derivative of inverse function

Theorem.  If the real function $f$ has an inverse function $f_{\leftarrow }$ and the derivative of $f$ at the point  $x=f_{\leftarrow }(y)$  is distinct from zero, then $f_{\leftarrow }$ is also differentiable at the point $y$ and

$$f_{\leftarrow }^{\prime }(y)=\frac{1}{f^{\prime }(x)}.$$(1)

That is, the derivatives of a function and its inverse function are inverse numbers of each other, provided that they have been taken at the points which correspond to each other.

{it Proof.Now we have

$$f(f_{\leftarrow }(y))=f(x)=y.$$

The derivatives of both sides must be equal:

$$\frac{d}{dy}\left[f(f_{\leftarrow }(y))\right]=\frac{d}{dy}y$$

Using the chain rule we get

$$f^{\prime }(f_{\leftarrow }(y))\cdot f_{\leftarrow }^{\prime }(y)=1,$$

whence

$$f_{\leftarrow }^{\prime }(y)=\frac{1}{f^{\prime }(f_{\leftarrow }(y))}.$$

This is same as the asserted (1).

Examples.  For simplicity, we express here the functions by symbols $y$ and the inverse functions by $x$.

1. $y=\tan x$,  $x=\arctan y$; $\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}=\frac{1}{1+{\tan }^{2}x}=\frac{1}{1+y^2}$

2. $y=\sin x$,  $x=\arcsin y$; $\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}=\frac{1}{\cos x}=\frac{1}{+\sqrt{1-{\sin }^{2}x}}=+\frac{1}{\sqrt{1-y^2}}$

3. $y=x^2$,   $x=\pm \sqrt{y}$; $\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}=\frac{1}{2x}=\frac{1}{\pm 2\sqrt{y}}$

If the variable symbol $y$ in those results is changed to $x$, the results can be written

$$\frac{d}{dx}\arctan x=\frac{1}{1+x^2},\frac{d}{dx}\arcsin x=\frac{1}{\sqrt{1-x^2}},\frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}.$$