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
\(\frac{dy}{dx}=(x+\frac{1}{x})^x\bigg[\frac{x^2-1}{x^2+1}+log(x+\frac{1}{x})\bigg]+x^{(1+\frac{1}{x})}(\frac{x+1-log\,x}{x^2})\)