Question

Differentiate $\sec h^{-1}x$ with respect to $x$, by first writing $x=\sec hy$

Answer 1

Given $\sec h^{-1}x=y\Rightarrow x=\sec hy$

$\Rightarrow x\cosh y=1$

Differentiate on both sides

$\Rightarrow x\sinh y\times \frac{dy}{dx}+\cos hy=0$

$\Rightarrow \frac{dy}{dx}=\frac{-\cos hy}{x\sin hy}$

$\Rightarrow \frac{dy}{dx}=\frac{-1}{x\tan hy}$ equation $\to 1$

$\tan h^{2}y=1-\sec h^{2}y$

$\Rightarrow \tanh y=\sqrt{1-x^2}$ equation $\to 2$

Substituting equation (2) in equation (1)

$\Rightarrow \frac{dy}{dx}=\frac{-1}{x\sqrt{1-x^2}}$

$\Rightarrow \frac{dy}{dx}=\frac{-1}{x\sqrt{1-x^2}}$