#### Please solve RD Sharma class 12 chapter Differentiation exercise 10.4 question 17 maths textbook solution

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

Hint:

Use quotient rule and properties of logarithm

Given:

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

Solution:

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

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

$\frac{1}{2} \log \left(x^{2}+y^{2}\right)=\tan ^{-1}\left(\frac{y}{x}\right) \quad\left[\because \log a^{m}=m \log a\right]$

Differentiate this above equation w.r.t x

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

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

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

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

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

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

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

$\frac{x}{x^{2}+y^{2}}+\frac{y}{x^{2}+y^{2}} \cdot \frac{d y}{d x}=\left[\frac{x}{x^{2}+y^{2}} \cdot \frac{d y}{d x}\right]-\frac{y}{x^{2}+y^{2}}$

$\frac{x}{x^{2}+y^{2}} \cdot \frac{d y}{d x}-\frac{y}{x^{2}+y^{2}} \cdot \frac{d y}{d x}=\frac{x}{x^{2}+y^{2}}+\frac{y}{x^{2}+y^{2}}$

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

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

Hence proved