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
\[ y = \sqrt{x^2 + \ln(5x)} \]
**Solution Approach:**
To differentiate this function, you need to use the chain rule and the derivatives of square root, quadratic, and logarithmic functions.
1. **Outer Function Derivative**: The outer function is the square root, \( \sqrt{u} \). The derivative of this with respect to \( u \) is \( \frac{1}{2\sqrt{u}} \).
2. **Inner Function Derivative**: The inner function is \( u = x^2 + \ln(5x) \).
- **Differentiating \( x^2 \)**: The derivative is \( 2x \).
- **Differentiating \( \ln(5x) \)**:
- First, apply the chain rule: differentiate \( \ln(5x) \) as \( \frac{1}{5x} \).
- Then, multiply by the derivative of \( 5x \), which is 5.
- This results in \( \frac{5}{5x} = \frac{1}{x} \).
- Combine these derivatives: \( u' = 2x + \frac{1}{x} \).
3. **Combining Using the Chain Rule**:
- The derivative of \( y \) with respect to \( x \) is given by:
\[
y' = \frac{1}{2\sqrt{x^2 + \ln(5x)}} \cdot (2x + \frac{1}{x})
\]
This derivative expression provides the rate of change of \( y \) with respect to \( x \), allowing you to analyze the behavior of the function at various points.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1e343709-1356-461b-bfde-23a6e93e0975%2F81b4c389-debb-4d25-9a1f-fe7cb8306e8c%2Ftyn0r9a_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Differentiate the function
\[ y = \sqrt{x^2 + \ln(5x)} \]
**Solution Approach:**
To differentiate this function, you need to use the chain rule and the derivatives of square root, quadratic, and logarithmic functions.
1. **Outer Function Derivative**: The outer function is the square root, \( \sqrt{u} \). The derivative of this with respect to \( u \) is \( \frac{1}{2\sqrt{u}} \).
2. **Inner Function Derivative**: The inner function is \( u = x^2 + \ln(5x) \).
- **Differentiating \( x^2 \)**: The derivative is \( 2x \).
- **Differentiating \( \ln(5x) \)**:
- First, apply the chain rule: differentiate \( \ln(5x) \) as \( \frac{1}{5x} \).
- Then, multiply by the derivative of \( 5x \), which is 5.
- This results in \( \frac{5}{5x} = \frac{1}{x} \).
- Combine these derivatives: \( u' = 2x + \frac{1}{x} \).
3. **Combining Using the Chain Rule**:
- The derivative of \( y \) with respect to \( x \) is given by:
\[
y' = \frac{1}{2\sqrt{x^2 + \ln(5x)}} \cdot (2x + \frac{1}{x})
\]
This derivative expression provides the rate of change of \( y \) with respect to \( x \), allowing you to analyze the behavior of the function at various points.
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