<s>[INST] Prove that $\frac{d}{dx}(\tan^{-1}(x))=\frac{1}{1+x^2}dx$ Prove that $$\frac{d}{dx}(\tan^{-1}(x))=\frac{1}{1+x^2}$$
 [/INST] Put $y=\tan^{-1}x$. Then $\tan y=x$. So
$$1 = \frac{d}{{dx}}\left( x \right) = \frac{d}{{dx}}\left( {\tan y} \right) = \frac{d}
{{dy}}\left( {\tan y} \right).\frac{d}{{dx}}\left( y \right) = \frac{1}{{{{\cos }^2}y}}.\frac{d}{{dx}}\left( {{{\tan }^{ - 1}}x} \right),$$
which gives
$$\frac{d}{{dx}}\left( {{{\tan }^{ - 1}}x} \right) = {\cos ^2}y = {\cos ^2}\left( {{{\tan }^{ - 1}}x} \right) = \frac{1}{{1 + \tan^2 \left( {{{\tan }^{ - 1}}x} \right)}} = \frac{1}{{1 + {x^2}}}.$$
Notice: In above we used the formula ${\cos ^2}\alpha  = \frac{1}{{1 + {{\tan }^2}\alpha }}$.
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