Question

If $y=(\log x{)}^x-x^{\log x}$ then find $\frac{dy}{dx}$.

Answer 1

Let $y_1=(\log x{)}^x$, $y_2=x^{\log x}$

$\log y_1=x\log \left(\log x\right)$, $\log y_2=\log x.\log x$

$\frac{1}{y_1}{y}_1^{\prime }=\log \left(\log x\right)+\frac{x}{\log x}\frac{1}{x}$, $\frac{y_2^{\prime }}{y_2}=\frac{2\log x}{x}$

$\Rightarrow y_1^{\prime }=(\log x{)}^x({\log }^{\log x}+\frac{1}{\log x})$

$y_2^1=x^{\log x}\left(\frac{2}{x}\log x\right)$

$\Rightarrow y^{\prime }=y_1^{\prime }-y_2^{\prime }$

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