To find $h'(x)$ where $h(x) = 1 + f(x)g(x)$, we can use the product rule of differentiation. The product rule states that if $h(x) = f(x)g(x)$, then $h'(x) = f'(x)g(x) + f(x)g'(x)$.

Given that $h(x) = 1 + f(x)g(x)$, the constant 1 has a derivative of 0. Therefore, we only need to differentiate $f(x)g(x)$.

So, $h'(x) = \frac{d}{dx}[1 + f(x)g(x)] = \frac{d}{dx}[f(x)g(x)]$.

Applying the product rule, we get: $h'(x) = f'(x)g(x) + f(x)g'(x)$

Now, we need to evaluate $h'(x)$ at $x = -3$.

Given: $f(-3) = 6$ $f'(-3) = 4$ $g(-3) = 5$ $g'(-3) = 2$

Substitute these values into the derivative formula: $h'(-3) = f'(-3)g(-3) + f(-3)g'(-3)$ $h'(-3) = (4)(5) + (6)(2)$ $h'(-3) = 20 + 12$ $h'(-3) = 32$

Thus, the value of $h'(-3)$ is $32$.

Answer: $h'(-3) = 32$

Would you like more details or have any questions about this solution?

Here are 5 related questions for further practice:

1. If $k(x) = 2 + f(x)g(x)$, how would you find $k'(-3)$ using the given values of $f$ and $g$ and their derivatives?
2. What is the derivative of $h(x) = f(x)g(x) + x$ at $x = -3$?
3. How would the product rule apply if $h(x) = f(x)g(x) \cdot h(x)$?
4. How can you verify the result of $h'(-3) = 32$ by using numerical differentiation techniques?
5. How would you apply the chain rule to differentiate $h(x) = f(g(x))$?

Tip: When differentiating a product of two functions, remember to apply the product rule carefully and substitute the known values at the specific point to find the derivative at that point.