According to a web page on the internet, the derivative of a pro derivatives: in other words, (fg)'(x) = f'(x)g'(x)` %3D Let f(x) = 8 x + 7', g(x) = 3 x + 4. f(x) = Preview g'(x) = 3 o Preview 'f(x) g'(x) = 24 o Preview (fg)(x)` = 24x^2+53x+28 %3D Preview (fg)'(x)` = 24x+53 X Preview
According to a web page on the internet, the derivative of a pro derivatives: in other words, (fg)'(x) = f'(x)g'(x)` %3D Let f(x) = 8 x + 7', g(x) = 3 x + 4. f(x) = Preview g'(x) = 3 o Preview 'f(x) g'(x) = 24 o Preview (fg)(x)` = 24x^2+53x+28 %3D Preview (fg)'(x)` = 24x+53 X Preview
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Can you help me with this the last part where I got wrong
![**Understanding the Derivative of a Product**
According to a web page on the internet, the derivative of a product is not simply the product of the derivatives. In mathematical notation, this is represented as:
\((fg)'(x) = f'(x)g(x) + f(x)g'(x)\).
In this exercise, we examine this concept using two functions, \( f(x) \) and \( g(x) \).
### Given Functions:
- \( f(x) = 8x + 7 \)
- \( g(x) = 3x + 4 \)
### Steps and Calculations:
1. **Calculate the derivative of \( f(x) \):**
\[ f'(x) = 8 \]
*(Correct as indicated by the checkmark.)*
2. **Calculate the derivative of \( g(x) \):**
\[ g'(x) = 3 \]
*(Correct as indicated by the checkmark.)*
3. **Calculate the product of the derivatives: \( f'(x) \cdot g'(x) \):**
\[ 8 \cdot 3 = 24 \]
*(Correct as indicated by the checkmark.)*
4. **Calculate the derivative of the product \( (fg)'(x) \) using the product rule:**
Substitute into the formula:
\[ (fg)'(x) = f'(x)g(x) + f(x)g'(x) \]
\[ = 8(3x + 4) + (8x + 7)(3) \]
\[ = 24x + 32 + 24x + 21 \]
\[ = 24x^2 + 53x + 28 \]
*(Correct as indicated by the checkmark.)*
5. **Verify if the student-calculated \((fg)'(x)\) is correct:**
The student wrote \( (fg)'(x) = 24x + 53 \). This is incorrect, as shown by the cross mark.
### Conclusion:
The correct application of the derivative of a product rule shows that the expression \((fg)'(x) = 24x^2 + 53x + 28\) is verified as correct. It emphasizes that the derivative of a product is not simply the product of the derivatives but must take into account](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2a9f1f77-996d-4872-99b9-7512aca31a68%2F3ad001be-579c-498f-846c-73fc32fdd84e%2Fy4myg9j_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Understanding the Derivative of a Product**
According to a web page on the internet, the derivative of a product is not simply the product of the derivatives. In mathematical notation, this is represented as:
\((fg)'(x) = f'(x)g(x) + f(x)g'(x)\).
In this exercise, we examine this concept using two functions, \( f(x) \) and \( g(x) \).
### Given Functions:
- \( f(x) = 8x + 7 \)
- \( g(x) = 3x + 4 \)
### Steps and Calculations:
1. **Calculate the derivative of \( f(x) \):**
\[ f'(x) = 8 \]
*(Correct as indicated by the checkmark.)*
2. **Calculate the derivative of \( g(x) \):**
\[ g'(x) = 3 \]
*(Correct as indicated by the checkmark.)*
3. **Calculate the product of the derivatives: \( f'(x) \cdot g'(x) \):**
\[ 8 \cdot 3 = 24 \]
*(Correct as indicated by the checkmark.)*
4. **Calculate the derivative of the product \( (fg)'(x) \) using the product rule:**
Substitute into the formula:
\[ (fg)'(x) = f'(x)g(x) + f(x)g'(x) \]
\[ = 8(3x + 4) + (8x + 7)(3) \]
\[ = 24x + 32 + 24x + 21 \]
\[ = 24x^2 + 53x + 28 \]
*(Correct as indicated by the checkmark.)*
5. **Verify if the student-calculated \((fg)'(x)\) is correct:**
The student wrote \( (fg)'(x) = 24x + 53 \). This is incorrect, as shown by the cross mark.
### Conclusion:
The correct application of the derivative of a product rule shows that the expression \((fg)'(x) = 24x^2 + 53x + 28\) is verified as correct. It emphasizes that the derivative of a product is not simply the product of the derivatives but must take into account
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