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

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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

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}