Question

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

Answer 1

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