To find $H'(7)$ where $H(x) = x \cdot g(x) \cdot f(x)$, we will apply the product rule for differentiation.

We have:

$H(x) = x \cdot g(x) \cdot f(x)$

To differentiate $H(x)$, we treat it as the product of three factors: $x$, $g(x)$, and $f(x)$.

Let’s apply the product rule step by step.

Step 1: Product Rule

For the product of three functions $u(x) = x$, $v(x) = g(x)$, and $w(x) = f(x)$, the derivative $H'(x)$ will be:

$H'(x) = \frac{d}{dx} \left( x \cdot g(x) \cdot f(x) \right)$ Using the product rule for three functions, this expands as:

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

Since $\frac{d}{dx}(x) = 1$, this simplifies to:

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

Step 2: Substitute Values

We are given the following information:

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

Substituting these values into the expression for $H'(7)$:

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

Now substitute the specific values:

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

Step 3: Simplify

Now, calculate each term:

$7 \cdot 33 = 231$ $7 \cdot (-4) \cdot 33 = -924$ $7 \cdot 7 \cdot (-5) = -245$

Now, sum them together:

$H'(7) = 231 + (-924) + (-245)$

$H'(7) = 231 - 924 - 245 = -938$

Thus, the exact value of $H'(7)$ is:

$H'(7) = -938$

---

Would you like more details or have any questions?

Here are 5 related questions:

1. How does the product rule change when differentiating more than two functions?
2. Can the product rule be extended to a chain of four or more functions?
3. How would you approach differentiating a product that includes exponential functions?
4. Why is the derivative of $x$ simply 1 in this context?
5. How can you apply the product rule to functions involving trigonometric identities?

Tip: When applying the product rule, make sure to keep track of the derivatives of each factor separately to avoid mistakes in complex functions.