Question

If $Y=\log x^x$, find $\frac{dy}{dx}$

Answer 1

Given that,

$y=\log x^x$

On differentiate and firstly taking log on both sides

$\log y=\log \left(\log x^x\right)$

$\log y=x\log \left(\log x\right)$

$\frac{1}{y}\frac{dy}{dx}=\log \left(\log x\right)+\frac{x}{\mathrm{logx}}\frac{d}{dx}\left(\log x\right)$

$\frac{dy}{dx}=y[\log \left(\log x\right)+\frac{x}{\log x}\times \frac{1}{x}]$

$\frac{dy}{dx}=y[\log \left(\log x\right)+\frac{1}{\log x}]$

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

This is the value of $\frac{dy}{dx}$