### Video Transcript

Differentiation of Inverse
Functions

In this video, we will learn how to
find the derivative of inverse functions. And weβll be covering a variety of
examples of how we can do so. Letβs start by recapping some
information about inverse functions.

Let π be a function with domain π
and range π, so π takes π to π. We call the function π, which
takes π to π, the inverse of π, if for all π¦ contained in π, π of π of π¦ is
equal to π¦. And for all π₯ contained in π, π
of π of π₯ is equal to π₯. What this definition is telling us
is that if some function π has an inverse and then we apply both π and the inverse
to some value. Then we will obtain the value
itself. Now, if the function π from this
definition exists, then we can say that π is invertible and π is the inverse of
π. We can also denote the inverse
function like this with π superscript negative one.

However, we must be careful not to
confuse this with π to the power of negative one. Since they are completely different
despite similar notation. If there is a superscript negative
one next to a function, it means the inverse. But if itβs next to a variable or a
constant, it means to the power of negative one. Another thing to note is that if π
is the inverse of π, then π is also the inverse of π. Letβs now consider the graph of
some function π.

Now, we know that we can find the
graph of the inverse of this function by reflecting it in the line π¦ equals π₯,
which is this line here. So our inverse function will look
something like this. What weβre aiming to do here is to
find the derivative of the inverse function. Now, the derivative is the function
of the slope of the graph. And one way to find the slope at a
given point is to find the tangent at that point and then find the slope of the
tangent. Letβs find the tangent to π at
some π₯-value, which we can call π. Now, the coordinates of the point
where we are taking the tangent will be π π of π. However, we can also call π of π
π such that the point at which weβre taking the tangent is at π, π.

Now, this gives us that π of π is
equal to π. Now, we somehow need to link this
point to the inverse function of π. If we apply π inverse to both
sides here, we will obtain that π inverse of π of π is equal to π inverse of
π. However, due to the definition of
an inverse function, we know that π inverse of π of π is equal to π. This gives us that π inverse of π
is equal to π. Letβs find this point on π inverse
of π₯. And we can find the tangent to π
inverse of π₯ at this point. Now, it looks as though our
tangents of π of π₯ at π, π and π inverse of π₯ at π, π are reflections of one
another in the line π¦ equals π₯. And this would make sense, since π
of π₯ is a reflection of π inverse of π₯ in the line π¦ equals π₯. And the point π, π is the
reflection of the point π, π in the line π¦ equals π₯.

Now, what weβre concerned with here
is this slope function of π inverse of π₯. So letβs consider the slope of
these two tangents. Letβs call the tangent of π of π₯
πΏ one and the tangent of π inverse of π₯ πΏ two. We can call the equation of πΏ one
π¦ is equal to ππ₯ plus π. Now, a reflection in the line π¦
equals π₯ corresponds to a mapping of π₯, π¦ going to π¦, π₯. Therefore, the equation of line πΏ
two is, π₯ is equal to ππ¦ plus π. And we can rearrange this equation
to make π¦ the subject, giving us that π¦ is equal to one over ππ₯ minus π over
π.

Another thing which weβre concerned
with here is the slopes of these two tangents. We can see that the slope of the
tangent to π of π₯ is π. And the slope of the tangent to π
inverse of π₯ is one over π. Since π is the slope of the
tangent at π, π, it can also be defined as the slope of π at π₯ equals π. And so, we have that π is equal to
π prime of π. One over π represents the slope of
the tangent to π inverse of π₯ at π, π. So itβs the slope of π inverse at
π₯ equals π. So we can say that the derivative
of π inverse at π is equal to one over π. However, we just found out that π
is equal to π prime of π. So this can be substituted in. And this gives us our result. And that is that if π of π is
equal to π, then the derivative of the inverse of π at π is equal to one over the
derivative of π at π. Which, of course, only makes sense
if π prime of π is nonzero.

Now, we havenβt proven this result
rigorously. Since we have only based the fact
that the tangents are reflections of one another off of intuition. This result can, however, be proven
using the chain rule. If we have some function π with an
inverse function π, then by the definition of the inverse function, π of π of π¦
is equal to π¦. Now, if we use the chain rule to
differentiate both sides of this equation with respect to π¦. Then we will obtain that π prime
of π of π¦ multiplied by π prime of π¦ is equal to one. Rearranging, we obtain our
result. Which is that π prime of π¦ is
equal to one over π prime of π of π¦. This is quite often also written in
Leibnizβs notation. Which tells us that dπ¦ by dπ₯ is
equal to one over dπ₯ by dπ¦. Letβs now apply these definitions
to some examples.

