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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![**Problem Statement:**
2) Differentiate \( f(y) = \tan(2y) \).
Simplify answer.
\[ f'(y) = \]
**Solution Explanation:**
To differentiate the function \( f(y) = \tan(2y) \), we need to use the chain rule of differentiation. Here's how it's done:
1. **Differentiate the outer function**: The derivative of \(\tan(u)\) with respect to \(u\) is \(\sec^2(u)\). Here, \(u = 2y\).
2. **Differentiate the inner function**: The derivative of \(2y\) with respect to \(y\) is 2.
3. **Apply the chain rule**: Multiply the derivative of the outer function by the derivative of the inner function:
\[ f'(y) = \sec^2(2y) \cdot 2 \]
4. **Simplify the expression**:
\[ f'(y) = 2\sec^2(2y) \]
This gives us the simplified derivative of the function.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0f39937e-08d5-4dd9-9b09-d4f7e1a5c538%2Fce3b921d-a033-409e-98f8-a9a962bfb4a0%2Fx0uz72q_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
2) Differentiate \( f(y) = \tan(2y) \).
Simplify answer.
\[ f'(y) = \]
**Solution Explanation:**
To differentiate the function \( f(y) = \tan(2y) \), we need to use the chain rule of differentiation. Here's how it's done:
1. **Differentiate the outer function**: The derivative of \(\tan(u)\) with respect to \(u\) is \(\sec^2(u)\). Here, \(u = 2y\).
2. **Differentiate the inner function**: The derivative of \(2y\) with respect to \(y\) is 2.
3. **Apply the chain rule**: Multiply the derivative of the outer function by the derivative of the inner function:
\[ f'(y) = \sec^2(2y) \cdot 2 \]
4. **Simplify the expression**:
\[ f'(y) = 2\sec^2(2y) \]
This gives us the simplified derivative of the function.
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