

equals 5y now let’s say we have 5y plus 2xy

equals x so i just move that over there and switch sides so y is equal to sorry

00:48

so we have x equals 5y plus 2xy so x equals y times 5 plus 2x

now divide by and i get y equals x over 5 plus 2x so this is my inverse function

and let’s write that up here so that was a 2 there 5 plus 2 2x

all right so there’s our inverse let’s write that back at the top

so here’s our original function [Music] and here’s our inverse function that we

just found here at the bottom it’s x over five plus two x

okay so let’s erase all that work that we

just did to find the inverse and we have the function

00:49

and we have the f inverse there now let’s go find the derivative of the inverse

so we have low derivative high which is one minus high times derivative low

all over five plus two x squared [Music]

and let’s see what does this simplify to

we have a 2x minus a 2x so end up with 5 over 5 plus 2x squared now

if i want to plug in 1 i would just simply plug in 1. so we get 5 over

1 so we get 7 squared so we get 5 over 49 and that’s what we got before

so the difference was this approach i had to go find the inverse first

00:50

so the first approach we just simply took f and found the derivative

but we had to go solve an equation so we solved the equation 1 equals 5x

over 1 minus 2x and the reason why is because we need that

x because that will be the f inverse of 1. so [Music]

in other words on this graph right here we have the point

on f is the point one seventh and one and on the inverse you have

the switch the x’s and y’s right so you have 1 and 1 7.

in any case this is the way to do it on the screen right now

without having to find the inverse even though you don’t have to find the

inverse you still have to solve an equation this approach here

00:51

i had to go find the inverse first and then i still had to go find the

derivative and then i had to plug in so which one is easier is debatable

um for simple problems like the prob probably the ones you’ll

come across either way is fine but let’s go on [Music]

all right inverse trigonometric functions now now we’re going to look at these

inverse functions here i’d like to just mention here

um in case this is your first episode watching here

that in the previous episode episode 5 we talked about derivatives of

exponentials and logarithmic functions so in this episode we’re going to

00:52

concentrate on inverse trigonometric functions so i’m going to

look at the arc sine and arc cosine in some detail and you know make sure that

we’re good on those before we go try to find the derivative of these

so these are functions that you should have seen in precalculus

but let me try to refresh your memory um and so let’s see here what happens if

we we do this here what if we go to our board here and say

all right here’s the sine function here it is remember what it looks like

so just something like this right it just keeps going on both both ways

so we have a height here of what one and down here minus one

00:53

here it’s pi over two we got pi we got three pi over two we got two pi

it just keeps repeating itself over and over again

now from what we said earlier well i mean we know sine is the function right but

you know from what we said earlier this doesn’t pass the horizontal line test

it does not have an inverse function hey so we’re done no inverse trig

functions no we want an inverse trig function so

actually what happens over here right this is minus pi over two

and what we’re going to do is try to restrict this

so that it is one to one right if i graph a horizontal line

it’s going to cross twice here and here it’s going to cross again and again

it’s going to cross sinfully many times you can find a horizontal line

across infinitely many times it fails the horizontal line test

00:54

particular sign is not a one to one function

if you plug in pi over two you get out one

if you plug in two pi plus pi over two then you get one again right so

you can see this function is not one to one

for this output i do not have a unique input

all right so what we’re going to do is restrict this to

so that it passes the horizontal line test so if i just erase this part of it

right here and i’ll just stop right here at pi over 2.

