Question

If $ y={{({{\sin }^{-1}}x)}^{2}}, $ then $ (1-{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}-x\frac{dy}{dx} $ is equal to

• A. $ 0 $
• B. $ -1 $
• C. $ -2 $
• D. $ 1 $
• E. $ 2 $

Answer 1

Answer: (E)

$ y={{({{\sin }^{-1}}x)}^{2}} $
Differentiating on both sides, $ \frac{dy}{dx}=2{{\sin }^{-1}}x.\frac{1}{\sqrt{1-{{x}^{2}}}} $
$ \Rightarrow $ $ \sqrt{1-{{x}^{2}}}=\frac{dy}{dx}=2{{\sin }^{-1}}x $
Differentiating on both sides, $ \sqrt{(1-{{x}^{2}})}\frac{{{d}^{2}}y}{d{{x}^{2}}}-\frac{2}{2\sqrt{1-{{x}^{2}}}}\frac{dy}{dx}=\frac{2}{\sqrt{1-{{x}^{2}}}} $ $ 2(1-{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}-2\frac{dy}{dx}.x=4 $
$ \Rightarrow $ $ (1-{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}-x\frac{dy}{dx}=2 $