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

Given $y = (\log x)^{\log x}$ where $x > 1$.

Taking log on both sides:

$\log y = \log[(\log x)^{\log x}]$

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

---

Differentiating both sides with respect to $x$:

$\frac{d}{dx}(\log y) = \frac{d}{dx}[(\log x) \cdot \log(\log x)]$

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

---

Finding the derivatives:

$\frac{d}{dx}[\log(\log x)] = \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x \log x}$

$\frac{d}{dx}[\log x] = \frac{1}{x}$

---

Substituting back:

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

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

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

---

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

Substituting $y = (\log x)^{\log x}$:

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