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  • explain solution RD Sharma class 12 chapter Differentiation exercise 10.2 question 64 maths

explain solution RD Sharma class 12 chapter Differentiation exercise 10.2 question 64 maths

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Answer: Proved

Hint: you must know the rules of derivative of inverse trigonometric functions.

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

Prove   \left(1-x^{2}\right) \frac{d y}{d x}=x+\frac{y}{x}

Solution:

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

Differentiate with respect to x,

\frac{d y}{d x}=\frac{d}{d x}\left(\frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}}\right)                [   quotient rule    ]

\frac{d y}{d x}=\frac{d}{d x}\left[\frac{\sqrt{1-x^{2}} \frac{d}{d x}\left(x \sin ^{-1} x\right)-\left(x \sin ^{-1} x\right) \frac{d}{d x}\left(\sqrt{1-x^{2}}\right)}{\left(\sqrt{1-x^{2}}\right)^{2}}\right] . \cdot \frac{d}{d x} u \cdot v=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}

\frac{d y}{d x}=\left[\frac{\sqrt{1-x^{2}}\left\{x \frac{d}{d x}\left(\sin ^{-1} x\right)+\left(\sin ^{-1} x\right) \frac{d}{d x}(x)\right\}-\left(x \sin ^{-1} x\right)\left(\frac{1}{2 \sqrt{1-x^{2}}}\right) \frac{d\left(1-x^{2}\right)}{d x}}{\left(\sqrt{1-x^{2}}\right)^{2}}\right]

\frac{d y}{d x}=\left[\frac{\sqrt{1-x^{2}}\left\{\frac{x}{\sqrt{1-x^{2}}}+\sin ^{-1} x\right\}-\frac{x \sin ^{-1} x(-2 x)}{2\left(\sqrt{1-x^{2}}\right)}}{\left(\sqrt{1-x^{2}}\right)^{2}}\right]

\frac{d y}{d x}=\left[\frac{x+\sqrt{1-x^{2}}\left(\sin ^{-1} x\right)+\frac{x^{2} \sin ^{-1} x}{\sqrt{1-x^{2}}}}{\left(1-x^{2}\right)}\right]

\left(1-x^{2}\right) \frac{d y}{d x}=x+\frac{\sqrt{1-x^{2}}\left(\sin ^{-1} x\right)}{1}+\frac{x^{2} \sin ^{-1} x}{\sqrt{1-x^{2}}}

\left(1-x^{2}\right) \frac{d y}{d x}=x+\left(\frac{\sin ^{-1} x-x^{2}\left(\sin ^{-1} x\right)+x^{2}\left(\sin ^{-1} x\right)}{\sqrt{1-x^{2}}}\right)

\left(1-x^{2}\right) \frac{d y}{d x}=x+\left(\frac{\sin ^{-1} x}{\sqrt{1-x^{2}}}\right)

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

\left(1-x^{2}\right) \frac{d y}{d x}=x+\frac{y}{x}

∴ Proved

 

 

Posted by

infoexpert26

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