JEE Main 201911 Jan, Morning shiftMathematicsEasyLimit, Continuity and Differentiability420× since 2002If xlog_e(log_ex) −-− x² + y² = 4(y > 0), then dydx{{dy} \over {dx}}dxdy at x = e is equal to :AOption A. (1+2e)24+e2{{\left( {1 + 2e} \right)} \over {2\sqrt {4 + {e^2}} }}24+e2(1+2e)BOption B. (1+2e)4+e2{{\left( {1 + 2e} \right)} \over {\sqrt {4 + {e^2}} }}4+e2(1+2e)COption C. (2e−1)24+e2{{\left( {2e - 1} \right)} \over {2\sqrt {4 + {e^2}} }}24+e2(2e−1)DOption D. e4+e2{e \over {\sqrt {4 + {e^2}} }}4+e2eCheckSkip · show solution
