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The given function is \( f(x) = x^{\sqrt{x}} \), and you are tasked with finding \( f'(9) \), the derivative of the function evaluated at \( x = 9 \).

### Step 1: Rewrite the function
We can express \( f(x) = x^{\sqrt{x}} \) in a more manageable form for differentiation by using logarithms. First, recall that:
\[
x^{\sqrt{x}} = e^{\sqrt{x} \ln(x)}
\]
Thus, \( f(x) = e^{\sqrt{x} \ln(x)} \).

### Step 2: Differentiate the function using the chain rule
We will now differentiate \( f(x) = e^{\sqrt{x} \ln(x)} \). Let \( g(x) = \sqrt{x} \ln(x) \), so:
\[
f(x) = e^{g(x)}
\]

To find \( f'(x) \), apply the chain rule:
\[
f'(x) = e^{g(x)} \cdot g'(x)
\]

### Step 3: Differentiate \( g(x) = \sqrt{x} \ln(x) \)
We need to compute \( g'(x) \). Use the product rule since \( g(x) \) is the product of \( \sqrt{x} \) and \( \ln(x) \). Recall that \( \sqrt{x} = x^{1/2} \), so we have:
\[
g(x) = x^{1/2} \ln(x)
\]

Using the product rule:
\[
g'(x) = \frac{1}{2}x^{-1/2} \ln(x) + x^{1/2} \cdot \frac{1}{x} = \frac{\ln(x)}{2\sqrt{x}} + \frac{1}{\sqrt{x}}
\]

### Step 4: Combine results
Now, we can express the derivative of \( f(x) \) as:
\[
f'(x) = e^{\sqrt{x} \ln(x)} \left( \frac{\ln(x)}{2\sqrt{x}} + \frac{1}{\sqrt{x}} \right)
\]
Since \( f(x) = e^{\sqrt{x} \ln(x)} = x^{\sqrt{x}} \), we can write:
\[
f'(x) = x^{\sqrt{x}} \left( \frac{\ln(x)}{2\sqrt{x}} + \frac{1}{\sqrt{x}} \right)
\]

### Step 5: Evaluate at \( x = 9 \)
Now substitute \( x = 9 \) into the derivative:
- \( \ln(9) = 2 \ln(3) \)
- \( \sqrt{9} = 3 \)
- \( 9^{\sqrt{9}} = 9^3 = 729 \)

Thus:
\[
f'(9) = 729 \left( \frac{\ln(9)}{2 \cdot 3} + \frac{1}{3} \right) = 729 \left( \frac{2 \ln(3)}{6} + \frac{1}{3} \right)
\]

This simplifies further to:
\[
f'(9) = 729 \left( \frac{\ln(3)}{3} + \frac{1}{3} \right) = 729 \cdot \frac{\ln(3) + 1}{3}
\]

Finally:
\[
f'(9) = 243 (\ln(3) + 1)
\]

Thus, the derivative at \( x = 9 \) is:
\[
f'(9) = 243 (\ln(3) + 1)
\]

Let me know if you would like further clarifications or details!

Here are five related questions:
1. How do we simplify expressions involving logarithms and powers?
2. What is the product rule for derivatives and how is it applied?
3. Can you differentiate functions with logarithmic and power terms using implicit differentiation?
4. What are some common derivative rules for exponential functions?
5. How would you evaluate logarithmic functions for specific values?

**Tip**: Always break down complex expressions into simpler components before differentiating, especially when logarithms or powers are involved.

Learn how to differentiate f(x) = x^√x and find f'(9) using logarithmic differentiation, chain rule, and product rule. This step-by-step solution covers exponential and logarithmic concepts, suitable for advanced high school or college-level calculus students.

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