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  • Please solve RD Sharma class 12 chapter Differentiation exercise 10.8 question 13 maths textbook solution

Please solve RD Sharma class 12 chapter Differentiation exercise 10.8 question 13 maths textbook solution

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Answer:   \frac{\sqrt{1-x^{2}}}{3\left(1+x^{2}\right)}

Hint:   \text { Let } u=\tan ^{-1}\left(\frac{x-1}{x+1}\right), v=\sin ^{-1}\left(3 x-4 x^{3}\right)


Given:  \tan ^{-1}\left(\frac{x-1}{x+1}\right) \text { w.r.t } \sin ^{-1}\left(3 x-4 x^{3}\right)

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

Explanation:  \text { Let } u=\tan ^{-1}\left(\frac{x-1}{x+1}\right)

\begin{aligned} &v=\sin ^{-1}\left(3 x-4 x^{3}\right) \\\\ &u=\tan ^{-1}\left(\frac{x-1}{x+1}\right) \\\\ &=\tan ^{-1} x-\tan ^{-1} 1 \\\\ &=\tan ^{-1} x-\frac{\pi}{4} \end{aligned}

\begin{aligned} &\frac{d u}{d x}=\frac{1}{1+x^{2}} \\\\ &v=\sin ^{-1}\left(3 x-4 x^{3}\right) \\\\ &\text { Let } x=\sin \theta \end{aligned}

\begin{aligned} &\frac{-1}{2}<x<\frac{1}{2} \\\\ &\frac{-1}{2}<\sin \theta<\frac{1}{2} \\\\ &-\frac{\pi}{6}<\theta<\frac{\pi}{6} \end{aligned}

\begin{aligned} &v=\sin ^{-1}\left(3 \sin \theta-4 \sin ^{3} \theta\right) \\\\ &=\sin ^{-1}(\sin 3 \theta) \\\\ &=3 \theta \\\\ &v=3 \sin ^{-1} x \end{aligned}

\begin{aligned} &\frac{d v}{d x}=\frac{3}{\sqrt{1-x^{2}}} \\\\ &\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{\frac{1}{1+x^{2}}}{\frac{3}{\sqrt{1-x^{2}}}}=\frac{\sqrt{1-x^{2}}}{3\left(1+x^{2}\right)} \end{aligned}

 

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