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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### Example Problem: Calculus and Function Composition
**Problem Statement:**
If \( f(x) = 5x - 1 \), \( g(0) = 2 \), \( g'(0) = 4 \), and \( h(x) = f(g(x)) \), then what is \( h'(0) \)?
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### Explanation:
To solve this problem, we need to use the chain rule for differentiation. The chain rule states that if we have two functions \( u(x) \) and \( v(x) \), and if \( h(x) = u(v(x)) \), then the derivative of \( h(x) \) is given by:
\[ h'(x) = u'(v(x)) \cdot v'(x) \]
Here, \( h(x) = f(g(x)) \). So, let's identify \( u(x) \) and \( v(x) \):
- \( u(x) = f(x) \)
- \( v(x) = g(x) \)
First, we need to find \( f'(x) \):
Given \( f(x) = 5x - 1 \), we differentiate it with respect to \( x \):
\[ f'(x) = 5 \]
Next, using the chain rule, we find \( h'(x) \):
\[ h'(x) = f'(g(x)) \cdot g'(x) \]
We need to find \( h'(0) \), which means we need to evaluate at \( x = 0 \). Therefore:
\[ h'(0) = f'(g(0)) \cdot g'(0) \]
We are given \( g(0) = 2 \) and \( g'(0) = 4 \). Substitute these values into the equation:
\[ h'(0) = f'(2) \cdot 4 \]
Since \( f'(x) = 5 \) (as it is a constant),
\[ f'(2) = 5 \]
Therefore:
\[ h'(0) = 5 \cdot 4 = 20 \]
So, \( h'(0) = 20 \).
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This problem demonstrates the application of the chain rule in calculus for differentiating composite functions. Understanding and applying such rules are crucial for solving many real-world problems involving rates of change.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1011249f-80d3-45e3-a646-295dd331971a%2Fa2e1dcfe-6c62-4801-a321-2d9cf2196c2b%2Fe9coije_processed.jpeg&w=3840&q=75)
Transcribed Image Text:---
### Example Problem: Calculus and Function Composition
**Problem Statement:**
If \( f(x) = 5x - 1 \), \( g(0) = 2 \), \( g'(0) = 4 \), and \( h(x) = f(g(x)) \), then what is \( h'(0) \)?
---
### Explanation:
To solve this problem, we need to use the chain rule for differentiation. The chain rule states that if we have two functions \( u(x) \) and \( v(x) \), and if \( h(x) = u(v(x)) \), then the derivative of \( h(x) \) is given by:
\[ h'(x) = u'(v(x)) \cdot v'(x) \]
Here, \( h(x) = f(g(x)) \). So, let's identify \( u(x) \) and \( v(x) \):
- \( u(x) = f(x) \)
- \( v(x) = g(x) \)
First, we need to find \( f'(x) \):
Given \( f(x) = 5x - 1 \), we differentiate it with respect to \( x \):
\[ f'(x) = 5 \]
Next, using the chain rule, we find \( h'(x) \):
\[ h'(x) = f'(g(x)) \cdot g'(x) \]
We need to find \( h'(0) \), which means we need to evaluate at \( x = 0 \). Therefore:
\[ h'(0) = f'(g(0)) \cdot g'(0) \]
We are given \( g(0) = 2 \) and \( g'(0) = 4 \). Substitute these values into the equation:
\[ h'(0) = f'(2) \cdot 4 \]
Since \( f'(x) = 5 \) (as it is a constant),
\[ f'(2) = 5 \]
Therefore:
\[ h'(0) = 5 \cdot 4 = 20 \]
So, \( h'(0) = 20 \).
---
This problem demonstrates the application of the chain rule in calculus for differentiating composite functions. Understanding and applying such rules are crucial for solving many real-world problems involving rates of change.
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