How Do You Find the Derivative of an Inverse Function (Examples and Theory)

Video Series: Calculus 1 (Explore, Discover, Learn) Series

(D4M) — Here is the video transcript for this video.

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

00:01
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]

00:02
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

00:03
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

00:04
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

00:05
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

00:06
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

00:07
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

00:08
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

00:09
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

00:10
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

00:11
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

00:12
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

00:13
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

00:14
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

00:16
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

00:17
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

00:18
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

00:20
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

00:21
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

00:22
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

00:23
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

00:24
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

00:25
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
you want to keep up with the videos that i come out with
and so i hope that you subscribe or like this video
and i’ll see you in the next one if you like this video
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About The Author
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David A. Smith (Dave)

Mathematics Educator

David A. Smith is the CEO and founder of Dave4Math. His background is in mathematics (B.S. & M.S. in Mathematics), computer science, and undergraduate teaching (15+ years). With extensive experience in higher education and a passion for learning, his professional and academic careers revolve around advancing knowledge for himself and others. His work helps others learn about subjects that can help them in their personal and professional lives.

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