Question

Find $\frac{dy}{dx}$, if y=sin-1x+sin-1$\sqrt{1-x^2}$, -1≤x≤1

Answer 1

It is given that ,y=sin-1x+sin-1$\sqrt{1-x^2}$
∴$\frac{dy}{dx}$=$\frac{d}{dx}$(sin-1x+sin-1$\sqrt{1-x^2}$)
$\Rightarrow$ $\frac{dy}{dx}$=$\frac{d}{dx}$(sin-1x)+$\frac{d}{dx}$(sin-1$\sqrt{1-x^2}$)
$\Rightarrow$ $\frac{dy}{dx}$=$\frac{1}{\sqrt{1-x^2}}$+$\frac{1}{1-(\sqrt{1-x^2})^2}$.$\frac{d}{dx}$($\sqrt{1-x^2}$)
$\Rightarrow$ $\frac{dy}{dx}$=$\frac{1}{\sqrt{1-x^2}}$+$\frac{1}{x}$.\frac{1}{2}$$\sqrt{1-x^2}.$\frac{d}{dx}$(1-x2)
$\Rightarrow$ $\frac{dy}{dx}$=$\frac{1}{\sqrt{1-x^2}}$+$\frac{1}{2x\sqrt{1-x^2}(-2x)}$
$\Rightarrow$ $\frac{dy}{dx}$=$\frac{1}{\sqrt{1-x^2}}$-$\frac{1}{\sqrt{1-x^2}}$
∴$\frac{dy}{dx}$=0