Mathematics · 2019
JEE Main · 11 January 2019, Shift 1 · Q71
If x log_e( log_e x)-x^2+y^2=4(y>0), then (d y)/(d x) at x=e is equal to:
If $\displaystyle x \log _e\left(\log _e x\right)-x^2+y^2=4(y>0)$, then $\displaystyle \frac{d y}{d x}$ at $\displaystyle x=e$ is equal to :
Official answer
(1)
$\displaystyle \frac{(2 e-1)}{2 \sqrt{4+e^2}}$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.