Answer
$\displaystyle \frac{d}{dx}[\tan^{-1}x]=\frac{1}{1+x^{2}}$
Work Step by Step
$y=\tan^{-1}x$
$x=\tan y\qquad $.... differentiate,
$\displaystyle \frac{d}{dx}[x]=(\sec^{2}y)\frac{dy}{dx}$
$1=[\displaystyle \sec(\tan^{-1}x)]^{2}\cdot\frac{dy}{dx}$
$\displaystyle \frac{dy}{dx}=\frac{1}{[\sec(\tan^{-1}x)]^{2}}$
To simplify $\sec(\tan^{-1}x)$,
sketch a right triangle with legs 1 and x.
The hypotenuse is $\sqrt{1+x^{2}}$
The angle opposite x is $y=\tan^{-1}x$ (the adjacent leg to y is 1)
$\displaystyle \sec y=\frac{\sqrt{1+x^{2}}}{1}$,
$\sec(\tan^{-1}x)=\sqrt{1+x^{2}}$
So,
$\displaystyle \frac{dy}{dx}=\frac{d}{dx}[\tan^{-1}x]=\frac{1}{\left[\sqrt{1+x^{2}}\right]^{2}}=\frac{1}{1+x^{2}}$
