Mathematics

Problem

Match the following.

Column I	Column II
(A) $y = \sin^{-1}\left(\frac{2x}{1+x^2}\right)$ , then $\frac{dy}{dx} = -\frac{2}{1+x^2}$	(p) for $x < 0$
(B) $y = \cos^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)$ , then $\frac{dy}{dx} = \frac{1}{1+x^2}$	(q) for $x > 1$
(C) $y = e^{ x } - e$ , then $\frac{dy}{dx} > 0$	(r) for $x < -1$
(D) $u = \log 2x$ , $v = \tan^{-1} x$ , then $\frac{du}{dv} > 2$	(s) for $-1 < x < 0$