Derivative of Arctangent Function

Theorem
Let $x \in \R$.

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

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

Proof
Let $y = \arctan x$.

Then $x = \tan y$.

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

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

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

Hence the result.