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JEE Main 2019 · 11 Jan · Shift I

JEE Main 2019 Mathematics question on Limits Continuity and Differentiability

If xlog⁡e(log⁡ex)−x2+y2=4x\log_e(\log_e x) - x^2 + y^2 = 4 (y>0)(y > 0), then dydx\frac{dy}{dx} at x=ex = e is equal to
  1. A.
    e4+e2\frac{e}{\sqrt{4+e^2}}
  2. B.
    (2e−1)24+e2\frac{(2e-1)}{2\sqrt{4+e^2}}
  3. C.
    (1+2e)4+e2\frac{(1+2e)}{\sqrt{4+e^2}}
  4. D.
    (1+2e)24+e2\frac{(1+2e)}{2\sqrt{4+e^2}}
Show answer

Correct option: B

[JEE Main 2019 · 11 Jan · Shift I]

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