Question:
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 :
Correct Option: , 2
Solution:
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),
$0+1-2 e+2 \sqrt{4+e^{2}} \frac{d y}{d x}=0$
$\frac{d y}{d x}=\frac{2 e-1}{2 \sqrt{4+e^{2}}}$.