Derivative of Arctangent Function

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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

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

$\blacksquare$


Proof 2

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

$\blacksquare$


Also defined as

This result can also be reported as:

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


Also see


Sources