arccosh_x_미분_증명

arcsinh_x_미분_증명과 마찬가지 방법.

Theorem:
$\displaystyle {d\over dx}(\cosh^{-1}x)=\frac1{\sqrt{x^2-1}}$

Proof:
$\displaystyle y=\cosh^{-1}x$
$\displaystyle \cosh y=x$
$\displaystyle \frac{dy}{dx}=\frac1{\;\frac{dx}{dy}\;}=\frac1{\sinh y}$
$\displaystyle =\frac1{\sqrt{\cosh^2y-1}}$
$\displaystyle =\frac1{\sqrt{x^2-1}}$