If $\tan \left( \frac{x+y}{x-y} \right) = k$, then $\frac{dy}{dx}$ is equal to:

$\frac{y}{x}$

The correct answer is Option (2) → $\frac{y}{x}$ ##

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

Differentiating both sides w.r.t. $x$:

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

$\frac{(x-y)\left(1 + \frac{dy}{dx}\right) - (x+y)\left(1 - \frac{dy}{dx}\right)}{(x-y)^2} = 0$

$(x-y+x+y)\frac{dy}{dx} - 2y = 0$

$\frac{dy}{dx} = \frac{2y}{2x}$

$= \frac{y}{x}$