If $y = {\left( {1 + \frac{1}{x}} \right)^x}$, then $\frac{{dy}}{{dx}} = $

Question

Accepted Answer

${\left( {1 + \frac{1}{x}} \right)^x}\left[ {\log \left( {1 + \frac{1}{x}} \right) - \frac{1}{{1 + x}}} \right]$

Answer

${\left( {1 + \frac{1}{x}} \right)^x}\left[ {\log \left( {1 + \frac{1}{x}} \right)} \right]$

Answer

${\left( {x + \frac{1}{x}} \right)^x}\left[ {\log (x - 1) - \frac{x}{{x + 1}}} \right]$

Answer

${\left( {1 + \frac{1}{x}} \right)^x}\left[ {\log \left( {1 + \frac{1}{x}} \right) + \frac{1}{{1 + x}}} \right]$

differentiation