MCQ
If $y = {e^{x + {e^{x + {e^{x + ....\infty }}}}}}$, then ${{dy} \over {dx}} = $
- ✓${y \over {1 - y}}$
- B${1 \over {1 - y}}$
- C${y \over {1 + y}}$
- D${y \over {y - 1}}$
==> $\frac{1}{y}\frac{{dy}}{{dx}} = 1 + \frac{{dy}}{{dx}}$
==> $ \frac{{dy}}{{dx}} = \frac{y}{1-y} $.
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