# What is the derivative of f(x)=cot^-1(x) ?

Sep 24, 2014

$f ' \left(x\right) = - \frac{1}{1 + {x}^{2}}$

Let us look at some details.

By replacing $f \left(x\right)$ by $y$,

$y = {\cot}^{- 1} x$

by rewriting in terms of cotangent,

$R i g h t a r r o w \cot y = x$

by implicitly differentiating with respect to x,

$R i g h t a r r o w - {\csc}^{2} y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

by dividing by $- {\csc}^{2} y$,

$R i g h t a r r o w \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{{\csc}^{2} y}$

by the trig identity ${\csc}^{2} y = 1 + {\cot}^{2} y = 1 + {x}^{2}$,

$R i g h t a r r o w \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{1 + {x}^{2}}$

Hence,

$f ' \left(x\right) = - \frac{1}{1 + {x}^{2}}$