Question

Differentiate the function w.r.t. $x$.
${(x+\frac{1}{x})}^x+x^{(1+\frac{1}{x})}$

Answer 1

Let $y={(x+\frac{1}{x})}^x+x^{(1+\frac{1}{x})}$
Also, let $u={(x+\frac{1}{x})}^x$ and $v=x^{(1+\frac{1}{x})}$
$\therefore y=u+v$
$\Rightarrow \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$ ...(1)
Then, $u={(x+\frac{1}{x})}^x$
$\Rightarrow \log u=\log {(x+\frac{1}{x})}^x$
$\Rightarrow \log u=x\log (x+\frac{1}{x})$
Differentiating both sides with respect to $x$, we obtain
$\frac{1}{u}.\frac{du}{dx}=\frac{d}{dx}\left(x\right)\times \log (x+\frac{1}{x})+x\times \frac{d}{dx}\left[\log (x+\frac{1}{x})\right]$
$\Rightarrow \frac{1}{u}.\frac{du}{dx}=1\times \log (x+\frac{1}{x})+x\times \frac{1}{(x+\frac{1}{x})}.\frac{d}{dx}(x+\frac{1}{x})$
$\Rightarrow \frac{du}{dx}=u[\log (x+\frac{1}{x})+\frac{x}{(x+\frac{1}{x})}\times (1-\frac{1}{x^2})]$
$\Rightarrow \frac{du}{dx}={(x+\frac{1}{x})}^x[\log (x+\frac{1}{x})+\frac{(x-\frac{1}{x})}{(x+\frac{1}{x})}]$
$\Rightarrow \frac{du}{dx}={(x+\frac{1}{x})}^x[\log (x+\frac{1}{x})+\frac{x^2-1}{x^2+1}]$
$\Rightarrow \frac{du}{dx}={(x+\frac{1}{x})}^x[\frac{x^2-1}{x^2+1}+\log (x+\frac{1}{x})]$ ...(2)
$v=x^{(1+\frac{1}{x})}$
$\Rightarrow \log v=\log \left[x^{(1+\frac{1}{x})}\right]$
$\Rightarrow \log v=(1+\frac{1}{x})\log x$
Differentiating both sides with respect to $x$, we obtain
$\frac{1}{v}.\frac{dv}{dx}=\left[\frac{d}{dx}(1+\frac{1}{x})\right]\times \log x+(1+\frac{1}{x}).\frac{d}{dx}\log x$
$\Rightarrow \frac{1}{v}.\frac{dv}{dx}=\left[\frac{d}{dx}(1+\frac{1}{x})\right]\times \log x+(1+\frac{1}{x}).\frac{d}{dx}\log x$
$\Rightarrow \frac{1}{v}.\frac{dv}{dx}=-\frac{\log x}{x^2}+\frac{1}{x}+\frac{1}{x^2}$
$\Rightarrow \frac{dv}{dx}=v[-\frac{\log x+x+1}{x^2}]$
$\Rightarrow \frac{dv}{dx}=x^{(1+\frac{1}{x})}[-\frac{x+1-\log x}{x^2}]$ ...(3)
Therefore, from (1), (2) and (3), we obtain
$\Rightarrow \frac{dy}{dx}={(x+\frac{1}{x})}^x[\frac{x^2-1}{x^2+1}+\log (x+\frac{1}{x})]+x^{(1+\frac{1}{x})}[-\frac{x+1-\log x}{x^2}]$