Differentiating inverse functions

​​In a nutshell

One application of the chain rule for differentiation is differentiating inverse functions. This will help you find the derivative of functions you don't know how to differentiate directly, but that you do know how to differentiate the function's inverse.

Inverse function reminder

Let $$f(x)$$ be a function. Then its inverse is denoted by $$f^{-1}(x)$$. Examples of functions and their inverse includes $$e^x$$ and $$\ln(x)$$, $$x^2$$ and $$\sqrt{x}$$, and $$\sin(x)$$ and $$\sin^{-1}(x)$$.

Example 1

Find the inverse function of $$g(x)=4x-3$$.

To find the inverse function, rearrange to make $$x$$ the subject:

$$x=\frac{g(x)+3}{4}$$​​

Next, replace the $$x$$ on the left with $$g^{-1}(x)$$ and replace $$g(x)$$ with $$x$$:

$$\underline{g^{-1}(x)=\frac{x+3}{4}}$$​​

Using the chain rule

The chain rule essentially gives you a method to manipulate derivatives. For example, it says that

​$$\dfrac{\text dy}{\text dx}=\dfrac{\text dy}{\text du}\times\dfrac{\text du}{\text dx}$$​​

You can approach statements like this as you would to a fraction. Notice that it is as if you are cancelling the $$\text du$$.

Note: It is not actually as straightforward as simply cancelling, but this is a helpful way to look at it.

For $$y=f(x)$$, you have that the derivative of $$f$$ is $$\dfrac{\text dy}{\text dx}$$.​

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The inverse is given by $$x=f^{-1}(y)$$ and thus the derivative of the inverse function is $$\dfrac{\text dx}{\text dy}$$.

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Now consider that by the chain rule, the following is true:

​$$1=\dfrac{\text dy}{\text dx}\times\dfrac{\text dx}{\text dy}$$​​

Thus you have the following result:

​$$\boxed{\dfrac{\text dx}{\text dy}=\dfrac1{\frac{\text dy}{\text dx}}}$$​​

This is the derivative of the inverse function. Equivalently:

​$$\dfrac{\text dy}{\text dx}=\dfrac1{\frac{\text dx}{\text dy}}$$​​

Example 2

Using the derivative of the inverse function rule, show that $$\frac{\text d}{\text dx}(\ln(x))=\frac1x$$.

Let $$y=\ln(x)$$. You seek $$\dfrac{\text dy}{\text dx}$$. The function can be re-expressed as $$e^y=x$$. This is the inverse function and you know how to differentiate it:

​$$\dfrac{\text dx}{\text dy}=e^y$$​​

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Thus:

​$$\begin{aligned}\dfrac{\text dy}{\text dx}&=\dfrac1{\frac{\text dx}{\text dy}}\\&=\frac1{e^y}\\&=\underline{\frac1x}\end{aligned}$$​​

Example 3

Use the derivative of the inverse function rule to differentiate $$\sin^{-1}(x)$$ with respect to $$x$$.

Let $$y=\sin^{-1}(x)$$​. You seek $$\dfrac{\text dy}{\text dx}$$. The function can be rearranged to give $$x=\sin(y)$$. This is the inverse function. Thus:

$$\dfrac{\text dx}{\text dy}=\cos(y)$$​​

Hence 

$$\frac{\text dy}{\text dx}=\frac1{\cos(y)}$$​​

Now you need a way to express $$\cos(y)$$ in terms of $$x$$. To do this, use that $$x=\sin(y)$$ and the identity $$1=\sin^2(y)+\cos^2(y)$$:

​$$\begin{aligned}x&=\sin(y)\\x^2&=\sin^2(y)\\1-x^2&=1-\sin^2(y)\\1-x^2&=\cos^2(y)\\\sqrt{1-x^2}&=\cos(y)\end{aligned}$$​​

Thus:

$$\underline{\frac{\text dy}{\text dx}=\frac1{\sqrt{1-x^2}}}$$​​