Given that π₯ is equal to π to the
power of π¦, find dπ¦ by dπ₯, giving your answer in terms of π₯.

We can start by differentiating π₯
with respect to π¦. Using the fact that the
differential of an exponential is just the exponential, we obtain that dπ₯ by dπ¦ is
equal to π to the power of π¦. Now, weβre trying to find the
differential of the reciprocal function, so thatβs π¦, with respect to π₯. So thatβs dπ¦ by dπ₯. And in order to do this, we can use
the fact that the derivative of an inverse of a function is equal to the reciprocal
of the derivative of the function. Giving us that dπ¦ by dπ₯ is equal
to one over dπ₯ by dπ¦. In order to use this, we must
ensure that the denominator of our fraction is nonzero. So thatβs dπ₯ by dπ¦.

Weβve just found that dπ₯ by dπ¦ is
equal to π to the power of π¦. Since π to the π¦ is an
exponential, we know that π to the power of π¦ is going to be greater than zero for
all values of π¦. Therefore, it is nonzero. And so, weβre able to use this
formula. And so, we obtain that dπ¦ by dπ₯
is equal to one over π to the power of π¦. However, the question has asked us
to give our answer in terms of π₯. In order to get our answer in terms
of π₯, we can use the fact that π₯ is equal to π to the power of π¦ and substitute
π₯ in for π to the power of π¦. From here, we reach our solution,
which is that dπ¦ by dπ₯ is equal to one over π₯.

Letβs quickly stop to think about
what weβre shown here. Our original function is π₯ is
equal to π to the power of π¦. We can make π¦ the subject of this
equation. We simply take natural logs of both
sides. This gives us that π¦ is equal to
the natural logarithm of π₯. And this is the inverse of the
function given in the question. Now, we have found dπ¦ by dπ₯. Since π¦ is equal to the natural
logarithm of π₯, dπ¦ by dπ₯ is the differential of the natural logarithm of π₯ with
respect to π₯. Therefore, we have just shown that
differential of the natural logarithm of π₯ with respect to π₯ is equal to one over
π₯.

We will now consider another
example.

Given that π₯ is equal to π¦ to the
power of five plus the square root of π¦ plus the cube root of π¦ squared, find dπ¦
by dπ₯.

We can find dπ¦ by dπ₯ by using the
formula for the derivative of the inverse function. Which is that dπ¦ by dπ₯ is equal
to one over dπ₯ by dπ¦. So we start by differentiating π₯
with respect to π¦. Letβs start by rewriting some of
the terms in π₯. We can rewrite the square root of
π¦ as π¦ to the power of a half and the cube root of π¦ squared as π¦ to the power
of two over three. Now, we can use the power rule for
differentiation in order to differentiate π₯ with respect to π¦, term by term.

We multiply by the power and
decrease the power by one. This gives us that dπ₯ by dπ¦ is
equal to five π¦ to the power of four plus one-half π¦ to the power of negative
one-half plus two-thirds π¦ to the power of negative one-third. We can rewrite that fractional
powers of π¦ back in their surd form. And then, we can combine these
three terms into one fraction by creating a common denominator of six π¦. Our first term becomes 30π¦ to the
power of five over six π¦. Our second term becomes three
multiplied by the square root of π¦ over six π¦. And our third term becomes four
multiplied by the cube root of π¦ squared over six π¦. We obtain that dπ₯ by dπ¦ is equal
to 30 multiplied by π¦ to the power of five plus three multiplied by the square root
of five plus four multiplied by the cube root of π¦ squared all over six π¦.

Now, we can apply the formula for
the derivative of the inverse function. And this gives us our solution that
dπ¦ by dπ₯ is simply the reciprocal of dπ₯ by dπ¦.

Sometimes, we may be asked to find
the derivative of the inverse function at a given point. We must be careful about which
point we evaluate the function, as the next example will show.

Let π of π₯ be equal to one-half
π₯ cubed plus one-half π₯ squared plus five π₯ minus four and let π be the inverse
of π. Given that π of two is equal to
12, what is π prime of 12?

In order to help us find π prime
of 12, we can use the formula for derivatives of inverse functions. This tells us that if π is the
inverse function of π, then π prime of π¦ is equal to one over π prime of π of
π¦. Letβs start by finding π prime of
π₯, the derivative of π with respect to π₯. We can see that π is a
polynomial. Therefore, in order to find its
derivative, we can differentiate it term by term using the power rule for
differentiation. We simply multiply by the power and
decrease the power by one. This gives us that π prime of π₯
is equal to three over two π₯ squared plus π₯ plus five.

