Q. If $xy$=$\tan^{- 1}(xy) + \cot^{- 1}(xy)$, then $\frac{dy}{dx}\$is equal to
A) $\frac{y}{x}$
B) $- \frac{y}{x}$
C) $\frac{x}{y}$
D) $- \frac{x}{y}$
Explanation
(b)
$xy = \tan^{- 1}{(xy) + \cot^{- 1}{(xy)} = \frac{\pi}{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$
$\Rightarrow \ xy = \frac{\pi}{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$
On differentiating, we get
$\therefore\ x\frac{dy}{dx} + y = 0$
$\Rightarrow \ \ \frac{dy}{dx} = \ \frac{- y}{x}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.