To find the derivative of the function $f(x) = \sqrt{\ln x} + \frac{1}{x} \log x$, we will use the rules of differentiation. Here is the step-by-step solution:

Function:

$f(x) = \sqrt{\ln x} + \frac{1}{x} \log x$

Step 1: Derivative of $\sqrt{\ln x}$

Using the chain rule: $\frac{d}{dx} \sqrt{\ln x} = \frac{1}{2\sqrt{\ln x}} \cdot \frac{d}{dx} (\ln x) = \frac{1}{2\sqrt{\ln x}} \cdot \frac{1}{x} = \frac{1}{2x\sqrt{\ln x}}$

Step 2: Derivative of $\frac{1}{x} \log x$

Using the product rule: $\frac{d}{dx} \left( \frac{1}{x} \log x \right) = \frac{d}{dx} \left( \frac{\log x}{x} \right)$ Let $u = \log x$ and $v = \frac{1}{x}$. Then: $\frac{d}{dx} \left( \frac{\log x}{x} \right) = u'v + uv'$ where $u' = \frac{1}{x}$ and $v' = -\frac{1}{x^2}$. Substituting these: $\frac{d}{dx} \left( \frac{\log x}{x} \right) = \frac{1}{x} \cdot \frac{1}{x} + (\log x) \cdot \left( -\frac{1}{x^2} \right)$ $\frac{d}{dx} \left( \frac{\log x}{x} \right) = \frac{1}{x^2} - \frac{\log x}{x^2} = \frac{1 - \log x}{x^2}$

Final Answer:

Combining both parts: $f'(x) = \frac{1}{2x\sqrt{\ln x}} + \frac{1 - \log x}{x^2}$

Let me know if you want further clarification or detailed steps for a specific part!

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Here are 5 questions to further your understanding:

1. Can you derive $\frac{d}{dx} \sqrt{\ln x}$ independently using the chain rule?
2. How does the product rule simplify differentiation in $\frac{\log x}{x}$?
3. Why does $\log x$ have a derivative of $\frac{1}{x}$?
4. What happens to the derivative when $x \to \infty$?
5. How does the presence of $\ln x$ and $\log x$ in the function affect its rate of change?

Tip: Always check the domain of the function before differentiating, especially