Method of Differentiation_JEE MAIN_11-01-2019_S1 - Competishun

JEE MAIN 2019

11-01-2019 S1

Question

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

Select the correct option:

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

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

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

$\frac{\mathrm{e}}{\sqrt{4+\mathrm{e}^2}}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

Question Tags

JEE Main

Mathematics

Easy

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