JEE Main 201816 Apr, Morning shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002If x=2cosec−1x = \sqrt {{2^{\cos e{c^{ - 1}}}}}x=2cosec−1 and y=2sec−1t (∣t∣≥1),y = \sqrt {{2^{se{c^{ - 1}}t}}} \,\,\left( {\left| t \right| \ge 1} \right),y=2sec−1t(∣t∣≥1), then dydx{{dy} \over {dx}}dxdy is equal to :AOption A. yx{y \over x}xyBOption B. xy{x \over y}yxCOption C. −-− yx{y \over x}xyDOption D. −-− xy{x \over y}yxCheckSkip · show solution
