# 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