JEE Main 201912 Jan, Morning shiftMathematicsEasyLimit, Continuity and Differentiability420× since 2002For x > 1, if (2x)2y = 4e2x−-−2y, then (1 + log_e 2x)² dydx{{dy} \over {dx}}dxdy is equal to :AOption A. x loge2x−loge2x{{x\,{{\log }_e}2x - {{\log }_e}2} \over x}xxloge2x−loge2BOption B. log_e 2xCOption C. x log_e 2xDOption D. x loge2x+loge2x{{x\,{{\log }_e}2x + {{\log }_e}2} \over x}xxloge2x+loge2CheckSkip · show solution
