If y = ( 1 + 1 x )^x , then dy dx = - Vidyadip

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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