**Proof.**Let y = \tan^{-1}(x). Then x = \tan(y) and -\frac{\pi}{2} < y < \frac{\pi}{2 }. We want to differentiate with respect to x, so we get:

\begin{align*} \frac{d}{dx} x = \frac{d}{dx} \tan(x) &\iff 1 = \frac{d(\tan(y))}{dy} \frac{dy}{dx} \\ &\iff 1 = \sec^2(y) \frac{dy}{dx} \\ &\iff \frac{dy}{dx} = \frac{1}{\sec^2(y)}, \end{align*}

\begin{align*} \sec^2(y) = 1 + \tan^2(y), \end{align*}

\begin{align*} \frac{d}{dx} \arctan(x) = \frac{d}{dx} \tan^{-1}(x) = \frac{dy}{dx} = \frac{1}{\sec^2(y)} = \frac{1}{1 + \tan^2(y)} = \frac{1}{1 + x^2}. \end{align*}