Question

Differentiate the following function with respect to $x$:
$(\log x{)}^x+x^{\log x}$.

Answer 1

Let $y=(\log x{)}^x+x^{\log x}$
Then $\frac{dy}{dx}=\frac{d\left(\right(\log x{)}^x+x^{\log x})}{dx}$

$=\frac{d(\log x{)}^x}{dx}+\frac{d\left(x^{\log x}\right)}{dx}$

$=(\log x{)}^x\frac{d\left(x\log \right(\log x\left)\right)}{dx}+x^{(}\log x)\frac{d\left(\log x\log x\right)}{dx}$

From $(\frac{d\left(u^v\right)}{dx}=u^v\frac{d\left(v\log u\right)}{dx})$

$=(\log x{)}^x\left\{x(\frac{1}{\log x}\frac{1}{x}+\log (\log x\left)\right)\right\}+x^{\log x}\left(2\right(\log x\left)\frac{1}{x}\right)$

Therefore, $(\frac{d\left(\log x\log x\right)}{dx}=\frac{d(\log x{)}^2}{dx})$

$=(\log x{)}^x\{\frac{1}{\log x}+\log (\log x\left)\right\}+2\left(\frac{\log x}{x}{x}^{\log x}\right)$