Next, weβll observe the fact that
weβre trying to find π prime of 12. And so, we can substitute π¦ equals
12 into our formula for π prime of π¦. This gives us that π prime of 12
is equal to one over π prime of π of 12. Now, we do not know what π of 12
is. However, we have been given in the
question that π of two is equal to 12. And since π is the inverse
function of π, we can apply π to both sides here. And weβll obtain that π of 12 is
equal to two. This is because of the way inverse
functions work. If we take π of π of two, then
weβll simply get two.

We can now substitute this value of
π of 12 back into our equation for π prime of 12. And we obtain that itβs equal to
one over π prime of two. Now, weβve already found π prime
of π₯. So we can simply substitute π₯
equals two in order to find π prime of two. And we obtain that π prime of two
is equal to 13. And substituting the value of π
prime of two back into π prime of 12, we obtain that π prime of 12 is equal to one
over 13.

Now, our original definition for
the derivative of an inverse function assumed that the inverse existed. We will now cover what is called
the inverse function theorem, which is more powerful than our original
definition. Since it guarantees the existence
and continuity of the inverse of a function when it is continuously differentiable
with a nonzero derivative.

The Inverse Function Theorem

Let π be a continuously
differentiable function with a nonzero derivative at a point π. Then the inverse function theorem
tells us that: One, π is invertible in a neighborhood of π. Two, π has a continuously
differentiable inverse in a neighborhood of π. Three, the derivative of the
inverse of the point π is equal to π of π is equal to the reciprocal of the
derivative of π at π. That is that the derivative of the
inverse of π of π is equal to one over the derivative of π of π.

All we require to use this theorem
is for π to be continuously differentiable and to have a nonzero derivative at some
point π. Now, the proof of this theorem is
beyond the scope of this video. So we will not be covering it
here.

We will now move on and look at
some more examples.

If π of two π is equal to
negative one, π prime of two π is equal to one, and π is equal to negative one,
find the derivative of the inverse of π at π.

Weβll be using the fact that the
derivative of the inverse of π at π is equal to one over π prime of π inverse of
π. In our case, π is equal to
negative one. We need to start by finding π
inverse of negative one. Weβre given in the question that π
of two π is equal to negative one. Since we know that π inverse is
the inverse function of π, this tells us that π inverse of negative one is equal
to two π. So we can substitute this into our
equation. And now, we have that the
derivative of π inverse at negative one is equal to one over π prime of two
π. And we can see that weβve actually
been given π prime of two π in the question. And itβs equal to one.

So we can substitute this in. And weβve reached our solution,
which is that the derivative of the inverse function of π at negative one is equal
to one.

We will now look at one final
example.

Let π be the inverse of π. Using the table below, find π
prime of zero.

In order to find π prime of zero,
weβll be using the formula for finding the derivative of an inverse of a
function. Which tells us that π prime of π¦
is equal to one over π prime of π of π¦. Weβre trying to find π prime of
zero. So we can substitute in zero for
π¦. This gives us that π prime of zero
is equal to one over π prime of π of zero. From the table, we can see that
when π₯ is equal to zero, π is equal to negative one. And so, we have that π of zero is
equal to negative one. Which we can substitute in to give
us that π prime of zero is equal to one over π prime of negative one.

And now, we can simply read π
prime of negative one off from the table. Since when π₯ is equal to negative
one, π prime is equal to one-third. Giving us that π prime of negative
one is equal to one-third. And again, this can be substituted
in. And so, we will obtain that π
prime of zero is equal to the reciprocal of one-third. And here, we reach our solution of
three.

We have now learned about
derivatives of inverse functions. And weβve seen a variety of
examples of how they work. Letβs recap some key points of this
video.

Key Points

Given a continuously differentiable
function π with a nonzero derivative at a point π. The derivative of the inverse of
the function at π, which is equal to π of π, is the derivative of the inverse of
π at π is equal to one over the derivative of π at π. This is often written in Leibnizβs
notation as dπ¦ by dπ₯ is equal to one over dπ₯ by dπ¦. We need to be careful about which
points weβre using.

When using these equations, we can
find derivatives of many familiar inverse functions, such as the natural
logarithm. The inverse function theorem
guarantees the existence of the inverse of a continuous function around points with
nonzero derivatives. Using this theorem, we can find the
derivatives of inverse functions. Even when we are unable to find an
explicit formula for the inverse.