Question

If $f\left(x\right)=\frac{{\sin }^{-1}x}{\sqrt{1-x^2}}$, then the value of $(1-x^2)$ $f^{\prime }\left(x\right)-xf\left(x\right)$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$
• E. $4$

Answer 1

Answer: (B)

$1$

Let $y=f\left(x\right)$ then $y=\frac{{\sin }^{-1}x}{\sqrt{1-x^2}}$

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

On differentiating w.r.t $x$ we get

$\Rightarrow y.\frac{1}{2\sqrt{1-x^2}}(-2x)+\sqrt{1-x^2}\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}$

$\Rightarrow -xy+(1-x^2)\frac{dy}{dx}=1$

$\therefore (1-x^2)f^{\prime }\left(x\right)-xf\left(x\right)=1$