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

Correct Option: , 3

Solution:

Differentiating with respect to x,

$x \cdot \frac{1}{\ln x} \cdot \frac{1}{x}+\ln (\ln x)-2 x+2 y \cdot \frac{d y}{d x}=0$

at $x=$ e we get

$1-2 e+2 y \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{2 e-1}{2 y}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2 \mathrm{e}-1}{2 \sqrt{4+\mathrm{e}^{2}}}$ as $\mathrm{y}(\mathrm{e})=\sqrt{4+\mathrm{e}^{2}}$