Question

If $y=\frac{{\sin }^{-1}x}{\sqrt{1-x^2}}$, prove that $\left(1-x^2\right)\frac{dy}{dx}-xy=1$

Answer 1

Consider the given function.

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

On differentiating both sides w.r.t $x$, we get

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

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

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

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

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

Hence, proved