# Implicit differentiation 2

# Derivatives of inverse functions.

- Let \(f(x)\) be a function. Recall that its inverse is defined in a way that \(y=f^{-1}(x)\) means \(f(y)=x\).
- Suppose that \(f(x)\) is differentiable on its domain. To find the derivative of its inverse, we can implicit derivate the equation \(f(y)=x\). We get \(f'(y)y'=1\), which shows \[
(f^{-1}(x))'=\frac{1}{f'(y)}.
\]

- We can use this to derivate inverse trigonometric functions.
- Let \(y=\sin^{-1}(x)\). Then applying the formula, we get \[
(\sin^{-1}(x))'=\frac{1}{\cos(\sin^{-1}(x))}.
\]
- Since \(-\frac{\pi}{2}\le y\le\frac{\pi}{2}\), we have \(\cos y\ge1\), and thus \[
\cos y=\sqrt{1-\sin(y)^2}.
\] This gives the following. \[
(\sin^{-1}(x))'=\frac{1}{\sqrt{1-\sin(y)^2}}=\frac{1}{\sqrt{1-x^2}}.
\]
- Exercise. Find the derivative of \(y=\tan(x)\).
- Exercises. 3.5: 50, 52, 58, 66, 68, 70, 76, 80