Question
यदि $y = {\left( {1 + \frac{1}{x}} \right)^x}$, तो $\frac{{dy}}{{dx}} = $
$\Rightarrow \log y = x\log \left( {1 + \frac{1}{x}} \right)$
$ \Rightarrow \frac{1}{y}\frac{{dy}}{{dx}} = \log \left( {1 + \frac{1}{x}} \right) - \frac{1}{{1 + x}}$
==> $\frac{{dy}}{{dx}} = {\left( {1 + \frac{1}{x}} \right)^x}\left[ {\log \left( {1 + \frac{1}{x}} \right) - \frac{1}{{1 + x}}} \right]$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.