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  • Need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.8 question 11

Need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.8 question 11

Answers (1)

Answer: 2

Hint: \text { Let } u=\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right), v=\tan ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right)
 

Given:  \sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right) \text { w.r.t } \tan ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right)

 

Explanation:

\text { Let } u=\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)

\begin{aligned} &v=\tan ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right) \\\\ &\text { Let } x=\sin \theta \\\\ &u=\sin ^{-1}\left(2 \sin \theta \sqrt{1-\sin ^{2} \theta}\right) \end{aligned}

    \begin{aligned} &=\sin ^{-1}(2 \cos \theta \sin \theta) \\\\ &=\sin ^{-1}(\sin 2 \theta) \end{aligned}

\begin{aligned} &v=\tan ^{-1}\left(\frac{\sin \theta}{\sqrt{1-\sin ^{2} \theta}}\right) \\\\ &=\tan ^{-1}\left(\frac{\sin \theta}{\cos \theta}\right)=\tan ^{-1}(\tan \theta) \end{aligned}

Now,

\begin{aligned} &\frac{-1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}} \\\\ &\frac{-1}{\sqrt{2}}<\sin \theta<\frac{1}{\sqrt{2}} \end{aligned}

\begin{aligned} &\frac{-\pi}{4}<\theta<\frac{\pi}{4} \\\\ &\frac{-\pi}{2}<2 \theta<\frac{\pi}{2} \end{aligned}

\begin{aligned} &u=\sin ^{-1}(\sin 2 \theta) \\\\ &=\sin ^{-1}(\sin 2 \theta) \\\\ &=2 \theta \\\\ &=2 \sin ^{-1} x \end{aligned}

\begin{aligned} &\frac{d u}{d x}=\frac{2}{\sqrt{1-x^{2}}} \\\\ &v=\tan ^{-1}(\tan \theta) \\\\ &=\theta \\\\ &v=\sin ^{-1} x \end{aligned}

\begin{aligned} &\frac{d v}{d x}=\frac{1}{\sqrt{1-x^{2}}} \\\\ &\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}} \end{aligned}

=\frac{\frac{2}{\sqrt{1-x^{2}}}}{\frac{1}{\sqrt{1-x^{2}}}}=2

Posted by

infoexpert26

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