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3 Let f(x) = 2 + In(bx), where b is a positive constant. Find f'(x).

3 Let f(x) = 2 + In(bx), where b is a positive constant. Find f'(x).

Calculus: Early Transcendentals
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:** Let \( f(x) = \frac{3}{x} + \ln(bx) \), where \( b \) is a positive constant. Find \( f'(x) \). **Options:** A) \( f'(x) = -\frac{3}{x^2} + \frac{b}{x} \) B) \( f'(x) = \frac{3}{x^2} - \frac{b}{x} \) C) \( f'(x) = -\frac{3}{x^2} + \frac{1}{x} \) D) \( f'(x) = -\frac{3}{x^2} - \frac{1}{x} \) E) \( f'(x) = \frac{3}{x^2} - \frac{b}{x} \) F) Answer not listed **Solution:** The correct answer is option: C) \( f'(x) = -\frac{3}{x^2} + \frac{1}{x} \) **Explanation:** To find the derivative \( f'(x) \), apply the derivatives of \( \frac{3}{x} \) and \( \ln(bx) \): - The derivative of \( \frac{3}{x} \) is \( -\frac{3}{x^2} \). - The derivative of \( \ln(bx) \) is \( \frac{b}{bx} = \frac{1}{x} \). So, \( f'(x) = -\frac{3}{x^2} + \frac{1}{x} \).
Transcribed Image Text:**Problem Statement:** Let \( f(x) = \frac{3}{x} + \ln(bx) \), where \( b \) is a positive constant. Find \( f'(x) \). **Options:** A) \( f'(x) = -\frac{3}{x^2} + \frac{b}{x} \) B) \( f'(x) = \frac{3}{x^2} - \frac{b}{x} \) C) \( f'(x) = -\frac{3}{x^2} + \frac{1}{x} \) D) \( f'(x) = -\frac{3}{x^2} - \frac{1}{x} \) E) \( f'(x) = \frac{3}{x^2} - \frac{b}{x} \) F) Answer not listed **Solution:** The correct answer is option: C) \( f'(x) = -\frac{3}{x^2} + \frac{1}{x} \) **Explanation:** To find the derivative \( f'(x) \), apply the derivatives of \( \frac{3}{x} \) and \( \ln(bx) \): - The derivative of \( \frac{3}{x} \) is \( -\frac{3}{x^2} \). - The derivative of \( \ln(bx) \) is \( \frac{b}{bx} = \frac{1}{x} \). So, \( f'(x) = -\frac{3}{x^2} + \frac{1}{x} \).
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