Question

If $log(x^2+y^2)=2{\tan }^{-1}\left(\frac{y}{x}\right)$, then show that $\frac{dy}{dx}=\frac{x+y}{x-y}$.

Answer 1

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

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

$\Rightarrow \frac{1}{x^2+y^2}\times (2x+2y\frac{dy}{dx})=2\times \frac{x^2}{x^2+y^2}\times \frac{x\frac{dy}{dx}-y}{x^2}$

$\Rightarrow (2x+2y\frac{dy}{dx})=2\times (x\frac{dy}{dx}-y)$ or

$x+y\frac{dy}{dx}y$ or $\frac{dy}{dx}=\frac{x+y}{x-y}HP$