# What is the derivative of y=cot^{-1}(x)?

Sep 14, 2014

The answer is $y ' = - \frac{1}{1 + {x}^{2}}$

We start by using implicit differentiation:

$y = {\cot}^{- 1} x$
$\cot y = x$
$- {\csc}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} = 1$
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{{\csc}^{2} y}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{1 + {\cot}^{2} y}$ using trig identity: $1 + {\cot}^{2} \theta = {\csc}^{2} \theta$
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{1 + {x}^{2}}$ using line 2: $\cot y = x$

The trick for this derivative is to use an identity that allows you to substitute $x$ back in for $y$ because you don't want leave the derivative as an implicit function; substituting $x$ back in will make the derivative an explicit function.