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  • Need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.8 question 15

Need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.8 question 15

Answers (1)

Answer: 1

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

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

            -1<x<1

Explanation:

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

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

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

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

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

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