<s>[INST] 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?
 [/INST] $$x=\tan y\implies x^2=\tan^2y$$$$\implies 1+x^2=1+\tan^2y=\sec^2y$$$$\implies \frac{dy}{dx}=\frac{1}{\sec^2y}=\frac{1}{1+x^2}.$$
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