A) \[\frac{1}{1-{{e}^{x}}}\]
B) \[-\frac{1}{1+{{e}^{x}}}\]
C) \[-\frac{1}{1-{{e}^{x}}}\]
D) None of these
Correct Answer: D
Solution :
\[\frac{d}{dx}\log \left( \frac{{{e}^{x}}}{1+{{e}^{x}}} \right)=\frac{1+{{e}^{x}}}{{{e}^{x}}}\times \frac{d}{dx}\left( \frac{{{e}^{x}}}{1+{{e}^{x}}} \right)\] \[=\frac{1+{{e}^{x}}}{{{e}^{x}}}\times \frac{{{e}^{x}}}{{{(1+{{e}^{x}})}^{2}}}=\frac{1}{1+{{e}^{x}}}\].
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