Question

If $y={\sin }^{-1}\left(x\sqrt{1-x}-\sqrt{x}\sqrt{1-x^2}\right\}\mathrm{and}0<x<1,\mathrm{then}\frac{\mathrm{dy}}{\mathrm{dx}}$is

• A. $\frac{1}{\sqrt{1-x^2}}-\frac{1}{2\sqrt{x}\sqrt{1-x}}$
• B. $\frac{1}{\sqrt{1-x^2}}-\frac{\sqrt{x}}{2\sqrt{1-x}}$
• C. $\frac{\sqrt{x}}{\sqrt{1-x^2}}-\frac{1}{2\sqrt{1-x}}$
• D. $\frac{1}{\sqrt{1-x^2}}-\frac{1}{2\sqrt{1-x}}$

Answer 1

The correct option is A $\frac{1}{\sqrt{1-x^2}}-\frac{1}{2\sqrt{x}\sqrt{1-x}}$

Given that,$y={\sin }^{-1}\left(x\sqrt{1-x}-\sqrt{x}\sqrt{1-x^2}\right\}.....\left(1\right]\mathrm{and}0<x<1,\mathrm{then}\frac{\mathrm{dy}}{\mathrm{dx}}\mathrm{at}x=\frac{1}{2}\mathrm{Let}x=\sin A\mathrm{and}\sqrt{x}=\sin B;A,B\in \left(0,\frac{\pi }{2}\right)\left(1\right]\mathrm{can}\mathrm{be}\mathrm{expressed}\mathrm{as}y={\sin }^{-1}\left(\sin A\sqrt{1-{\sin }^{2}B}-\sin B\sqrt{1-{\sin }^{2}A}\right\}\Rightarrow y={\sin }^{-1}\left(\sin A\cos B-\sin B\cos A\right\}\Rightarrow y={\sin }^{-1}\left(\sin (A-B)\right\}\Rightarrow y=A-B\left(\because A-B\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)\right)\Rightarrow y={\sin }^{-1}x-{\sin }^{-1}\sqrt{x}\mathrm{Differentiating}w.r.t.x,\mathrm{we}\mathrm{get}\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\sqrt{1-x^2}}-\frac{1}{2\sqrt{x}\sqrt{1-x}}$