If $y = (\log x)^{\log x}, x > 1$ then $\frac{dy}{dx}$ is equal to

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

The correct answer is Option (4) → $(\log x)^{\log x} \left[\frac{1 + \log(\log x)}{x}\right], x > 1$

Let $u=\log x$ so $y=u^{\,u}$.

$\log y = u\log u$

Differentiate w.r.t. $x$:

$\frac{1}{y}\frac{dy}{dx} = \frac{du}{dx}\big(\log u + 1\big)$

Since $\frac{du}{dx}=\frac{1}{x}$,

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

Back-substitute $y=u^{u}$ and $u=\log x$:

$\displaystyle \frac{dy}{dx} = (\log x)^{\log x}\,\frac{1+\log(\log x)}{x}$