# Derivative of Arctangent Function

## Theorem

Let $x \in \R$.

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

Then:

$\dfrac {\mathrm d \left({\arctan x}\right)} {\mathrm d x} = \dfrac 1 {1 + x^2}$

### Corollary

$\dfrac {\mathrm d \left({\arctan \left({\frac x a}\right) }\right)} {\mathrm d x} = \dfrac a {a^2 + x^2}$

## Proof 1

 $\displaystyle y$ $=$ $\displaystyle \arctan x$ $\displaystyle \implies \ \$ $\displaystyle x$ $=$ $\displaystyle \tan y$ Definition of Arctangent $\displaystyle \implies \ \$ $\displaystyle \frac {\mathrm d x} {\mathrm d y}$ $=$ $\displaystyle \sec^2 y$ Derivative of Tangent Function $\displaystyle$ $=$ $\displaystyle 1 + \tan^2 y$ Difference of Squares of Secant and Tangent $\displaystyle$ $=$ $\displaystyle 1 + x^2$ Definition of $x$ $\displaystyle \implies \ \$ $\displaystyle \frac {\mathrm d y} {\mathrm d x}$ $=$ $\displaystyle \frac 1 {1 + x^2}$ Derivative of Inverse Function

$\blacksquare$

## Proof 2

 $\displaystyle \frac{\mathrm d\left( \arctan\left(x\right) \right)}{\mathrm dx}$ $=$ $\displaystyle \lim_{h \to 0} \frac{\arctan\left(x+h\right) - \arctan\left(x\right)} h$ Definition of Derivative of Real Function at Point $\displaystyle$ $=$ $\displaystyle \lim_{h \to 0} \frac{\arctan(x+h) + \arctan(-x)} h$ Arctangent Function is Odd $\displaystyle$ $=$ $\displaystyle \lim_{h \to 0} \frac 1 h \arctan\left(\frac {x+h-x}{1+x\left(x+h\right)}\right)$ Sum of Arctangents $\displaystyle$ $=$ $\displaystyle \lim_{h \to 0} \frac 1 h \arctan\left(\frac h {1+x^2+hx}\right)$ $\displaystyle$ $=$ $\displaystyle \lim_{h \to 0} \frac 1 h \left(\frac h {1+x^2+hx} - \frac 1 3 \left(\frac h {1+x^2+hx}\right)^3 + \frac 1 5\left(\frac h {1+x^2+hx}\right)^5 + \mathcal O\left(h^7\right)\right)$ Definition of Arctangent $\displaystyle$ $=$ $\displaystyle \lim_{h \to 0} \left(\frac 1 {1+x^2+hx} - \frac {h^2} {3\left(1+x^2+hx\right)^3} + \frac{h^4} {5\left(1+x^2+hx\right)^5} + \mathcal O\left(h^6\right)\right)$ $\displaystyle$ $=$ $\displaystyle \frac 1 {1+x^2 + 0x} - \frac {0^2} {3\left(1+x^2+0x\right)^3} + \frac{0^4} {5\left(1+x^2+0x\right)^5}$ $\displaystyle$ $=$ $\displaystyle \frac 1 {1+x^2}$

$\blacksquare$

## Also defined as

This result can also be reported as:

$\dfrac {\mathrm d \left({\arctan x}\right)} {\mathrm d x} = \dfrac 1 {x^2 + 1}$