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#### Provide solution for RD Sharma maths class 12 chapter 10 Differentiation exercise Multiple choice question 6

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

Hint:

Differentiate the function w.r.t x

Given:

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

Solution:

Let

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

Taking log on both sides

\begin{aligned} &\log y=x \log \left(1+\frac{1}{x}\right) \\\\ &\frac{1}{y} \frac{d y}{d x}=x \frac{d}{d x} \log \left(1+\frac{1}{x}\right)+\log \left(1+\frac{1}{x}\right) \frac{d}{d x}(x) \end{aligned}

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

\begin{aligned} &=x \times \frac{x}{x+1}\left(\frac{-1}{x^{2}}\right)+\log \left(1+\frac{1}{x}\right) \\\\ &=\frac{x^{2}}{x+1} \times \frac{-1}{x^{2}}+\log \left(1+\frac{1}{x}\right) \end{aligned}

\begin{aligned} &\frac{1}{y} \frac{d y}{d x}=\frac{-1}{x+1}+\log \left(1+\frac{1}{x}\right) \\\\ &\frac{d y}{d x}=y\left[\frac{-1}{x+1}+\log \left(1+\frac{1}{x}\right)\right] \end{aligned}

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