${(\log x)^x} + {x^{\log x}}$

Let $y =$ ${(\log x)^x} + {x^{\log x}}$

$\Rightarrow$ $y = u + v,$ where$u = {(\log x)^x},v = {x^{\log x}}$

$\Rightarrow$ $\cfrac{{dy}}{{dx}} = \cfrac{{du}}{{dx}} + \cfrac{{dv}}{{dx}}$ ….(i)

Now, $u = {(\log x)^x}$

By taking log on both sides , we get
therefore, $\log u = x\log (\log x)$

Differentiating w.r.t. x, we get

$\cfrac{1}{u} \cdot \cfrac{{du}}{{dx}} = x\cfrac{d}{{dx}}\log (\log x) + \log (\log x)$

$= x \cdot \cfrac{1}{{\log x}} \cdot \cfrac{1}{x} + \log (\log x) = \cfrac{1}{{\log x}} + \log (\log x)$

therefore, $\cfrac{{du}}{{dx}} = {(\log x)^x}\left[ {\cfrac{1}{{\log x}} + \log (\log x)} \right]$ …(ii)

Also, $v = {x^{\log x}}$

By taking log on both sides , we get

$\log v = \log x\log x = {(\log x)^2}$

Differentiating w.r.t. x, we get

$\cfrac{1}{v}\cfrac{{dv}}{{dx}} = 2\log x \cdot \cfrac{1}{x}.$

therefore, $\cfrac{{dv}}{{dx}} = {x^{\log x}}\left[ {\cfrac{{2\log x}}{x}} \right]$ …(iii)

From (i), (ii) \& (iii), we get

$\cfrac{{dy}}{{dx}} = {(\log x)^x}\left[ {\cfrac{1}{{\log x}} + \log (\log x)} \right] + {x^{\log x}}\left[ {\cfrac{{2\log x}}{x}} \right]$

$= {(\log x)^{x - 1}}[1 + \log x \cdot \log (\log x)] + 2{x^{\log x - 1}} \cdot \log x$