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We are tasked with finding \( g'(4) \), given the following information:
- \( f(4) = -1 \)
- \( f'(4) = 7 \)
- \( g(x) = \sqrt{x} f(x) \)

To solve for \( g'(4) \), we need to differentiate \( g(x) = \sqrt{x} f(x) \) using the product rule since \( g(x) \) is a product of two functions: \( \sqrt{x} \) and \( f(x) \).

### Differentiation:
Let \( g(x) = \sqrt{x} f(x) = x^{1/2} f(x) \).

Using the product rule:
\[
g'(x) = \frac{d}{dx} \left( \sqrt{x} \right) f(x) + \sqrt{x} \frac{d}{dx} \left( f(x) \right)
\]
1. Derivative of \( \sqrt{x} = x^{1/2} \) is:
\[
\frac{d}{dx} \left( \sqrt{x} \right) = \frac{1}{2} x^{-1/2}
\]
2. The derivative of \( f(x) \) is \( f'(x) \).

Now applying the product rule:
\[
g'(x) = \frac{1}{2} x^{-1/2} f(x) + \sqrt{x} f'(x)
\]

### Find \( g'(4) \):
Substitute \( x = 4 \), \( f(4) = -1 \), and \( f'(4) = 7 \):
\[
g'(4) = \frac{1}{2} (4)^{-1/2} (-1) + \sqrt{4} (7)
\]
\[
g'(4) = \frac{1}{2} \times \frac{1}{2} \times (-1) + 2 \times 7
\]
\[
g'(4) = -\frac{1}{4} + 14
\]
\[
g'(4) = 13.75
\]

So, \( g'(4) = 13.75 \).

Would you like more details or have any questions?

Here are five related questions:
1. How does the product rule work in differentiation?
2. What is the significance of \( f'(x) \) in calculus?
3. How would the answer change if \( g(x) \) involved a different function of \( x \)?
4. What other rules of differentiation could be useful here?
5. How would you compute higher-order derivatives of \( g(x) \)?

**Tip:** When applying the product rule, make sure to differentiate each part separately and sum them correctly.

Find g'(4) given that f(4) = -1, f'(4) = 7, and g(x) = √x f(x).

Learn how to find the derivative g'(4) for the function g(x) = √x f(x) using the product rule and given values of f(4) = -1 and f'(4) = 7. This problem demonstrates key calculus concepts such as the product rule and differentiation of square root functions, suitable for high school calculus students.

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