MCQ
${d \over {dx}}\left\{ {\log \left( {{{{e^x}} \over {1 + {e^x}}}} \right)} \right\} = $
- A${1 \over {1 - {e^x}}}$
- B$ - {1 \over {1 + {e^x}}}$
- C$ - {1 \over {1 - {e^x}}}$
- ✓None of these
$ = \frac{{1 + {e^x}}}{{{e^x}}} \times \frac{{{e^x}}}{{{{(1 + {e^x})}^2}}} = \frac{1}{{1 + {e^x}}}$.
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