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

The derivative of $(\log x)^x$ with respect to $\log x$ needs to be found.

This requires differentiating with respect to $(\log x)$, not with respect to $x$.

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Using the chain rule:

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

Since $\frac{d(\log x)}{dx} = \frac{1}{x}$, we have:

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

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Let $y = (\log x)^x$

Taking the natural logarithm of both sides:

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

Differentiating both sides with respect to $x$:

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

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

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Solving for $\frac{dy}{dx}$:

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

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

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Converting to the derivative with respect to $\log x$:

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

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

$\frac{dy}{d(\log x)} = x \cdot (\log x)^{x-1}[1 + \log x \cdot \ln(\log x)]$

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Therefore, the answer is:

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