2) DIfferenticate fcyz tan (24). Simplify aswer.

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.
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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Differentiating the function

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