Introduction

This topic will be going over a further application of the chain rule, that of differentiating inverse functions. As mentioned in the Topic 3.1 article, Unit 3 is all about applying the chain rule to differentiate different types of function. Before we get directly into differentiating inverse functions though I want to go over a review of inverse functions in general.

Inverse Functions

Let’s first familiarize ourselves with inverse functions again. In a normal function, we express it as $f(x)=y$. However, an inverse function is $f(y)=x$, and the proper notation is $f^{-1}(x)$. Notice that the $x$ and $y$ variables swapped places. This is because the input of $f(x)$ becomes the result of the inverse, and the output of $f(x)$ becomes the input of the inverse. 

An inverse function can also be defined as the following:

For a function $f(x)$, there exists an inverse function $g(x)$ for which $f(g(x))=x$ and $g(f(x))=x$ for all $x$. This means when you plug a value $x$ into a function and get an output value $y$. Then you can plug this output value $y$ into the inverse function to get back the original input value $x$.

Inverse functions are commonly notated as $f^{-1}(x)$ and that is the notation I will be using for this topic, i.e. $\arcsin (x)={\sin }^{-1}(x)$.

Finding Inverse Functions

I am going to go over a refresher of how to find the inverse of a function. Let's say we are given a function $f(x)=3x-2$ or $y=3x-2$. To find the inverse function it’s as simple as swapping the $y$ and $x$ variables’ position and then solving for $y$.

$y=3x-2$

$x=3y-2$

$x+2=3y$

$\frac{x+2}{3}=y$

$y=\frac{x+2}{3}$

$f^{-1}(x)=\frac{x+2}{3}$

It is important to note that in Calculus AB you may be given functions which are extremely hard or even impossible to calculate the inverse of.

Differentiating Inverse Functions

It is actually relatively simple to figure out a rule for the inverse function. First of all, let’s take the equation for an inverse function.

$x=f(y)$

Now, let’s differentiate it. The key here is to notice that this is an inverse function, with the inner function being $y$ and the outer function being $f(x)$. So we just use implicit differentiation to differentiate this term. Take note that because $f(y)$ has the $y$ variable, you’re going to have to multiply the term with $\frac{dy}{dx}$.

$1=f’(y)\cdot \frac{dy}{dx}$

Now like solving for a derivative using implicit differentiation we need to isolate the $y’$ term.

$\frac{1}{f’(y)}=\frac{dy}{dx}$

We can call this our answer, but we may not always have access to the derivative of the inverse, or even the inverse. If we have an extremely complex function (which is often the case in AP Calculus), the inverse would be incredibly complex and difficult to calculate.

However, there is something to take note of. Remember that $x=f(y)$, and if we take the inverse function of both sides, we get $f^{-1}(x)=y$, and so we can substitute this and get a better looking formula.

$\frac{dy}{dx}=\frac{1}{f’(f^{-1}(x))}$

This is the standard formula used for differentiating inverse functions. While you could have just memorized that formula on its own, it’s always useful to see how a formula is derived.

Example of Differentiating an Inverse Function

Now that we have our formula for differentiating inverse functions I will go over an example to show how to apply it. I will go over an AP style question where this formula would be applied.

Let $f(x)=\ln (x)$ and let $g$ be the inverse function of $f$.

Find the value of $g’(2)$.

Note our formula from earlier: $\frac{dy}{dx}=\frac{1}{f’(f^{-1}(x))}$. This can be rewritten for this situation as: $g’(x)=\frac{1}{f’(g(x))}$

In this situation $x=2$ so we can plug that into the equation. $g’(2)=\frac{1}{f’(g(2))}$

To find $g(2)$ we need to find $g(x)$ so we will solve for $g(x)$ using $f(x)$.

$f(x)=\ln (x)$

$x=\ln (y)$

$e^x=y$

$g(x)=e^x$

Now we can figure out that $g(2)=e^2$ and we can plug that in.

$g’(2)=\frac{1}{f’(e^2)}$

Now we need to solve for $f’(x)$ which we know the derivative of $\ln (x)$ to just be $\frac{1}{x}$ so we can just plug that in.

$g’(2)=\frac{1}{\left(\frac{1}{e^2}\right)}$

$g’(2)=e^2$

And that is our answer, $g’(2)=e^2$. That is the process for differentiating inverse functions from an equation. There is another method of differentiating inverse functions that is tested on the AP exam, and that is doing it from a table of values.

Differentiating Inverse Functions from Tables

A type of question you will see on the AP Exam relating to inverse functions is a Differentiating Inverse Functions from Tables questions. These questions look like this:

Let $f$ and $g$ be inverse functions.

The following table lists values of $f$, $g$, and $g’$.

$\begin{array}{cccc}x& f(x)& g(x)& g^{\prime }(x)\\ 1& 2& 8& -1\\ 2& 7& 1& -\frac{1}{2}\end{array}$

Note that $f$ and $g$ are inverses.

What is the value of $f’(1)$?

To approach this question we need the same formula we used for the last type of question, $\frac{dy}{dx}=\frac{1}{f’(f^{-1}(x))}$, and we can just tread f(x) as the inverse function and we get:

$f’(x)=\frac{1}{g’(f(x))}$

Since the problem is asking for $f’(1)$ so we need to plug in $1$ for $x$.

$f’(2)=\frac{1}{g’(f(1))}$

We are told in the table that $f(1)=2$.

$f’(2)=\frac{1}{g’(2)}$

We are also told that $g’(2)=-\frac{1}{2}$

$f’(2)=\frac{1}{\left(-\frac{1}{2}\right)}$

$f’(2)=-2$

And then we get our final answer of $f’(2)=-2$. This type of question is a bit strange but it is important to know how to solve it since a question like it may very well show up on your AP Exam. For solving both this type of question and when solving from equations the key thing to remember is $\frac{dy}{dx}=\frac{1}{f’(f^{-1}(x))}$. Though as I have shown, you can derive it using the chain rule and a bit of implicit differentiation it is worth memorizing.

Notes on Differentiating Inverse Functions

Practice Questions