Question

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

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

Answer 1

Answer: (B)

$1+xy$

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

$\sqrt{1-x^2}.y=si{n}^{-1}x$

Differentiating both sides w.r.t.x, we get

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

$\sqrt{1-x^2}\frac{dy}{dx}-\frac{xy}{\sqrt{1-x^2}}=\frac{1}{\sqrt{1-x^2}}$

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

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