We are asked to find $H'(7)$, where $H(x) = x g(x) f(x)$.

We are given the following information:

$f(7) = 33, \quad f'(7) = -5, \quad g(7) = 7, \quad g'(7) = -4$

Step 1: Apply the Product Rule

Since $H(x) = x g(x) f(x)$, we need to differentiate this product of three functions. We'll use the product rule.

The derivative of $H(x) = x g(x) f(x)$ can be computed as:

$H'(x) = \frac{d}{dx} \left( x g(x) f(x) \right)$

We apply the product rule:

$H'(x) = \frac{d}{dx}(x) \cdot g(x) f(x) + x \cdot \frac{d}{dx}(g(x) f(x))$

First, the derivative of $x$ is 1, so:

$H'(x) = g(x) f(x) + x \cdot \frac{d}{dx}(g(x) f(x))$

Now, apply the product rule again to $g(x) f(x)$:

$\frac{d}{dx}(g(x) f(x)) = g'(x) f(x) + g(x) f'(x)$

Thus,

$H'(x) = g(x) f(x) + x \cdot \left( g'(x) f(x) + g(x) f'(x) \right)$

Step 2: Substitute the values at $x = 7$

Now substitute the given values $f(7) = 33$, $f'(7) = -5$, $g(7) = 7$, and $g'(7) = -4$ into the expression:

$H'(7) = g(7) f(7) + 7 \cdot \left( g'(7) f(7) + g(7) f'(7) \right)$

Substitute the values:

$H'(7) = 7 \cdot 33 + 7 \cdot \left( (-4) \cdot 33 + 7 \cdot (-5) \right)$

Step 3: Simplify the expression

First, calculate the individual parts:

$7 \cdot 33 = 231$ $(-4) \cdot 33 = -132$ $7 \cdot (-5) = -35$

Now sum the terms inside the parentheses:

$-132 + (-35) = -167$

Multiply by 7:

$7 \cdot (-167) = -1169$

Finally, add to the first term:

$H'(7) = 231 + (-1169) = -938$

Final Answer:

$H'(7) = -938$

Would you like further clarification on any step, or should I break down any details further?

Here are 5 related questions:

1. How does the product rule work for three functions?
2. Can the chain rule be applied in cases like this?
3. What are common mistakes to avoid when using the product rule?
4. How can this problem be solved using Leibniz's notation for differentiation?
5. How does this method extend to implicit differentiation problems?

Tip: Always double-check your arithmetic when applying the product rule to avoid simple mistakes.