#### Please solve RD Sharma class 12 chapter Differentiation exercise 10.4 question 9 maths textbook solution

Answer:$\left[-\left(\frac{x}{y}\right)\right]$

Hint:

Use chain rule and $\frac{d\left(\tan ^{-1} x\right)}{d x}=\frac{1}{1+x^{2}}$

Given:

$\tan ^{-1}\left(x^{2}+y^{2}\right)=a$

Solution:

Differentiate the given equation w.r.t x

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

$\frac{d\left(\tan ^{-1}\left(x^{2}+y^{2}\right)\right)}{d\left(x^{2}+y^{2}\right)} \times \frac{d\left(x^{2}+y^{2}\right)}{d x}=0$                [Using chain rule] $\left[\because \frac{d(\operatorname{cons} \tan t)}{d x}=0\right]$

$\frac{1}{1+\left(x^{2}+y^{2}\right)^{2}} \times\left(\frac{d x^{2}}{d x}+\frac{d y^{2}}{d x}\right)=0$                                        $\left[\because \frac{d\left(\tan ^{-1} x\right)}{d x}=\frac{1}{1+x^{2}}\right]$

$\frac{1}{1+\left(x^{2}+y^{2}\right)^{2}} \times\left(2 x+\frac{d y^{2}}{d y} \times \frac{d y}{d x}\right)=0$

$\frac{1}{1+\left(x^{2}+y^{2}\right)^{2}} \times\left(2 x+2 y \frac{d y}{d x}\right)=0$

$\frac{2 x}{1+\left(x^{2}+y^{2}\right)^{2}}+\frac{2 y}{1+\left(x^{2}+y^{2}\right)^{2}} \frac{d y}{d x}=0$

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

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

$\frac{d y}{d x}=\frac{-2 x}{2 y}=-\frac{x}{y}$

$\frac{d y}{d x}=-\frac{x}{y}$

Hence $\frac{d y}{d x}=-\frac{x}{y}$ is the required answer