#### need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.4 question 19

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

Hint:

Use differentiation formulas

Given:

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

Solution:

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

$\tan a=\frac{x^{2}-y^{2}}{x^{2}+y^{2}}$

\begin{aligned} &\left(x^{2}+y^{2}\right) \tan a=x^{2}-y^{2} \\ &x^{2}-y^{2}=(\tan a)\left(x^{2}+y^{2}\right) \end{aligned}

Differentiate the given equation w.r.t x

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

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

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

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

$2 x-2 y \frac{d y}{d x}=2 x \tan a+2 y \tan a \frac{d y}{d x}$

$2 y \tan a \frac{d y}{d x}+2 y \frac{d y}{d x}=2 x-2 x \tan a$

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

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

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

Thus, proved