[[arcsinh_x_미분_증명]]과 마찬가지 방법.

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

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