and i’ll just stop right here at minus pi over two

so i’m going to restrict this so i’m gonna say the x’s are between

pi over two and minus pi over two so quadrant one and quadrant four

00:55

and the outputs the y’s i’m going to say are between minus one and one still

so i’m looking at this but in a restricted sense and so now if i go try to

you know if we take our sign and we just switch the x’s and y’s and

just do it like a relation instead of going this way now we’re

going to go this way as you can see this doesn’t pass the vertical line test

[Music] so what we’re going to do is take the restriction the corresponding

restriction so now we have minus pi over 2 and pi over 2. so this is x

these are going to be the y’s and now we’re going to have

so this was the domain right here [Music] and this was the range

00:56

the outputs are between minus one and one right we could write it like this and

here we can write it like this minus pi over two to pi over two

close they’re both closed because they hit those points

and now we’re going to restrict so we’re going to take this and switch the x’s

and y’s so we’re going to look at this part right here

and this part right here and now where we cut off

at so this is not pi over 2 anymore because we switched the x’s and y’s

this is 1 and now this is pi over two and this is minus pi over two

and this right here is minus one so here we’re going to be between

that was pi over 2 sorry so here the x’s are between the 1 and -1 here

00:57

here it was the y’s now it’s the x’s that are between minus one and one

and here it’s the y’s that are between minus pi over two and pi over two well

what’s the y over here now this is called this arc sign

so we’ll just say arc sign okay so the domain and range has switched

and these hit right here just like these did over here

these are filled in all right so nice nice and curvy just like it would

be on a sign it’s just part of a sine graph all right so that’s the cos

uh sine sine one so we can look at the domain and range there of arc sine and

you can see where we get that domain and range from and we can do the

00:58

same thing for cosine so just very briefly let’s do that for cosine also

so let’s see here we have y equals cosine and those signs start to appear

and now we’re hitting it what pi over two pi three pi over two

and then it hits back up here at one again at two pi it’s just a height of one

and this is say minus one here now as you can see

doesn’t pass the horizontal line test so we’re going to make a restriction

of course it keeps going on it keeps going on but we’re going to restrict it

so it is passing the horizontal line test so i’m

going to erase all that right there and i’m just going to look at this part

right here erase that [Music] so i’m restricting the angle here

00:59

to be between 0 and pi quadrant 1 and 2.

