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