Question

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

• A. $\frac{dy}{dx}=\frac{x+y}{x-y}$
• B. $\frac{dy}{dx}=\frac{2x+y}{x-2y}$
• C. $\frac{dy}{dx}=\frac{x-y}{x+y}$
• D. None of these

Answer 1

Answer: (A)

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

Differentiating both sides w.r.t. x, we get
$\frac{d}{dx}\left\{\log (x^2+y^2)\right\}=2\frac{d}{dx}\left\{{\tan }^{-1}\left(\frac{y}{x}\right)\right\}$
or $\frac{1}{x^2+y^2}\times \frac{d}{dx}(x^2+y^2)=2\frac{1}{1+{\left(\frac{y}{x}\right)}^2}\times \frac{d}{dx}\left(\frac{y}{x}\right)$
or $\frac{1}{x^2+y^2}\{\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(y^2\right)\}=2\times \frac{x^2}{x^2+y^2}\left\{\frac{x\frac{dy}{dx}-y\times 1}{x^2}\right\}$
or $\frac{1}{x^2+y^2}\{2x+2y\frac{dy}{dx}\}=\frac{2}{x^2+y^2}\{x\frac{dy}{dx}-y\}$
or $2\{x+y\frac{dy}{dx}\}=2\{x\frac{dy}{dx}-y\}$
or $x+y\frac{dy}{dx}=x\frac{dy}{dx}-y$
or $\frac{dy}{dx}(y-x)=-(x+y)$
or $\frac{dy}{dx}=\frac{x+y}{x-y}$