If $\tan \left( \frac{x+y}{x-y} \right) = k$, then $\frac{dy}{dx}$ is equal to:
Answer & explanation
Correct answer: option 2
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. $x$:
$\frac{d}{dx} \left( \frac{x+y}{x-y} \right) = \frac{d}{dx} (\tan^{-1} k)$
$\frac{(x-y)\left(1 + \frac{dy}{dx}\right) - (x+y)\left(1 - \frac{dy}{dx}\right)}{(x-y)^2} = 0$
$(x-y+x+y)\frac{dy}{dx} - 2y = 0$
$\frac{dy}{dx} = \frac{2y}{2x}$
$= \frac{y}{x}$