Question

$y={(1+\frac{1}{x})}^x+x^{1+\frac{1}{x}}.$
Differentiate

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

Answer 1

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

Let $y=y_1+y_2$
$\Rightarrow \frac{dy}{dx}=\frac{d{y}_1}{dx}+\frac{d{y}_2}{dx}$
Now $y_1={(1+\frac{1}{x})}^x\Rightarrow \log y_1=x\log (1+\frac{1}{x})$

Differentiating both side w.r.t $x$

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

$\Rightarrow \frac{d{y}_1}{dx}=y_1\left[\log \right(1+\frac{1}{x})-\frac{1}{1+x}]$
And $y_2=x^{1+\frac{1}{x}}\Rightarrow \log y_2=(1+\frac{1}{x})\log x$

Differentiating both sides,

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

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