The problem you uploaded asks to find the derivative of the function:

$f(x) = x^{\sqrt{x}}$

and evaluate it at $x = 9$.

Step-by-step solution:

1. Rewrite the function using logarithms: $f(x) = x^{\sqrt{x}} = e^{\ln(x^{\sqrt{x}})} = e^{\sqrt{x} \ln(x)}$

   So now we have $f(x) = e^{\sqrt{x} \ln(x)}$.

2. Differentiate using the chain rule: Let $u(x) = \sqrt{x} \ln(x)$, then $f(x) = e^{u(x)}$.

   The derivative of $f(x)$ is: $f'(x) = e^{u(x)} \cdot u'(x)$

   Now, let's differentiate $u(x)$: $u(x) = \sqrt{x} \ln(x) = x^{1/2} \ln(x)$

   To differentiate $u(x)$, use the product rule: $u'(x) = \frac{d}{dx}\left( x^{1/2} \ln(x) \right) = \left( \frac{1}{2}x^{-1/2} \ln(x) \right) + \left( x^{1/2} \cdot \frac{1}{x} \right)$ Simplifying: $u'(x) = \frac{1}{2\sqrt{x}} \ln(x) + \frac{1}{\sqrt{x}}$

3. Substitute back into the derivative of $f(x)$: $f'(x) = e^{\sqrt{x} \ln(x)} \cdot \left( \frac{1}{2\sqrt{x}} \ln(x) + \frac{1}{\sqrt{x}} \right)$

4. Evaluate at $x = 9$: First, calculate the components at $x = 9$: $\sqrt{9} = 3, \quad \ln(9) = 2 \ln(3)$ So: $u(9) = 3 \cdot 2 \ln(3) = 6 \ln(3)$ $e^{u(9)} = e^{6 \ln(3)} = 3^6 = 729$ Now, calculate $u'(9)$: $u'(9) = \frac{1}{6} \cdot 2 \ln(3) + \frac{1}{3} = \frac{\ln(3)}{3} + \frac{1}{3}$ Therefore: $f'(9) = 729 \left( \frac{\ln(3)}{3} + \frac{1}{3} \right) = 729 \cdot \frac{\ln(3) + 1}{3}$ Simplify: $f'(9) = 243 (\ln(3) + 1)$

Final answer:

$f'(9) = 243 (\ln(3) + 1)$

Would you like any further details or clarifications?

Follow-up questions:

1. How does the product rule apply in this situation, and how does it simplify the derivative?
2. What is the chain rule, and why is it used when differentiating $e^{\sqrt{x} \ln(x)}$?
3. Can you explain the logarithmic property $\ln(x^{a}) = a \ln(x)$ and how it helps simplify the function?
4. How can we generalize this approach for other exponentials of the form $x^{g(x)}$?
5. How does knowing the natural logarithm of common numbers like $\ln(3)$ aid in solving problems like this?

Tip:

Whenever you encounter functions with exponents involving variables, applying logarithmic differentiation often simplifies the process of finding the derivative.