Q40 / 90 · JEE Main 2019 (11 Jan Shift 1) · Mathematics
If $x \log _{e}\left(\log _{e} x\right)-x^{2}+y^{2}=4(y>0),$ then $\frac{d y}{d x}$ at $x=e$ is
equal to :
- A$\frac{(1+2 e)}{2 \sqrt{4+e^{2}}}$
- Correct answerB$\frac{(2 e-1)}{2 \sqrt{4+e^{2}}}$
- C$\frac{(1+2 e)}{\sqrt{4+e^{2}}}$
- D$\frac{e}{\sqrt{4+e^{2}}}$
Solution
Consider the equation,
$x \log _{e}\left(\log _{e} x\right)-x^{2}+y^{2}=4$
Differentiate both sides w.r.t. $x$
$\log _{e}\left(\log _{e} x\right)+x \cdot \frac{1}{x \cdot \log _{e} x}-2 x+2 y \frac{d y}{d x}=0$
$\log _{e}\left(\log _{e} x\right)+\frac{1}{\log _{e} x}-2 x+2 y \frac{d y}{d x}=0$
When $x=e, y=\sqrt{4+e^{2}}$. Put these values in (1),
$0+1-2 e+2 \sqrt{4+e^{2}} \frac{d y}{d x}=0$
$\frac{d y}{d x}=\frac{2 e-1}{2 \sqrt{4+e^{2}}}$
Hence, the correct answer is option: 'B'.