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STD 12 - 5. continuity and differentiation — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 5. continuity and differentiation1 Mark
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
✓

Answer

Correct option: D.
None of these
d
(d) $\frac{d}{{dx}}\log \left( {\frac{{{e^x}}}{{1 + {e^x}}}} \right) = \frac{{1 + {e^x}}}{{{e^x}}} \times \frac{d}{{dx}}\left( {\frac{{{e^x}}}{{1 + {e^x}}}} \right)$

$ = \frac{{1 + {e^x}}}{{{e^x}}} \times \frac{{{e^x}}}{{{{(1 + {e^x})}^2}}} = \frac{1}{{1 + {e^x}}}$.

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