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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.