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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![### Calculus: Finding Derivatives
1. **Find each of the following derivatives.**
a. \( \frac{d}{dx} \left( 4^{\sin x} \right) \)
b. \( \frac{d}{dx} \left[ \ln(2x^4 - 3x) \right] \)
#### Detailed Explanation:
**Part a:**
To find the derivative of \( 4^{\sin x} \), we can use the chain rule and the fact that the derivative of \( a^u \) (where \( a \) is a constant and \( u \) is a function of \( x \)) is given by:
\[ \frac{d}{dx} \left( a^u \right) = a^u \ln(a) \frac{du}{dx} \]
In this case, \( a = 4 \) and \( u = \sin x \). Thus:
\[ \frac{d}{dx} \left( 4^{\sin x} \right) = 4^{\sin x} \ln(4) \cos(x) \]
**Part b:**
To find the derivative of \( \ln(2x^4 - 3x) \), we will use the chain rule again, along with the derivative of the natural logarithm function, which states that \( \frac{d}{dx} \left( \ln(u) \right) = \frac{1}{u} \frac{du}{dx} \).
In this case, \( u = 2x^4 - 3x \). First, we need to find \( \frac{du}{dx} \):
\[ \frac{d}{dx} \left( 2x^4 - 3x \right) = 8x^3 - 3 \]
Then, applying the chain rule, we get:
\[ \frac{d}{dx} \left[ \ln(2x^4 - 3x) \right] = \frac{1}{2x^4 - 3x} (8x^3 - 3) \]
Therefore:
\[ \frac{d}{dx} \left[ \ln(2x^4 - 3x) \right] = \frac{8x^3 - 3}{2x^4 -](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6a885dd9-3ed1-489d-bae2-5a33154ef1ff%2Fde10597f-1415-48bb-875f-a4a4aa69668f%2F1yjaxkz_processed.png&w=3840&q=75)
Transcribed Image Text:### Calculus: Finding Derivatives
1. **Find each of the following derivatives.**
a. \( \frac{d}{dx} \left( 4^{\sin x} \right) \)
b. \( \frac{d}{dx} \left[ \ln(2x^4 - 3x) \right] \)
#### Detailed Explanation:
**Part a:**
To find the derivative of \( 4^{\sin x} \), we can use the chain rule and the fact that the derivative of \( a^u \) (where \( a \) is a constant and \( u \) is a function of \( x \)) is given by:
\[ \frac{d}{dx} \left( a^u \right) = a^u \ln(a) \frac{du}{dx} \]
In this case, \( a = 4 \) and \( u = \sin x \). Thus:
\[ \frac{d}{dx} \left( 4^{\sin x} \right) = 4^{\sin x} \ln(4) \cos(x) \]
**Part b:**
To find the derivative of \( \ln(2x^4 - 3x) \), we will use the chain rule again, along with the derivative of the natural logarithm function, which states that \( \frac{d}{dx} \left( \ln(u) \right) = \frac{1}{u} \frac{du}{dx} \).
In this case, \( u = 2x^4 - 3x \). First, we need to find \( \frac{du}{dx} \):
\[ \frac{d}{dx} \left( 2x^4 - 3x \right) = 8x^3 - 3 \]
Then, applying the chain rule, we get:
\[ \frac{d}{dx} \left[ \ln(2x^4 - 3x) \right] = \frac{1}{2x^4 - 3x} (8x^3 - 3) \]
Therefore:
\[ \frac{d}{dx} \left[ \ln(2x^4 - 3x) \right] = \frac{8x^3 - 3}{2x^4 -
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