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  • Please solve RD Sharma class 12 chapter Differentiation exercise 10.8 question 17 maths textbook solution

Please solve RD Sharma class 12 chapter Differentiation exercise 10.8 question 17 maths textbook solution

Answers (1)

Answer: \frac{1}{2}

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


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

            -\frac{1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}

Explanation:

u=\tan ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right)

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

\begin{aligned} &u=\tan ^{-1}\left(\frac{\sin \theta}{\sqrt{1-\sin ^{2} \theta}}\right), \\\\ &u=\tan ^{-1}\left(\frac{\sin \theta}{\sqrt{\cos ^{2} \theta}}\right) \\\\ &u=\tan ^{-1}(\tan \theta) \end{aligned}                    \begin{aligned} &v=\sin ^{-1}\left(2 \sin \theta \sqrt{1-\sin ^{2} \theta}\right) \\\\ &v=\sin ^{-1}(2 \sin \theta \cos \theta) \\\\ &v=\sin ^{-1}(\sin 2 \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{array}{ll} u=\tan ^{-1}(\tan \theta)=\theta & \theta \in\left(-\frac{\pi}{4}, \frac{\pi}{4}\right) \\\\ v=\sin ^{-1}(\sin 2 \theta)=2 \theta & 2 \theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \end{array}

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

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

\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{\frac{1}{\sqrt{1-x^{2}}}}{\frac{2}{\sqrt{1-x^{2}}}}=\frac{1}{2}

\frac{d u}{d v}=\frac{1}{2}

 

Posted by

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