arctan_x(아크탄젠트)_미분_증명

Theorem:
$\displaystyle \frac{\operatorname{d}}{\operatorname{d}x}(\tan^{-1}x)=\frac{1}{1+x^2}\;(x\in\mathbb{R})$

Proof

$\displaystyle y=\tan^{-1} x$ 라 하면,
$\displaystyle \tan y = x$ 이고, 양변을 $\displaystyle x$ 에 대해 미분하면
$\displaystyle (\sec y)^2\cdot \frac{\mathrm{d}y}{\mathrm{d}x}=1$
$\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=\frac1{\sec^2 y}=\frac1{1+\tan^2 y}=\frac1{1+x^2}$

Example

Q:
$\displaystyle \frac{\rm d}{\mathrm{d}x}\left(\tan^{-1}(x^2+1)\right)$
A:
$\displaystyle =\frac{2x}{1+(x^2+1)^2}$
$\displaystyle =\frac{2x}{x^4+2x^2+2}$