Question

Differentiate ${\sin }^{-1}x$.

Answer 1

Differentiation.

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

$\therefore \sin y=x$

Differentiate the above equation with respect to $x$.

$$\begin{gathered} \therefore \cos y·\frac{dy}{dx}=1 \\ \Rightarrow \frac{dy}{dx}=\frac{1}{\cos y} \\ \Rightarrow \frac{dy}{dx}=\frac{1}{\sqrt{1-{\sin }^{2}y}}\left[\because {\sin }^{2}y+{\cos }^{2}y=1\right] \\ \Rightarrow \frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}\left[\because \sin y=x\right] \end{gathered}$$

Hence, $\frac{d}{dx}{\sin }^{-1}x=\frac{1}{\sqrt{1-x^2}},-1<x<1$