Derivatives of inverse trigonometric functions for JEE

Formula

\[\begin{aligned}\frac{d}{dx}(\sin^{-1}x)&=\frac1{\sqrt{1-x^2}},\\ \frac{d}{dx}(\cos^{-1}x)&=-\frac1{\sqrt{1-x^2}},\\ \frac{d}{dx}(\tan^{-1}x)&=\frac1{1+x^2},\\ \frac{d}{dx}(\cot^{-1}x)&=-\frac1{1+x^2},\\ \frac{d}{dx}(\sec^{-1}x)&=\frac1{|x|\sqrt{x^2-1}},\\ \frac{d}{dx}(\csc^{-1}x)&=-\frac1{|x|\sqrt{x^2-1}}\end{aligned}\]

Variables: \(x\) is real in the domain of each inverse function.

Conditions: For \(\sin^{-1}x,\cos^{-1}x\): \(|x|<1\); for \(\sec^{-1}x,\csc^{-1}x\): \(|x|>1\).

Standard derivative evaluation and chain rule problems.