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.