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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#10 can you show me what I’ve done wrong? I’ve attached my work so you can see how my teacher is teaching this.
![**Find the derivative of the function using the definition of derivative.**
Given function:
\[ f(x) = x^3 - 5x + 9 \]
Incorrect attempt at derivative:
\[ f'(x) = 3x - 5 \] (marked with a red cross)
**Additional Materials:**
- [eBook](#) (represented by an icon of a book)
The task is to find the derivative of the function \( f(x) = x^3 - 5x + 9 \) correctly using the definition of the derivative.
The incorrect attempt \( f'(x) = 3x - 5 \) does not account for the derivative of each term properly. Using the rules of differentiation, the correct derivative should be calculated.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5f440a4a-a87c-43e8-9c30-e4890b6a6760%2F4cd1d7c3-e7bc-4dc1-9153-fa84965c2157%2Fc2ahdv_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Find the derivative of the function using the definition of derivative.**
Given function:
\[ f(x) = x^3 - 5x + 9 \]
Incorrect attempt at derivative:
\[ f'(x) = 3x - 5 \] (marked with a red cross)
**Additional Materials:**
- [eBook](#) (represented by an icon of a book)
The task is to find the derivative of the function \( f(x) = x^3 - 5x + 9 \) correctly using the definition of the derivative.
The incorrect attempt \( f'(x) = 3x - 5 \) does not account for the derivative of each term properly. Using the rules of differentiation, the correct derivative should be calculated.
![### Calculus: Derivative Calculation
#### Problem 9
To find the derivative of a given function, we have:
Given the function:
\[ f(x) = 5(x+h)^2 - 5x^2 - 8 \]
Using the limit definition of a derivative:
\[ f'(x) = \lim_{{h \to 0}} \frac{{5(x^2 + 2xh + h^2) - 8 - (5x^2 - 8)}}{h} \]
Simplifying, we get:
\[ = \lim_{{h \to 0}} \frac{{5x^2 + 10xh + 5h^2 - 8 - 5x^2 + 8}}{h} \]
\[ = \lim_{{h \to 0}} \frac{{10xh + 5h^2}}{h} \]
Cancel out \( h \):
\[ = \lim_{{h \to 0}} (10x + 5h) \]
As \( h \to 0 \), we find:
\[ f'(x) = 10x \]
#### Problem 10
Given the function:
\[ f(x) = x^3 - 5x + 19 \]
To find the derivative \( f'(x) \), we have:
\[ f'(x) = \lim_{{h \to 0}} \frac{{(x+h)^3 - 5(x+h) + 19 - (x^3 - 5x + 19)}}{h} \]
Expand and simplify:
\[ (x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3 \]
\[ -5(x+h) = -5x - 5h \]
Combine like terms:
\[ = \lim_{{h \to 0}} \frac{{x^3 + 3x^2h + 3xh^2 + h^3 - 5x - 5h + 19 - x^3 + 5x - 19}}{h} \]
\[ = \lim_{{h \to 0}} \frac{{3x^2h + 3xh^2 + h^3 - 5h}}{h} \]
Simplify by canceling \( h \):
\[ =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5f440a4a-a87c-43e8-9c30-e4890b6a6760%2F4cd1d7c3-e7bc-4dc1-9153-fa84965c2157%2Fuo6ugmr_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Calculus: Derivative Calculation
#### Problem 9
To find the derivative of a given function, we have:
Given the function:
\[ f(x) = 5(x+h)^2 - 5x^2 - 8 \]
Using the limit definition of a derivative:
\[ f'(x) = \lim_{{h \to 0}} \frac{{5(x^2 + 2xh + h^2) - 8 - (5x^2 - 8)}}{h} \]
Simplifying, we get:
\[ = \lim_{{h \to 0}} \frac{{5x^2 + 10xh + 5h^2 - 8 - 5x^2 + 8}}{h} \]
\[ = \lim_{{h \to 0}} \frac{{10xh + 5h^2}}{h} \]
Cancel out \( h \):
\[ = \lim_{{h \to 0}} (10x + 5h) \]
As \( h \to 0 \), we find:
\[ f'(x) = 10x \]
#### Problem 10
Given the function:
\[ f(x) = x^3 - 5x + 19 \]
To find the derivative \( f'(x) \), we have:
\[ f'(x) = \lim_{{h \to 0}} \frac{{(x+h)^3 - 5(x+h) + 19 - (x^3 - 5x + 19)}}{h} \]
Expand and simplify:
\[ (x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3 \]
\[ -5(x+h) = -5x - 5h \]
Combine like terms:
\[ = \lim_{{h \to 0}} \frac{{x^3 + 3x^2h + 3xh^2 + h^3 - 5x - 5h + 19 - x^3 + 5x - 19}}{h} \]
\[ = \lim_{{h \to 0}} \frac{{3x^2h + 3xh^2 + h^3 - 5h}}{h} \]
Simplify by canceling \( h \):
\[ =
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