Question

Differentiate the function with respect to \(x\)
\((log\,x)^x+x^{log\,x}\)

Answer 1

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