The derivative of (log x)^x with respect to log x is

The correct answer is Option (3) → $x(\log x)^{x-1} [1+ \log x.\log(\log x)]$

Given function: $y = (\log x)^x$

Taking $\ln$:

$\ln y = x \ln(\log x)$

Differentiate w.r.t. $x$:

$\frac{1}{y}\frac{dy}{dx} = \ln(\log x) + x\cdot \frac{1}{\log x}\cdot \frac{1}{x}$

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

Now derivative w.r.t. $\log x$:

$\frac{dy}{d(\log x)} = x\cdot \frac{dy}{dx}$

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

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