and this passes the horizontal line test and the cosine of course is always

between one and minus one so the y’s are between one and minus one

and the x’s are between zero and pi and so now we can say arc cosine

will be the just switch the x’s and y’s okay so you know we’re going to be

looking something like this it’s going to go like this but i’m going to restrict

so the x’s were between 0 and pi now the y’s will between

0 and pi and the x’s will be between the zero and one

01:00

and so this will be instead of pi this will be a one sorry this will be a um

yeah so we have here the y which let’s just go and write that as arc cosine

arc cosine that’s the y it’s between zero and pi

and the x’s are between zero and minus one this is minus one and this is one

and let’s restrict it right there so it’s past the horizontal line test

x’s and y’s you know [Music] so there’s the graph nice and curved

like it goes along the cosine and so this height right here is pi

01:01

right there’s pi all right so there’s the arc cosine and

arc sine and the arc tangent isn’t much more difficult either

let’s just very briefly look at the arc tangent so let’s start with tangent

so tangent has so tangent think of tangent is sine over cosine

so where is cosine zero right and so we’re going to have a vertical asymptote

at x equals pi over two and minus pi over two that’s what makes the cosine zero

and those are vertical asymptotes we’re gonna go through zero when x is zero

we’re gonna get zero over one so we’re gonna get zero we’re gonna

go right through there at zero and it’s going to repeat this and as you can see

01:02

it’s not going to pass the horizontal line test and so what’s going to happen is

it’s not one to one it doesn’t have an inverse function

so we’re going to have to restrict tangent and we’re going to strict it to

this branch right here and we’re just going to ignore all the rest of them

so tangent we’re going to restrict tangent and we’re looking at minus pi over 2

less than strictly and pi over 2. these are isotopes we cannot plug in pi

over 2 into tangent norm minus pi over two so these are strict here

and then the y’s or the tangents um there’s no restriction on the range

the range just so we’ll just say positive infinity and minus infinity

01:03

so there’s the domain and here’s the range and now when we

look at the inverse of tangent when we look at arc tangent here

we’re going to be looking at the switching of the x’s and y’s

and so we’ll be able to get a graph that looks something like this

instead of a vertical tangent we get a horizontal tangent it’ll still

go through the point zero zero because when you invert zero zero you

still get zero zero so you get these horizontal isotopes

the vertical isotopes turn into horizontal isotopes

and this will be y equals minus pi over two

and see now you can see the domain will be all real numbers

just like the range was all real numbers over here

and then you can see the range here is bound between strictly bound between

the minus pi over two and the pi over two okay so that’s enough review

01:04

for that for those inverse functions there now let’s look at their derivatives

so the derivatives here the derivative of sine inverse the

derivative of cosine inverse the derivative of all six so what this

theorem is saying is that they’re differentiable on their domains

and here’s the formula here’s actually how you find their derivatives

so i want to go through two of these and make sure that you’re okay with them

the proofs for all six are very similar but i’ll just do two

so let’s look at the first one here y is the derivative of arc sine

1 over the square root of 1 minus x squared where is that square root coming

from why is there an x squared why is it on the left hand side the

01:05

derivative of an arc sine sine inverse and on the right side there’s no more

inverse function so you know where these rules come from here let’s look at that

all right so by what we said so far if we look at the proper restrictions

i can say that sine of sine inverse of x is equal to x these are inverses of

each other and when i say that i mean by looking at the proper restrictions

as we’ve shown on the previous previous uh slides here right we look at

the proper restrictions right there so when we do that we can say

sine of sine inverse of x is x in other words these are going to undo each other

when you put an x in and you can put an x any x here you want between minus one

and one when you get that output you’ll be able to put it into sine and you’ll

01:06

and you’ll get that x out okay so now you know before we do implicit

differentiation let’s let’s remember we know that sine is the

differentiable function and by our inverse theorem we know that

sine inverse is differentiable on its domain and so this new theorem that we’re

working on says yeah it’s differentiable but here’s an actual formula

so let’s go find that formula i’m going to use implicit differentiation

with resp take the derivative with respect to x

all right so the derivative of sine is cosine and i leave the inside alone

now times the derivative of the inside here so the derivative

of sine inverse and on the right hand side i get one

now let’s remember this is implicit differentiation

but what was the derivative rule that we just used derivative of sine is cosine

and i’ll leave the inside part alone times the derivative of the inside part

01:07

that’s the chain rule so now we’re trying to find all of this right here right

so let’s just leave that alone the derivative of sine inverse

or arc sine if you want is one over and let’s divide by all of that right

there so remember this is cosine of sine inverse that’s not cosine times

sine inverse it’s cosine of sine inverse all right so there we go the derivative

of sine inverse is equal to one over cosine of sine inverse

now i know what you might be thinking that’s not exactly what the formula

said the formula said the derivative of sine inverse is

1 over square root of 1 minus x squared so where does that come from

so we’re going to go on here and try to simplify this

and the way that we’re going to do that is some trig how do you do you remember

how to work with cosine of sine inverse well

01:08

i’m going to say that alpha is the angle sine inverse here

which means that sine of alpha is x again with the suitable restrictions

next year has to be between zero and one but um let’s go and just draw a quick

triangle here so let’s say this is on the unit circle right here

and so we have sine is at x which is say x over one

so what will be the length of the side right down here let’s call it a

let’s call it a y and we remember that x squared plus y squared equals one

so this y is in fact uh one minus x squared with the square root

and right so just solving for the y now really we need a plus or minus here but

actually if you think about what’s happening over here the x could be

01:09

um plus or um you know the x could be the x is between -1 and 1

but whatever quadrant we’re in quadrant 1 or 4

the y is going to be the same value right there so i’m looking for

1 over cosine of what do we call that alpha

right we just called sine inverse alpha what is cosine of alpha

we look at this right here cosine of alpha is y over one um

but let’s put that right here so we have one over

instead of using the y i’ll say square root of one minus x over one

you know one is the hypotenuse so i’ll just leave it like that

so there’s the derivative of arc sine it’s one over square root of one minus x

01:10

squared just using a little bit of trigonometry

here we can get rid of the cosine sine inverse which greatly simplifies it

this is very nice this is the derivative of sine inverse so now let’s go do the

derivative of cosine inverse so let’s try the same approach and see if it works

so cosine of cosine inverse of x has to be equal to x so if i put cosine inverse

of some input into cosine they will undo each other and we’ll get back to x

and this x is suitable for input into the cosine inverse

01:11

so x here is again excuse me okay so the x is between minus one and one again

okay so here we go implicit differentiation derivative of the cosine is as

minus sine and then leave the inside alone times the derivative of the inside

and then on the right side we have one so this is what we’re looking for so i’m

going to solve for it so dx so the derivative of cosine inverse

is one over and we’re going to multiply and divide all right so here we go minus

sine of cosine inverse of x okay and let’s do some trig and find out

01:12

what the sine of cosine inverse is so i’m going to use the same trick

before or not trick but technique i’m going to say alpha is the

cosine inverse it’s the angle and what that means is that cosine of alpha is x

so let me go look at a triangle here i usually like to draw it in the first

quadrant but cosine is adjacent so let’s say here’s alpha and here’s x

here’s the y what is the y well x squared plus y

squared is 1 so y is square root of 1 minus x squared plus or minus here

let’s see what’s going on this x is between -1 and 1

so this x is between minus one and one so

you know we could be in quadrant one or two but in either case the y will be

01:13

positive so i’ll use that as my y right there

so we have i’m to move this minus up here so this will be sine of cosine inverse

was alpha all right very good now what is sine of y reading off this

triangle here sorry what is sine of alpha reading off

this triangle here here’s alpha so sine is opposite over hypotenuse so

we get y over one so this will be y over one down here

but y over one here’s the y and that’s it so the derivative of cosine inverse is

the same as the derivative of sine inverse but instead we have a minus one

in fact if you look at that for the whole thing if you look at that the

derivative of the cofunctions all have those minus signs there

01:14

all right so that’s enough of that let’s go practice using those

find the derivative of this function right here so our function is um [Music]

