Prove that $\frac{d}{dx}(\tan^{-1}(x))=\frac{1}{1+x^2}dx$ Prove that $$\frac{d}{dx}(\tan^{-1}(x))=\frac{1}{1+x^2}$$
 A: As $\tan\circ\tan^{-1}  = id$,
\begin{align}
1 = id'(x) &= (\tan^{-1})'(x) \times \tan'(\tan^{-1}(x))
 \\&=  (\tan^{-1})'(x) \times  (1+\tan^2 (\tan^{-1}(x)))
\end{align}
now replace $y= \tan^{-1}(x)$ and you get your result.
A: Hint:
$$y = \arctan x \Rightarrow \tan y =  x \Rightarrow y' \sec^2 y = 1 $$
Use $\tan^2y + 1  = \sec^2 y$. 
A: Let $y=\tan^{-1}x$. Then $x=\tan y$.
So  $$[\tan^{-1}(x)]'=\frac{1}{\tan' y} = \frac{1}{\sec^2 y} =  \frac{1}{1+\tan^2 y}=  \frac{1}{1+x^2}$$
A: 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 ${\cos ^2}\alpha  = \frac{1}{{1 + {{\tan }^2}\alpha }}$.
