Question

State the whether given statement is true or false

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

• A. True
• B. False

Answer 1

The correct option is A True

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

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

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

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

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

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