Question:medium

If $x \cdot \log_e(\log_e x) - x^2 + y^2 = 4(y > 0)$, then $\frac{dy}{dx}$ at $x = e$ is

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Always find the value of the dependent variable ($y$) at the given $x$ before substituting into the derivative.
Updated On: Apr 30, 2026
  • $\frac{e}{\sqrt{4 + e^2}}$
  • $\frac{2e - 1}{2\sqrt{4 + e^2}}$
  • $\frac{1 + 2e}{\sqrt{4 + e^2}}$
  • $\frac{1 + 2e}{2\sqrt{4 + e^2}}$
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The Correct Option is D

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