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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
If $y = {\left( {1 + {1 \over x}} \right)^x}$, then ${{dy} \over {dx}} = $
  • ✓
    ${\left( {1 + {1 \over x}} \right)^x}\left[ {\log \left( {1 + {1 \over x}} \right) - {1 \over {1 + x}}} \right]$
  • B
    ${\left( {1 + {1 \over x}} \right)^x}\left[ {\log \left( {1 + {1 \over x}} \right)} \right]$
  • C
    ${\left( {x + {1 \over x}} \right)^x}\left[ {\log (x - 1) - {x \over {x + 1}}} \right]$
  • D
    ${\left( {1 + {1 \over x}} \right)^x}\left[ {\log \left( {1 + {1 \over x}} \right) + {1 \over {1 + x}}} \right]$
✓

Answer

Correct option: A.
${\left( {1 + {1 \over x}} \right)^x}\left[ {\log \left( {1 + {1 \over x}} \right) - {1 \over {1 + x}}} \right]$
a
(a) $y = {\left( {1 + \frac{1}{x}} \right)^x} $

$\Rightarrow \log y = x\log \left( {1 + \frac{1}{x}} \right)$

$ \Rightarrow \frac{1}{y}\frac{{dy}}{{dx}} = \log \left( {1 + \frac{1}{x}} \right) - \frac{1}{{1 + x}}$

==> $\frac{{dy}}{{dx}} = {\left( {1 + \frac{1}{x}} \right)^x}\left[ {\log \left( {1 + \frac{1}{x}} \right) - \frac{1}{{1 + x}}} \right]$.

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