Derivative of Arc tangent or tangent inverse

Let $y = f(x) = \tan ^{ - 1} x$

Then $x = \tan y$.

Differentiating both sides with respect to $x$, we get

$1 = (\sec ^2 y){{dy} \over {dx}}$

Note that we used the chain rule to differentiate $\tan y$ with respect to $x$ because $\tan y$ is a function of $y$, and $y$ in turn is a function of $x$.

or, ${{dy} \over {dx}} = {1 \over {\sec ^2 y}}$

or, ${{dy} \over {dx}} = {1 \over {1 + \tan ^2 y}}$

or, ${{dy} \over {dx}} = {1 \over {1 + x^2 }}$