If $\log \sqrt{x^{2}+y^{2}}=\tan ^{-1}\left(\frac{y}{x}\right)$, prove that $\frac{d y}{d x}=\frac{x+y}{x-y}$.
We are given with an equation $\log \sqrt{x^{2}+y^{2}}=\tan ^{-1}\left(\frac{y}{x}\right)$, we have to prove that $\frac{d y}{d x}=\frac{y+x}{y-x}$ by using the given
equation we will first find the value of $\frac{d y}{d x}$ and we will put this in the equation we have to prove, so by differentiating the equation on both sides with respect to $x$, we get,
$\log \left(x^{2}+y^{2}\right)=2 \tan ^{-1}\left(\frac{y}{x}\right)$
$\frac{2 x+2 y \frac{d y}{d x}}{x^{2}+y^{2}}=\frac{2}{1+\left(\frac{y}{x}\right)^{2}} \frac{x \frac{d y}{d x}-y(1)}{x^{2}}$
$x+y \frac{d y}{d x}=x \frac{d y}{d x}-y$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{x}+\mathrm{y}}{\mathrm{x}-\mathrm{y}}$
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