Question

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

• A. $1$
• B. $2$
• C. $\frac{2x}{\sqrt{x^2+y^2}}$
• D. $\frac{x+y}{x-y}$

Answer 1

The correct option is D $\frac{x+y}{x-y}$

Given,

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

Now, differentiating both sides w.r.to $x$ we get,

or, $\frac{1}{2}\frac{2x+2y\frac{dy}{dx}}{x^2+y^2}=\frac{x^2}{x^2+y^2}.(-\frac{y}{x^2}+\frac{\frac{dy}{dx}}{x})$

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

or, $(x-y)\frac{dy}{dx}=(x+y)$

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