Question

Differentiate the function with respect to $x$.
$(x+\frac{1}{x})^x+x^{(1+\frac{1}{x})}$

Answer 1

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