# Calculating Derivative – Computing a derivative of an inverse function – Exercise 2076

Exercise

Find the derivative of the inverse of the following function:

$$f(x)=\sin x$$

$$(f^{-1})'(x)=\frac{1}{\sqrt{1-x^2}}$$

Solution

Given the function:

$$f(x)=\sin x$$

Its inverse function is

$$f^{-1}(x)=\arcsin x$$

We use the formula to find the derivative of the inverse function and get:

$$(f^{-1})'(x)=(\arcsin x)'=$$

$$=\frac{1}{(\sin (\arcsin x))'}=$$

$$=\frac{1}{\cos (arcsin x)}=$$

Using the following trigonometric identity:

$$\cos x =\sqrt{1-\sin^2 x}$$

we get:

$$=\frac{1}{\sqrt{1-\sin^2 (\arcsin x)}}=$$

$$=\frac{1}{\sqrt{1-x^2}}$$

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