Let's work through the given problem using the information provided:

We are given:

• $f(5) = 1$
• $f'(5) = 6$
• $g(5) = -3$
• $g'(5) = 2$

We are asked to find the following derivatives:

(a) $(f + g)'(5)$

By the sum rule of differentiation: $(f + g)'(x) = f'(x) + g'(x)$ Thus, at $x = 5$: $(f + g)'(5) = f'(5) + g'(5) = 6 + 2 = 8$

(b) $(fg)'(5)$

By the product rule of differentiation: $(fg)'(x) = f'(x)g(x) + f(x)g'(x)$ Thus, at $x = 5$: $(fg)'(5) = f'(5)g(5) + f(5)g'(5) = 6(-3) + 1(2) = -18 + 2 = -16$

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

By the quotient rule of differentiation: $\left( \frac{f}{g} \right)'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}$ Thus, at $x = 5$: $\left( \frac{f}{g} \right)'(5) = \frac{f'(5)g(5) - f(5)g'(5)}{g(5)^2} = \frac{6(-3) - 1(2)}{(-3)^2} = \frac{-18 - 2}{9} = \frac{-20}{9}$

(d) $\left( \frac{g}{f} \right)'(5)$

Again, by the quotient rule of differentiation: $\left( \frac{g}{f} \right)'(x) = \frac{g'(x)f(x) - g(x)f'(x)}{f(x)^2}$ Thus, at $x = 5$: $\left( \frac{g}{f} \right)'(5) = \frac{g'(5)f(5) - g(5)f'(5)}{f(5)^2} = \frac{2(1) - (-3)(6)}{1^2} = \frac{2 + 18}{1} = 20$

Summary of Answers:

• (a) $(f + g)'(5) = 8$
• (b) $(fg)'(5) = -16$
• (c) $\left( \frac{f}{g} \right)'(5) = \frac{-20}{9}$
• (d) $\left( \frac{g}{f} \right)'(5) = 20$

Would you like more details on any of these steps?

Here are 5 related questions to further explore the concepts:

1. How does the product rule differ from the sum rule in calculus?
2. Can you apply the quotient rule for higher-order derivatives?
3. How would these results change if $f(5)$ and $g(5)$ were both zero?
4. What is the general form of the product rule for three functions?
5. How would you differentiate a function raised to a power, such as $(f(x)^2)$?

Tip: Always check if the denominator in the quotient rule is zero to avoid undefined values.