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
Given the function:
\[ y = (x^2 + 4)^x \]
Use Logarithmic Differentiation to find \(\frac{dy}{dx}\).
### Solution
To find the derivative \(\frac{dy}{dx}\) using logarithmic differentiation, follow these steps:
1. **Take the natural logarithm on both sides:**
\[ \ln y = \ln \left((x^2 + 4)^x\right) \]
2. **Use properties of logarithms to simplify:**
\[ \ln y = x \ln (x^2 + 4) \]
3. **Differentiate both sides with respect to \(x\):**
\[ \frac{d}{dx} (\ln y) = \frac{d}{dx} \left( x \ln (x^2 + 4) \right) \]
4. **Use the chain rule on the left-hand side and the product rule on the right-hand side:**
\[ \frac{1}{y} \frac{dy}{dx} = \ln (x^2 + 4) + x \cdot \frac{1}{x^2 + 4} \cdot \frac{d}{dx} (x^2 + 4) \]
5. **Simplify the derivative of \(x^2 + 4\):**
\[ \frac{d}{dx} (x^2 + 4) = 2x \]
6. **Substitute this back into the equation:**
\[ \frac{1}{y} \frac{dy}{dx} = \ln (x^2 + 4) + \frac{x \cdot 2x}{x^2 + 4} \]
\[ \frac{1}{y} \frac{dy}{dx} = \ln (x^2 + 4) + \frac{2x^2}{x^2 + 4} \]
7. **Multiply both sides by \(y\) to isolate \(\frac{dy}{dx}\):**
\[ \frac{dy}{dx} = y \left( \ln (x^2 + 4) + \frac{2x^2}{x^2 + 4} \right) \]
8. **Substitute \(y = (x^2 + 4)^x\) back](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4a3ba5c9-3580-45cd-ae24-18a07977dcd9%2F50e98681-e3ed-48f5-9cc2-3a13dea6c398%2Fb2bd1xs_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
Given the function:
\[ y = (x^2 + 4)^x \]
Use Logarithmic Differentiation to find \(\frac{dy}{dx}\).
### Solution
To find the derivative \(\frac{dy}{dx}\) using logarithmic differentiation, follow these steps:
1. **Take the natural logarithm on both sides:**
\[ \ln y = \ln \left((x^2 + 4)^x\right) \]
2. **Use properties of logarithms to simplify:**
\[ \ln y = x \ln (x^2 + 4) \]
3. **Differentiate both sides with respect to \(x\):**
\[ \frac{d}{dx} (\ln y) = \frac{d}{dx} \left( x \ln (x^2 + 4) \right) \]
4. **Use the chain rule on the left-hand side and the product rule on the right-hand side:**
\[ \frac{1}{y} \frac{dy}{dx} = \ln (x^2 + 4) + x \cdot \frac{1}{x^2 + 4} \cdot \frac{d}{dx} (x^2 + 4) \]
5. **Simplify the derivative of \(x^2 + 4\):**
\[ \frac{d}{dx} (x^2 + 4) = 2x \]
6. **Substitute this back into the equation:**
\[ \frac{1}{y} \frac{dy}{dx} = \ln (x^2 + 4) + \frac{x \cdot 2x}{x^2 + 4} \]
\[ \frac{1}{y} \frac{dy}{dx} = \ln (x^2 + 4) + \frac{2x^2}{x^2 + 4} \]
7. **Multiply both sides by \(y\) to isolate \(\frac{dy}{dx}\):**
\[ \frac{dy}{dx} = y \left( \ln (x^2 + 4) + \frac{2x^2}{x^2 + 4} \right) \]
8. **Substitute \(y = (x^2 + 4)^x\) back
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