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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 :

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

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

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

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

Correct Option: , 2


Consider the equation,

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

Differentiate both sides w.r.t. $x$,

$\log _{e}\left(\log _{e} x\right)+x \cdot \frac{1}{x \cdot \log _{e} x}-2 x+2 y \frac{d y}{d x}=0$

$\log _{e}\left(\log _{e} x\right)+\frac{1}{\log _{e} x}-2 x+2 y \frac{d y}{d x}=0$   .....(1)

When $x=e, y=\sqrt{4+e^{2}}$. Put these values in (1),

$0+1-2 e+2 \sqrt{4+e^{2}} \frac{d y}{d x}=0$

$\frac{d y}{d x}=\frac{2 e-1}{2 \sqrt{4+e^{2}}}$.


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