Use the information about values of the functions and their derivatives to calculate 𝐻′(7),H′(7), where 𝐻(𝑥)=𝑥𝑔(𝑥)𝑓(𝑥).H(x)=xg(x)f(x).
𝑓(7)f(7) 	𝑓′(7)f′(7) 	𝑔(7)g(7) 	𝑔′(7)g′(7)
33 	−5−5 	77 	−4−4

(Give an exact answer. Use symbolic notation and fractions where needed.)

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.

Use the information about values of the functions and their derivatives to calculate H'(7), where H(x) = xg(x)f(x). f(7) = 33, f'(7) = -5, g(7) = 7, g'(7) = -4. (Give an exact answer. Use symbolic notation and fractions where needed.)

Learn how to calculate H'(7) for H(x) = xg(x)f(x) using the product rule of differentiation. This problem involves given values of functions and their derivatives. Detailed step-by-step explanation suitable for students in Grades 11-12 studying advanced calculus.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation