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We are given the function \( k(x) = \frac{f(x)g(x)}{h(x)} \), and we need to find \( k'(-3) \). To do this, we will apply the quotient rule for differentiation.

### Step 1: Quotient Rule
The derivative of \( k(x) = \frac{f(x)g(x)}{h(x)} \) is given by the quotient rule:
\[
k'(x) = \frac{(f'(x)g(x) + f(x)g'(x))h(x) - f(x)g(x)h'(x)}{(h(x))^2}
\]
Now, substitute \( x = -3 \) and use the given values:
- \( f(-3) = -1 \)
- \( f'(-3) = -5 \)
- \( g(-3) = 8 \)
- \( g'(-3) = 5 \)
- \( h(-3) = -2 \)
- \( h'(-3) = -8 \)

### Step 2: Plug in the values into the derivative formula
Substitute the given values into the formula for \( k'(-3) \):
\[
k'(-3) = \frac{(f'(-3)g(-3) + f(-3)g'(-3))h(-3) - f(-3)g(-3)h'(-3)}{(h(-3))^2}
\]
Substitute the specific values:
\[
k'(-3) = \frac{((-5)(8) + (-1)(5))(-2) - (-1)(8)(-8)}{(-2)^2}
\]

### Step 3: Simplify the expression
First, simplify the terms inside the numerator:
\[
k'(-3) = \frac{((-40) + (-5))(-2) - (8)(8)}{4}
\]
\[
k'(-3) = \frac{(-45)(-2) - 64}{4}
\]
\[
k'(-3) = \frac{90 - 64}{4}
\]
\[
k'(-3) = \frac{26}{4} = 6.5
\]

Thus, \( k'(-3) = 6.5 \).

Let me know if you need further details or clarifications! Would you like me to expand on any part of the process?

Here are 5 related questions for further practice:
1. How do you apply the product rule in combination with the quotient rule?
2. How would the result change if \( h'(-3) \) had a different value?
3. What happens if \( h(x) \) is 0 at some point in the domain of \( k(x) \)?
4. How would you differentiate if \( k(x) = \frac{f(x)}{g(x)} \)?
5. How does the chain rule factor into more complex quotient rule problems?

**Tip:** When applying the quotient rule, be careful to apply it in the correct order, ensuring the subtraction in the numerator is maintained properly.

Let k(x) = (f(x)g(x)) / h(x). If f(-3) = -1, f'(-3) = -5, g(-3) = 8, g'(-3) = 5, h(-3) = -2, and h'(-3) = -8, what is k'(-3)?

Learn how to apply the quotient rule to find the derivative of the function k(x) = (f(x)g(x)) / h(x) at x = -3. This step-by-step solution demonstrates the substitution of values and careful simplification, focusing on calculus concepts like the quotient rule. Suitable for high school or college calculus students.

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