Question

Differentiate $y=(1+1/x{)}^x+x^{1+1/x}$

• A. ${(1+\frac{1}{x})}^x[log(1+\frac{1}{x})-\frac{1}{x+1}]+x^{1+1/x}\left[\frac{x+1-logx}{x^2}\right]$
• B. ${(1+\frac{1}{x})}^{-x}\left[log(1-\frac{1}{x}-\frac{1}{x+1})\right]+x^{1+1/x}\left[\frac{x+1+logx}{x^2}\right]$
• C. ${(-1+\frac{1}{x})}^x\left[log(1+\frac{1}{x}-\frac{1}{x+1})\right]-x^{1+1/x}\left[\frac{x+1-logx}{x^2}\right]$
• D. ${(1+\frac{1}{x})}^x\left[log(1-\frac{1}{x}-\frac{1}{x+1})\right]+x^{1+1/x}\left[\frac{x+1+logx}{x^2}\right]$

Answer 1

The correct option is A ${(1+\frac{1}{x})}^x[log(1+\frac{1}{x})-\frac{1}{x+1}]+x^{1+1/x}\left[\frac{x+1-logx}{x^2}\right]$

Let $y=y_1+y_2$

Differentiating both sides w.r.t $x$

$\Rightarrow \frac{dy}{dx}=\frac{d{y}_1}{dx}+\frac{d{y}_2}{dx}$

Now $y_1=(1+1/x{)}^x\Rightarrow \log y_1=x\log (1+1/x)$

Differentiating both sides w.r.t $x$

$\Rightarrow \frac{1}{y_1}\frac{d{y}_1}{dx}=\log (1+1/x)+x.\frac{1}{1+1/x}.(-1/x^2)$

$\Rightarrow \frac{d{y}_1}{dx}=y_1\left[\log \right(1+1/x)-\frac{1}{1+x}]$

And $y_2=x^{1+1/x}\Rightarrow \log y_2=(1+1/x)\log x$

Differentiating both sides,

$\Rightarrow \frac{1}{y_2}\frac{d{y}_2}{dx}=(1+1/x)\frac{1}{x}+\left(\log x\right)(-1/x^2)$

$\Rightarrow \frac{d{y}_2}{dx}=y_2\left[\frac{(1+x)-\log x}{x^2}\right]$

$\therefore \frac{dy}{dx}=(1+\frac{1}{x})\left[\log \right(1+1/x)-\frac{1}{1+x}]+x^{1+1/x}\left[\frac{(1+x)-\log x}{x^2}\right]$