If x=e^ y+e^ y+2+0 , x>0, then d y d x is
Mathematics · JEE Main · NTA Exams — Limit, Continuity and Differentiability
If \(x=e^{y+e^{y+2+0 \infty}}, x>0, \text { then } \frac{d y}{d x}\) is
- \(\frac{\mathbf{x}}{\mathbf{x}+1}\)
- \(\frac{1}{x}\)
- \(\frac{1-\mathbf{x}}{\mathbf{x}}\)
- \(\frac{1+\mathbf{x}}{\mathbf{x}}\)
Answer
(C) 1- x x
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