Q. The derivative of $\sin^{2}x$ with respect to $\cos^{2}x$ is
A) $\tan^{2}x$
B) $\tan x$
C) $- \tan x$
D) None of these
Explanation
(d)
Let $u = \sin^{2}{x\ and\ v = \cos^{2}x}$
On differentiating w.r.t. $x$, we get
$\frac{du}{dx} = 2\sin{x\cos{x = \sin{2x}}}$
and $\frac{dv}{dx} = - 2\cos{x\sin{x = - \sin{2x}}}$
$\therefore\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{\sin{2x}}{- \sin{2x}} = - 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.