# Suppose f(x) = sin(pi cos(x)). On any interval where the inverse function y = f^-1(x) exists,...

## Question:

Suppose {eq}f(x) = sin[\pi cos(x)] {/eq}. On any interval where the inverse function {eq}y = f^{-1}(x) {/eq} exists, what is the derivative of {eq}f^{-1}(x) {/eq} with respect to x ?

## Derivative of an Inverse:

Let {eq}f^{-1}(x) {/eq} represent the inverse of the function {eq}f(x) {/eq}.

The following formula provides us a way of computing the derivative of the inverse of a function:

{eq}\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}f^{-1}(x)= \frac{1}{f'(f^{-1}(x))} {/eq}

Apparently, we will first need to find the inverse of the indicated function.

## Answer and Explanation:

First, let's find the inverse by solving for {eq}y {/eq} after interchanging {eq}x {/eq} and {eq}y {/eq} as follows:

{eq}\begin{align*} \displaystyle x& = \sin(\pi \cos(y))\\ \displaystyle \sin^{-1}(x) & = \pi \cos(y)\\ \cos(y)& =\frac{\sin^{-1}(x)}{\pi} \\ y & = \cos^{-1} \left( \frac{\sin^{-1}(x)}{\pi}\right) \\ f^{-1}(x) & = \cos^{-1} \left( \frac{\sin^{-1}(x)}{\pi}\right) \\ \end{align*} {/eq}

The derivative of {eq}f(x) = \sin(\pi \cos(x)) {/eq} is calculated via the chain rule:

{eq}\begin{align*} \displaystyle f(x)& = \sin(\pi \cos(x))\\ \displaystyle f'(x)& =(-\pi \sin(x) ) \cos(\pi \cos(x))\\ \displaystyle f'(x)& =-\pi \sin(x) \cos(\pi \cos(x))\\ \end{align*} {/eq}

Defining {eq}f'(x) =-\pi \sin(x) \cos(\pi \cos(x)) {/eq} at {eq}x= \displaystyle f^{-1}(x) = \cos^{-1} \left( \frac{\sin^{-1}(x)}{\pi}\right) {/eq} and applying the formula {eq}\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}f^{-1}(x)= \frac{1}{f'(f^{-1}(x))} {/eq} yields the solution we seek:

{eq}\begin{align*} \displaystyle \frac{\mathrm{d}}{\mathrm{d}x}f^{-1}(x)&= \frac{1}{f'(f^{-1}(x))}\\ \displaystyle \frac{\mathrm{d}}{\mathrm{d}x}f^{-1}(x)&= \frac{1}{-\pi \sin(f^{-1}(x)) \cos(\pi \cos(f^{-1}(x)))}\\ \displaystyle \frac{\mathrm{d}}{\mathrm{d}x}f^{-1}(x)&=- \frac{1}{\pi \sin\left( \cos^{-1} \left( \frac{\sin^{-1}(x)}{\pi}\right)\right) \cos\left(\pi \cos\left( \cos^{-1} \left( \frac{\sin^{-1}(x)}{\pi}\right) \right)\right)}\\ \end{align*} {/eq}

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from Math 104: Calculus

Chapter 8 / Lesson 12