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To solve this, we can use the provided values for \( f(5) \), \( f'(5) \), \( g(5) \), and \( g'(5) \) and apply product, quotient, and chain rules as required for each part:

Given values:
- \( f(5) = 1 \)
- \( f'(5) = 5 \)
- \( g(5) = -6 \)
- \( g'(5) = 7 \)

### (a) \( (fg)'(5) \)

For the derivative of the product \( (fg)(x) = f(x) \cdot g(x) \), we use the **product rule**:
\[
(fg)'(x) = f'(x)g(x) + f(x)g'(x)
\]

Substituting \( x = 5 \):
\[
(fg)'(5) = f'(5) \cdot g(5) + f(5) \cdot g'(5)
\]
\[
= (5)(-6) + (1)(7)
\]
\[
= -30 + 7 = -23
\]

So, \( (fg)'(5) = -23 \).

---

### (b) \( \left( \frac{f}{g} \right)'(5) \)

For the derivative of the quotient \( \left( \frac{f}{g} \right)(x) = \frac{f(x)}{g(x)} \), we use the **quotient rule**:
\[
\left( \frac{f}{g} \right)'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}
\]

Substituting \( x = 5 \):
\[
\left( \frac{f}{g} \right)'(5) = \frac{f'(5) \cdot g(5) - f(5) \cdot g'(5)}{(g(5))^2}
\]
\[
= \frac{(5)(-6) - (1)(7)}{(-6)^2}
\]
\[
= \frac{-30 - 7}{36}
\]
\[
= \frac{-37}{36}
\]

So, \( \left( \frac{f}{g} \right)'(5) = -\frac{37}{36} \).

---

### (c) \( \left( \frac{g}{f} \right)'(5) \)

Similarly, we use the **quotient rule** for \( \left( \frac{g}{f} \right)(x) = \frac{g(x)}{f(x)} \):
\[
\left( \frac{g}{f} \right)'(x) = \frac{g'(x)f(x) - g(x)f'(x)}{(f(x))^2}
\]

Substituting \( x = 5 \):
\[
\left( \frac{g}{f} \right)'(5) = \frac{g'(5) \cdot f(5) - g(5) \cdot f'(5)}{(f(5))^2}
\]
\[
= \frac{(7)(1) - (-6)(5)}{(1)^2}
\]
\[
= \frac{7 + 30}{1}
\]
\[
= 37
\]

So, \( \left( \frac{g}{f} \right)'(5) = 37 \).

---

### Summary of Answers
- (a) \( (fg)'(5) = -23 \)
- (b) \( \left( \frac{f}{g} \right)'(5) = -\frac{37}{36} \)
- (c) \( \left( \frac{g}{f} \right)'(5) = 37 \)

Would you like further explanations on any part?

### Additional Related Questions
1. What is the product rule and how is it used in derivatives?
2. Can you explain the quotient rule and its application?
3. How do we interpret the derivative of a function at a given point?
4. How would these values change if \( f(5) \) or \( g(5) \) were zero?
5. What would the result be for a similar setup if we had \( f(x) = x^2 \) and \( g(x) = \sin(x) \) at a specific point?

### Tip
When working with derivatives of products or quotients, ensure all substitutions are accurate, as small errors can significantly change the results.

Suppose that f(5) = 1, f'(5) = 5, g(5) = -6, and g'(5) = 7. Find the following values: (a) (fg)'(5), (b) (f/g)'(5), (c) (g/f)'(5).

Learn how to find derivatives using the product and quotient rules for functions f and g at a specific point. Given f(5) = 1, f'(5) = 5, g(5) = -6, and g'(5) = 7, find (fg)'(5), (f/g)'(5), and (g/f)'(5). Step-by-step solutions provided.

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