Derivative of Arctangent Function

Theorem
Let $$x \in \R$$.

Let $$\arctan x$$ be the arctangent of $$x$$.

Then:
 * $$D_x \left({\arctan x}\right) = \frac 1 {1 + x^2}$$

Proof
Let $$y = \arctan x$$.

Then $$x = \tan y$$.

Then $$\frac {dx} {dy} = \sec^2 y$$ from Derivative of Tangent Function.

Thus from the corollary to Sum of Squares of Sine and Cosine:
 * $$\frac {dx} {dy} = 1 + \tan^2 y = 1 + x^2$$

Hence from Derivative of an Inverse Function:
 * $$\frac {dy} {dx} = \frac 1 {1 + x^2}$$

Hence the result.