Question

If $y=\frac{{\sin }^{-1}x}{\sqrt{1-x^2}},$ $-1<x<1$ then $(1-x^2)\frac{dy}{dx}$ equals to

• A. $x+y$
• B. $1+xy$
• C. $1-xy$
• D. $xy-2$

Answer 1

The correct option is B $1+xy$

$y=\frac{{\sin }^{-1}x}{\sqrt{1-x^2}}$

We have,

$\frac{d}{dx}\left(\frac{u}{v}\right)$ $=\frac{u\frac{d}{dx}\left(v\right)-\left(v\right)\frac{d}{dx}\left(u\right)}{v^2}$

$\therefore \frac{dy}{dx}=\frac{\sqrt{1-x^2}\times \frac{1}{\sqrt{1-x^2}}-\left({\sin }^{-1}x\right)\frac{1}{2\sqrt{1-x^2}}\times (-2x)}{(1-x^2)}$
$(1-x^2)\frac{dy}{dx}=1+\frac{{\sin }^{-1}x}{\sqrt{1-x^2}}\times x$
$(1-x^2)\frac{dy}{dx}=1+xy$