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need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.2 question 69

Answers (1)

Answer:Proved

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

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


Prove:   \frac{d y}{d x}=\sin ^{-1} x
 

Solution:

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

Differentiate with respect to x,use product rule

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

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

∴ Proved

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