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  • Need solution for RD Sharma maths class 12 chapter 10 Differentiation exercise Very short answers question 23

Need solution for RD Sharma maths class 12 chapter 10 Differentiation exercise Very short answers question 23

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Answer:

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Hint:

Using \frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}

Given:

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

Solution:  

u=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)        ..............(i)

Put x=\tan \theta

u=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right)

u=\sin ^{-1}(\sin 2 \theta) \quad\left[\because \frac{2 \tan \theta}{1+\tan ^{2} \theta}=\sin 2 \theta\right]

u=2 \theta \quad\left[\sin ^{-1}(\sin \theta)=\theta\right]

u=2 \tan ^{-1} x \quad\left[\because \theta=\tan ^{-1} x\right]

Differentiating it w.r.t x we get

\frac{d u}{d x}=\frac{2}{1+x^{2}}

Again we have,

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

v=\tan ^{-1}(\tan 2 \theta) \quad\left[\because \tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^{2} \theta}\right]

v=2 \theta \quad\left[\because \tan ^{-1}(\tan \theta)=\theta\right]

v=2 \tan ^{-1} x \quad\left[\because \theta=\tan ^{-1} x\right]

Differentiating it w.r.t x, we get

\frac{d v}{d x}=\frac{2}{1+x^{2}}

now,

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

 

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infoexpert26

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