Choose the correct alternative in the following:
Question:

Choose the correct alternative in the following:

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

A. $\left(1+\frac{1}{x}\right)^{x}\left\{\log \left(1+\frac{1}{x}\right)-\frac{1}{x+1}\right\}$

B. $\left(1+\frac{1}{x}\right)^{x} \log \left(1+\frac{1}{x}\right)$

C. $\left(1+\frac{1}{x}\right)^{x}\left\{\log (x+1)-\frac{x}{x+1}\right\}$

D. $\left(1+\frac{1}{x}\right)^{x}\left\{\log \left(x+\frac{1}{x}\right)+\frac{1}{x+1}\right\}$

Solution:

Given $y=\left(1+\frac{1}{x}\right)^{x}$

Taking log both sides we get

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

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

Differentiating w.r.t $x$ we get,

$\Rightarrow \frac{1}{y} \frac{d y}{d x}=1 \cdot \log \left(1+\frac{1}{x}\right)+\frac{1}{1+\frac{1}{x}} \cdot\left(-\frac{1}{x^{2}}\right) \cdot x$

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

Putting value of $y$, we get

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