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:
Mathematics · JEE Main · NTA Exams — Limit, Continuity and Differentiability
If x loge(logex) - x2 + y2 = 4 (y > 0), then \(\frac{d y}{d x}\) at x = e is equal to:
- \(\frac{(1+2 e)}{2 \sqrt{4+e^{2}}}\)
- \(\frac{(1+2 e)}{\sqrt{4+e^{2}}}\)
- \(\frac{(2 e-1)}{2 \sqrt{4+e^{2}}}\)
- \(\frac{e}{\sqrt{4+e^{2}}}\)
Answer
(A) (1+2 e) 2 4+e^ 2
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