#### Provide solution for RD Sharma maths class 12 chapter 10 Differentiation exercise Multiple choice question 14

1

Hint:

Differentiate the function w.r.t x

Given:

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

Solution:

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

Differentiating u w.r.t x then

$\frac{d y}{d x}=\frac{d}{d x}\left(\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\right)$

$=\frac{1}{\sqrt{1-\left(\frac{2 x}{1+x^{2}}\right)^{2}}}\left[\frac{\left(1+x^{2}\right) \frac{d}{d x}(2 x)-2 x \frac{d}{d x}\left(1+x^{2}\right)}{\left(1+x^{2}\right)^{2}}\right]$    $\left[\begin{array}{l} \frac{d}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^{2}}} \\\\ \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}} \end{array}\right]$

$=\frac{1}{\sqrt{\frac{\left(1+x^{2}\right)^{2}-(2 x)^{2}}{\left(1+x^{2}\right)^{2}}}}\left[\frac{\left(1+x^{2}\right) 2 \cdot \frac{d(x)}{d x}-2 x\left\{\frac{d}{d x}(1)+\frac{d}{d x}\left(x^{2}\right)\right\}}{\left(1+x^{2}\right)^{2}}\right]$    $\left[\begin{array}{l} \because \frac{d}{d x}\{f(x)+g(x)\}=\frac{d}{d x} f(x)+\frac{d}{d x} g(x) \\\\ \frac{d}{d x}\left(a x^{n}\right)=a \frac{d}{d x}\left(x^{n}\right) \end{array}\right]$

$=\frac{\sqrt{\left(1+x^{2}\right)^{2}}}{\sqrt{1+x^{4}+2 x^{2}-4 x^{2}}} \times\left[\frac{\left(1+x^{2}\right) 2 \cdot 1-2 x(0+2 x)}{\left(1+x^{2}\right)^{2}}\right]$    $\left[\because \frac{d}{d x}(\operatorname{cons} \tan t)=0, \frac{d}{d x}\left(x^{n}\right)=n x^{n-1}\right]$

$=\frac{\left(1+x^{2}\right)}{\sqrt{1+x^{2}-2 x^{2}}}\left[\frac{\left(1+x^{2}\right) 2-2 x(2 x)}{\left(1+x^{2}\right)^{2}}\right]$

$=\frac{\left(1+x^{2}\right)}{\sqrt{\left(1-x^{2}\right)^{2}}}\left[\frac{2+2 x^{2}-4 x^{2}}{\left(1+x^{2}\right)^{2}}\right] \quad\quad\quad\left[\because a^{2}+b^{2}-2 a b=(a-b)^{2}\right]$

\begin{aligned} &=\frac{\left(1+x^{2}\right)}{\left(1-x^{2}\right)} \times \frac{2-2 x^{2}}{\left(1+x^{2}\right)^{2}}=\frac{2\left(1-x^{2}\right)}{\left(1-x^{2}\right)\left(1+x^{2}\right)}=\frac{2}{1+x^{2}}\\\\ &\therefore \frac{d u}{d x}=\frac{2}{1+x^{2}} &\text { } \end{aligned}................(1)

Differentiating v w.r.t x then,

$\frac{d v}{d x}=\frac{d}{d x}\left(\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)\right)$

$=\frac{1}{1+\left(\frac{2 x}{1-x^{2}}\right)^{2}}\left[\frac{\left(1-x^{2}\right) \frac{d}{d x}(2 x)-\frac{d}{d x}\left(1-x^{2}\right) \cdot 2 x}{\left(1-x^{2}\right)^{2}}\right]$        $\left[\because \frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^{2}}, \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}\right]$

$=\frac{1}{\frac{\left(1-x^{2}\right)^{2}+(2 x)^{2}}{\left(1-x^{2}\right)^{2}}}\left[\frac{\left(1-x^{2}\right) 2 \frac{d}{d x}(x)-2 x\left\{\frac{d}{d x}(1)-\frac{d}{d x}\left(x^{2}\right)\right\}}{\left(1-x^{2}\right)^{2}}\right]$        $\left[\begin{array}{l} \frac{d}{d x}\{f(x)+g(x)\}=\frac{d}{d x} f(x)+\frac{d}{d x} g(x) \\\\ \frac{d}{d x}\left(a x^{n}\right)=a \frac{d}{d x}\left(x^{n}\right) \end{array}\right]$

$=\frac{\left(1-x^{2}\right)^{2}}{1+x^{4}-2 x^{2}+4 x^{2}}\left[\frac{\left(1-x^{2}\right) \cdot 2 \cdot 1-2 x(0-2 x)}{\left(1-x^{2}\right)^{2}}\right]$            $\left[\begin{array}{l} \because \frac{d}{d x}\left(x^{n}\right)=n x^{n-1}, \frac{d}{d x}(\operatorname{cons} \tan t)=0, \frac{d(x)}{d x}=1 \\\\ (a-b)^{2}=a^{2}+b^{2}-2 a b \end{array}\right]$

$=\frac{1}{1+x^{4}+2 x^{2}}\left[\frac{2-2 x^{2}+4 x^{2}}{1}\right]$

$=\frac{2+2 x^{2}}{\left(1+x^{2}\right)^{2}}=\frac{2\left(1+x^{2}\right)}{\left(1+x^{2}\right)^{2}} \quad\quad\quad\left[\because a^{2}+b^{2}+2 a b=(a+b)^{2}\right]$

$\therefore \frac{d v}{d x}=\frac{2}{1+x^{2}}$                ..............................(2)

Thus

$\frac{d y}{d x}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{d u}{d x} \times \frac{d x}{d v}$

$=\frac{2}{1+x^{2}} \times \frac{1+x^{2}}{2}$            [Using (1) and (2)]

$\therefore \frac{d y}{d x}=1$