Question

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

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

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

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

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

Answer 1

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

$$\begin{gathered} \mathrm{Let}y={\left(1+\frac{1}{x}\right)}^x \\ \mathrm{Taking}\log \mathrm{on}\mathrm{both}\mathrm{sides}, \\ \log y=x\log \left(1+\frac{1}{x}\right) \\ \Rightarrow \frac{1}{y}\frac{dy}{dx}=x\frac{d}{dx}\log \left(1+\frac{1}{x}\right)+\log \left(1+\frac{1}{x}\right)\frac{d}{dx}\left(x\right) \\ \Rightarrow \frac{1}{y}\frac{dy}{dx}=x\left(\frac{1}{1+{\displaystyle \frac{1}{x}}}\right)\frac{d}{dx}\left(1+\frac{1}{x}\right)+\log \left(1+\frac{1}{x}\right) \\ \Rightarrow \frac{1}{y}\frac{dy}{dx}=x\times \frac{x}{x+1}\left(-\frac{1}{x^2}\right)+\log \left(1+\frac{1}{x}\right) \\ \Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{x^2}{x+1}\times \frac{-1}{x^2}+\log \left(1+\frac{1}{x}\right) \\ \Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{-1}{x+1}+\log \left(1+\frac{1}{x}\right) \\ \Rightarrow \frac{dy}{dx}=y\left[\frac{-1}{x+1}+\log \left(1+\frac{1}{x}\right)\right] \\ \Rightarrow \frac{dy}{dx}={\left(1+\frac{1}{x}\right)}^x\left[\log \left(1+\frac{1}{x}\right)-\frac{1}{x+1}\right] \end{gathered}$$