Proof that $\frac{d}{dx} \tan^{-1} x = \frac{1}{1+x^2}$using implicit differentiation 
Proof that $\frac{d}{dx} \tan^{-1} x = \frac{1}{1+x^2}$ using implicit differentiation

My workings:
$y=\tan^{-1} x$,
$x= \tan y$
$\frac{d}{dx} (x) = \frac{d}{dx} \tan y$
$1 = \sec^2 y \frac{dy}{dx}$
$\frac{dy}{dx} = \frac{1}{\sec^2 y} = \cos^2 y$
How do I carry on from here?

Use the basic trigonometric identity
$$
y'=\cos^2y=\frac{\cos^2y}1=\frac{\cos^2y}{\cos^2y+\sin^2y}=\frac1{1+\tan^2y}=\frac1{1+x^2}
$$
