To solve for the derivative of the product and quotient of the functions $f$ and $g$ at $x = 5$, we will use the following differentiation rules:

1. Product Rule for $(fg)'$:
   $(fg)' = f'g + fg'$

2. Quotient Rule for $\left(\frac{f}{g}\right)'$:
   $\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}$

Now let's solve each part:

(a) $(fg)'(5)$

Using the Product Rule: $(fg)'(5) = f'(5)g(5) + f(5)g'(5)$ Substitute the given values: $(fg)'(5) = 7(-4) + 1(5) = -28 + 5 = -23$

Thus, $(fg)'(5) = -23$.

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

Using the Quotient Rule: $\left(\frac{f}{g}\right)'(5) = \frac{f'(5)g(5) - f(5)g'(5)}{g(5)^2}$ Substitute the given values: $\left(\frac{f}{g}\right)'(5) = \frac{7(-4) - 1(5)}{(-4)^2} = \frac{-28 - 5}{16} = \frac{-33}{16}$

Thus, $\left(\frac{f}{g}\right)'(5) = \frac{-33}{16}$.

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

Using the Quotient Rule again for $\left(\frac{g}{f}\right)'$: $\left(\frac{g}{f}\right)'(5) = \frac{g'(5)f(5) - g(5)f'(5)}{f(5)^2}$ Substitute the given values: $\left(\frac{g}{f}\right)'(5) = \frac{5(1) - (-4)(7)}{(1)^2} = \frac{5 + 28}{1} = 33$

Thus, $\left(\frac{g}{f}\right)'(5) = 33$.

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Final Answers:

(a) $(fg)'(5) = -23$

(b) $\left(\frac{f}{g}\right)'(5) = \frac{-33}{16}$

(c) $\left(\frac{g}{f}\right)'(5) = 33$

Let me know if you want further details on any step. Here are some follow-up questions to practice related concepts:

1. What is the product rule for three functions?
2. How do you derive the quotient rule?
3. What happens to the quotient rule if $g(5) = 0$?
4. Can you apply the chain rule with these functions? If so, how?
5. How would these derivatives change if we had higher-order derivatives?

Tip: The product rule is useful even when dealing with more complex combinations of functions, like compositions or multi-variable cases.