> [!NOTE] > Let $f:\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]\to [-1,1]$ be the function $x\mapsto \sin x,$ where $\sin$ denotes the [[Sine|sine]], and $g$ its [[Function Inverse|inverse]]. Then $g$ is [[Fréchet Differentiation|differentiable]] and its derivative is given by $g'(x)=\frac{1}{\sqrt{ 1-x^{2} }}$ **Proof**: By [[Derivative of Inverse of Strictly Monotonic Differentiable Real Function]], $g'(x)=\frac{1}{\cos(\sin^{-1}(x))}.$ Now using [[Pythagorean Trigonometric Identity]], $\cos(\sin^{-1}(x)) =\sqrt{ 1- \sin ^{2}(\sin^{-1} (x)) }= \sqrt{ 1-x^{2} }.$