If $x{\log _e}\left( {{{\log }_e}x} \right) - {x^2} + {y^2} = 4\left( {y > 0} \right)\,$, then $\frac{{dy}}{{dx}}$ at $x = e$ is equal to

$\frac{e}{{\sqrt {4 + {e^2}} }}$

$\frac{{\left( {1 + 2e} \right)}}{{\sqrt {4 + {e^2}} }}$

$\frac{{\left( {1 + 2e} \right)}}{{2\sqrt {4 + {e^2}} }}$

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

x{\log _e}\left( {{{\log }_e}x} \right) - {x^2} + {y^2} = 4\left( {y > 0} \right)\,

, then

\frac{{dy}}{{dx}}

x = e

is equal to

View answer and explanation

Correct option: B

\frac{{\left( {2e - 1} \right)}}{{2\sqrt {4 + {e^2}} }}

Option B (

\frac{{\left( {2e - 1} \right)}}{{2\sqrt {4 + {e^2}} }}

) is correct because it is the value of the derivative at x=e, obtained by implicit differentiation and substitution.