Differentiate f (x) = cos (In x) e“.

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:**

Differentiate the function \( f(x) = \cos(\ln x) \cdot e^x \).

**Solution:**

To solve this problem, we need to apply the product rule of differentiation. The product rule states that if you have a function \( f(x) = u(x) \cdot v(x) \), then the derivative \( f'(x) \) is given by:

\[ f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) \]

**Step-by-Step Explanation:**

1. **Identify \( u(x) \) and \( v(x) \):**

   Let \( u(x) = \cos(\ln x) \) and \( v(x) = e^x \).

2. **Find \( u'(x) \):**

   - The derivative of \( \cos(\ln x) \) with respect to \( x \) requires the chain rule.
   - Let \( g(x) = \ln x \), then \( u(x) = \cos(g(x)) \).
   - The derivative \( u'(x) = -\sin(\ln x) \cdot \frac{d}{dx}[g(x)] \).
   - Since \( \frac{d}{dx}[\ln x] = \frac{1}{x} \), we have \( u'(x) = -\sin(\ln x) \cdot \frac{1}{x} \).

3. **Find \( v'(x) \):**

   - The derivative of \( e^x \) is \( e^x \).

4. **Apply the Product Rule:**

   Substitute \( u(x) \), \( u'(x) \), \( v(x) \), and \( v'(x) \) into the product rule formula:

   \[
   f'(x) = (-\sin(\ln x) \cdot \frac{1}{x}) \cdot e^x + \cos(\ln x) \cdot e^x
   \]

5. **Simplify the Expression:**

   - Combine the terms:
   
   \[
   f'(x) = -\sin(\ln x) \cdot \frac{e^x}{x} + \cos(\ln x) \cdot e^x
   \]

This
Transcribed Image Text:**Problem Statement:** Differentiate the function \( f(x) = \cos(\ln x) \cdot e^x \). **Solution:** To solve this problem, we need to apply the product rule of differentiation. The product rule states that if you have a function \( f(x) = u(x) \cdot v(x) \), then the derivative \( f'(x) \) is given by: \[ f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) \] **Step-by-Step Explanation:** 1. **Identify \( u(x) \) and \( v(x) \):** Let \( u(x) = \cos(\ln x) \) and \( v(x) = e^x \). 2. **Find \( u'(x) \):** - The derivative of \( \cos(\ln x) \) with respect to \( x \) requires the chain rule. - Let \( g(x) = \ln x \), then \( u(x) = \cos(g(x)) \). - The derivative \( u'(x) = -\sin(\ln x) \cdot \frac{d}{dx}[g(x)] \). - Since \( \frac{d}{dx}[\ln x] = \frac{1}{x} \), we have \( u'(x) = -\sin(\ln x) \cdot \frac{1}{x} \). 3. **Find \( v'(x) \):** - The derivative of \( e^x \) is \( e^x \). 4. **Apply the Product Rule:** Substitute \( u(x) \), \( u'(x) \), \( v(x) \), and \( v'(x) \) into the product rule formula: \[ f'(x) = (-\sin(\ln x) \cdot \frac{1}{x}) \cdot e^x + \cos(\ln x) \cdot e^x \] 5. **Simplify the Expression:** - Combine the terms: \[ f'(x) = -\sin(\ln x) \cdot \frac{e^x}{x} + \cos(\ln x) \cdot e^x \] This
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