Question

$\frac{d}{dx}\left({\sec }^{-1}x\right)=$ ?

Answer 1

Let $y={\sec }^{-1}x$

$\sec y=x\Rightarrow \cos y=\frac{1}{x}$

Differentiate on both sides w.r.t. x

$\sec y\tan y\frac{dy}{dx}=1$

$\frac{dy}{dx}=\frac{1}{\sec y\tan y}$

$=\frac{{\cos }^{2}y}{\sin y}$

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

$\frac{dy}{dx}=\frac{1}{\left|x\right|\sqrt{x^2-1}}$