# Derivative of Arctangent Function

## Theorem

Let $x \in \R$.

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

Then:

$\dfrac {\map \d {\arctan x} } {\d x} = \dfrac 1 {1 + x^2}$

### 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 \leadsto \ \$ $\displaystyle x$ $=$ $\displaystyle \tan y$ Definition of Real Arctangent $\displaystyle \leadsto \ \$ $\displaystyle \frac {\d x} {\d y}$ $=$ $\displaystyle \sec^2 y$ 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 \leadsto \ \$ $\displaystyle \frac {\d y} {\d x}$ $=$ $\displaystyle \frac 1 {1 + x^2}$ Derivative of Inverse Function

$\blacksquare$

## Proof 2

 $\displaystyle \frac {\map \d {\map \arctan x} } {\d x}$ $=$ $\displaystyle \lim_{h \mathop \to 0} \frac {\map \arctan {x + h} - \map \arctan x} h$ Definition of Derivative of Real Function at Point $\displaystyle$ $=$ $\displaystyle \lim_{h \mathop \to 0} \frac {\arctan {x + h} + \map \arctan {-x} } h$ Arctangent Function is Odd $\displaystyle$ $=$ $\displaystyle \lim_{h \mathop \to 0} \frac 1 h \map \arctan {\frac {x + h - x} {1 + x \paren {x + h} } }$ Sum of Arctangents $\displaystyle$ $=$ $\displaystyle \lim_{h \mathop \to 0} \frac 1 h \map \arctan {\frac h {1 + x^2 + h x} }$ $\displaystyle$ $=$ $\displaystyle \lim_{h \mathop \to 0} \frac 1 h \paren {\frac h {1 + x^2 + h x} - \frac 1 3 \paren {\frac h {1 + x^2 + h x} }^3 + \frac 1 5 \paren {\frac h {1 + x^2 + h x} }^5 + \map \OO {h^7} }$ Definition of Real Arctangent $\displaystyle$ $=$ $\displaystyle \lim_{h \mathop \to 0} \paren {\frac 1 {1 + x^2 + h x} - \frac {h^2} {3 \paren {1 + x^2 + h x}^3} + \frac {h^4} {5 \paren {1 + x^2 + h x}^5} + \map \OO {h^6} }$ $\displaystyle$ $=$ $\displaystyle \frac 1 {1 + x^2 + 0 x} - \frac {0^2} {3 \paren {1 + x^2 + 0 x}^3} + \frac {0^4} {5 \paren {1 + x^2 + 0 x}^5}$ $\displaystyle$ $=$ $\displaystyle \frac 1 {1 + x^2}$

$\blacksquare$

## Also defined as

This result can also be reported as:

$\dfrac {\map \d {\arctan x} } {\d x} = \dfrac 1 {x^2 + 1}$