Find the derivative of y = (log x)^(log x), assuming log x means ln x.

Given $y = (\log x)^{\log x}$.
Taking the natural logarithm on both sides:

$$\ln y = \ln \left( (\log x)^{\log x} \right)$$

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

Note that $\log x$ typically refers to $\log_{10} x$ in general contexts, but in calculus, it often implies $\ln x$ (natural logarithm). However, given the explicit use of $\ln$ in the solution, it's safer to assume $\log x$ means $\log_e x$ or $\ln x$. If it were $\log_{10} x$, the derivative would involve an extra factor of $\frac{1}{\ln 10}$. Let's proceed assuming $\log x = \ln x$.
So, the equation becomes:

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

The derivative of $\ln y$ with respect to $x$ is $\frac{1}{y} \frac{dy}{dx}$.
For the right side, use the product rule $\frac{d}{dx}(uv) = u'v + uv'$ where $u = \ln x$ and $v = \ln(\ln x)$.

$$\frac{du}{dx} = \frac{d}{dx}(\ln x) = \frac{1}{x}$$ $$\frac{dv}{dx} = \frac{d}{dx}(\ln(\ln x))$$

Using the chain rule for $v$, let $w = \ln x$. Then $v = \ln w$.

$$\frac{dv}{dx} = \frac{d}{dw}(\ln w) \cdot \frac{dw}{dx} = \frac{1}{w} \cdot \frac{1}{x} = \frac{1}{\ln x} \cdot \frac{1}{x}$$

Now apply the product rule:

$$\frac{d}{dx} ((\ln x) \ln(\ln x)) = \left(\frac{1}{x}\right) \ln(\ln x) + (\ln x) \left(\frac{1}{x \ln x}\right)$$ $$\frac{d}{dx} ((\ln x) \ln(\ln x)) = \frac{\ln(\ln x)}{x} + \frac{1}{x}$$

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

Multiply both sides by $y$:

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

Substitute $y = (\ln x)^{\ln x}$ back into the equation:

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

The final answer is $(\ln x)^{\ln x} \left( \frac{1 + \ln(\ln x)}{x} \right)$.