sine inverse times cosine inverse so our derivative will be product rule here

derivative of sine inverse which is 1 over 1 minus x squared times the second

function cosine inverse of x plus now leave the first function alone

times the derivative of cosine inverse so that’ll be minus one over square root

i guess we could put parentheses around all those

01:15

and there’s our derivative right there so we can put those together slightly

in the numerator just say cosine inverse minus

sine inverse and that’s all there is to do there

so that’s just a product rule with those derivative formulas there let’s look at

another one what about the product of tangent inverse times

cos cotangent inverse so here we go product rule again

derivative of tangent inverse so i’ll just write that out derivative

of tangent inverse times cotangent inverse plus tangent inverse

times the derivative of cotangent uh sorry um

01:16

yeah times the derivative of cotangent inverse

so what is the derivative of tangent inverse that is 1 over 1 plus x squared

and what’s the derivative of cotangent inverse the same thing but with a minus

sign so i’ll say minus tangent inverse over 1 plus x squared

and then now we can put them together they have a common denominator

so we have cotangent inverse minus tangent inverse

all right looks good let’s do another one what about if we have tangent inverse

minus a x times secant inverse so let’s see if we can do this we have

01:17

the product rule so y prime will be derivative of tangent inverse which is

one over one plus x squared minus now we got the product rule here

so remember this is minus times the products that minus is going to go

for both terms so we’ll say derivative of the x is one

so times secant inverse and then minus and then now leave the x alone and then

times the inverse the derivative of secant inverse which

is one over absolute value and then one over then x squared minus one

all right very good all right what about sine inverse over cosine inverse

01:18

so let’s look at the quotient rule here so y prime will be low derivative

of the numerator what’s the derivative of sine inverse

it’s 1 over square root of 1 minus x squared minus sine inverse

times the derivative of cosine inverse which will be minus 1 over

and then all over cosine inverse of x squared okay so we have low

derivative high minus high times derivative low

and let’s see what does that simplify any

so we’re going to get a plus in between there

01:19

so we’re going to get sine inverse plus cosine inverse on top

then we have square root of 1 minus x squared

and that’s the same denominator for both so i’m going to move the denominator

right down here in the denominator keep that denominator the same

and then i guess the last thing to think about here

right is sine inverse plus cosine inverse that is just pi

over two so i’ll write it like this so remember these are angles

and the those angles mu because they’re cofunctions those angles must add up to

90 degrees so in other words the numerator is pi over two so i wrote pi

in the numerator and two in the denominator so it’s pi over two

all right so we can look how that’s written up

01:20

so first thing is i want to recall the cofunction theorem because i want to run

into that sine inverse plus cosine inverse so if

you have alpha is sine cosine inverse and beta is sine inverse and you know

that those are equal if and only if their angles are the same

all right so using the quotient rule and then simplifying

and then realizing that that sum right there is always pi over two

so that doesn’t simplify very much but it’s a good step it simplifies a little

bit especially if you want to take the second derivative

okay so let’s see what’s next okay so now we’re going to talk about

some exercises and these exercises are for anyone wanting to

01:21

make a request please do so in the comments below

but i’ll work any of these problems out for you

so here are two problems right here the first set is

is these two are really calculus problems but

you know they can help you with your reviewing make sure that you’re

familiar with all the ideas so one through four there and exercise one

just find out if they pass the horizontal line test determine if

they’re one to one or not and exercise two is find the inverses of each other

so those aren’t really calculus problems but they’re good review you need to

understand all that information and practice it same thing with number

three you just need to find the inverse and then problem number four is working

with some inverse functions exponentials and logs which we talked

about in our previous video i put it here for review all right and

so then here’s some really good media problems for us to work on

01:22

here’s the function f and you can go take the derivative of that

and you can go find the inverse of that and then you can go take the derivative

of the inverse of that you don’t need to do all of that because

we have the inverse theorem so you can take either route you want

and you can get either side of the equation in the inverse theorem

but there’s two exercises there for you to work on

and then here are some exploring exercises

um i hope that you give those a try and see what you can

and see what you can do and then last but not least

we have some practicing the derivative with the inverse

trigonometric functions and the chain rule and so there are some problems

right there to practice on if you get stuck with any of those

please leave a comment below and so now i want to say thank you for watching

and hope you’re getting some good value out of these

videos here in the next episode we’re going to talk about

01:23

linearization and differentiables so we’re going to see how the tangent line

gives us a tangent line approximation and so i’m

i’m looking forward to sharing that episode with you um in the um

description below is the link to the playlist that i’m talking about for the

for the series and social media links if

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