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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![### Finding the Derivative of a Function
#### Problem Statement:
Compute the derivative of the given function:
\[ g(x) = \sin(x^5 + 3x - 1) \]
#### Solution:
To find the derivative of \( g(x) \), we need to apply the chain rule. The chain rule states that if we have a composition of functions \( h(u(x)) \), then the derivative \( h'(u(x)) \cdot u'(x) \).
1. Let \( u = x^5 + 3x - 1 \).
2. Then the original function \( g(x) = \sin(u) \).
We need to find:
\[ g'(x) = \frac{d}{dx} \sin(u) \cdot \frac{d}{dx} u \]
First, compute the derivative of \( u \):
\[ u = x^5 + 3x - 1 \]
\[ \frac{d}{dx} u = 5x^4 + 3 \]
Next, compute the derivative of \( \sin(u) \):
\[ \frac{d}{dx} \sin(u) = \cos(u) \cdot \frac{d}{dx} u \]
Putting it all together, we get:
\[ g'(x) = \cos(x^5 + 3x - 1) \cdot (5x^4 + 3) \]
Thus, the derivative of the function \( g(x) = \sin(x^5 + 3x - 1) \) is:
\[ g'(x) = \cos(x^5 + 3x - 1) \cdot (5x^4 + 3) \]
### Explanation:
- **Step 1**: Identify the inner function \( u \) and the outer function \( \sin(u) \).
- **Step 2**: Use the chain rule to differentiate the outer function \( \sin(u) \) and then multiply it by the derivative of the inner function \( u \).
This problem illustrates the application of the chain rule in calculus to find the derivative of a composite function.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F129dcb8f-8f8c-4338-8228-2ca9a753b007%2F3d911814-f5ba-4cf4-9aeb-8a488d1bea33%2Fhs4gr1e_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Finding the Derivative of a Function
#### Problem Statement:
Compute the derivative of the given function:
\[ g(x) = \sin(x^5 + 3x - 1) \]
#### Solution:
To find the derivative of \( g(x) \), we need to apply the chain rule. The chain rule states that if we have a composition of functions \( h(u(x)) \), then the derivative \( h'(u(x)) \cdot u'(x) \).
1. Let \( u = x^5 + 3x - 1 \).
2. Then the original function \( g(x) = \sin(u) \).
We need to find:
\[ g'(x) = \frac{d}{dx} \sin(u) \cdot \frac{d}{dx} u \]
First, compute the derivative of \( u \):
\[ u = x^5 + 3x - 1 \]
\[ \frac{d}{dx} u = 5x^4 + 3 \]
Next, compute the derivative of \( \sin(u) \):
\[ \frac{d}{dx} \sin(u) = \cos(u) \cdot \frac{d}{dx} u \]
Putting it all together, we get:
\[ g'(x) = \cos(x^5 + 3x - 1) \cdot (5x^4 + 3) \]
Thus, the derivative of the function \( g(x) = \sin(x^5 + 3x - 1) \) is:
\[ g'(x) = \cos(x^5 + 3x - 1) \cdot (5x^4 + 3) \]
### Explanation:
- **Step 1**: Identify the inner function \( u \) and the outer function \( \sin(u) \).
- **Step 2**: Use the chain rule to differentiate the outer function \( \sin(u) \) and then multiply it by the derivative of the inner function \( u \).
This problem illustrates the application of the chain rule in calculus to find the derivative of a composite function.
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