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

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

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

JEE MATHEMATICS MCQ: DIFFERENTIATION

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Application +4 | -1

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

, then

\frac{d y}{d x}

\mathrm{x}=\mathrm{e}

is equal to :

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because it is the result of differentiating the given equation implicitly with respect to x and substituting x=e and y=

\sqrt{4+e^{2}}

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