Find the derivative of the function. y = Xx x + 3

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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I need help solving this chain rule problem. I tried to solve it, but I'm a little confused on how to continue

The image contains a step-by-step differentiation process using the chain rule and quotient rule. Here's a transcription and explanation:

1. **Starting Expression:**
   \[
   y = \left( \frac{x}{x+3} \right)^{1/2} = u^{1/2}
   \]

2. **Substitution:**
   \[
   u = \frac{x}{x+3}
   \]

3. **Functions Definition:**
   \[
   f(u) = u^{1/2} \quad  g(x) = \frac{x}{x+3}
   \]

4. **Expression for y:**
   \[
   y = \left( \frac{x}{x+3} \right)^{1/2}
   \]

5. **Derivative Start:**
   \[
   y' = \frac{d}{dx} \left( \left( \frac{x}{x+3} \right)^{1/2} \right)
   \]

6. **Chain Rule Application:**
   \[
   y' = \frac{1}{2} \left( u \right)^{-1/2} \cdot \frac{d}{dx} \left( \frac{x}{x+3} \right)
   \]
   - Use the quotient rule for \(\frac{x}{x+3}\).

7. **Quotient Rule Application:**
   \[
   y' = \frac{1}{2} \cdot u^{-1/2} \cdot \frac{(x+3)(1) - (x)(1)}{(x+3)^2}
   \]

8. **Simplified Derivative:**
   \[
   y' = \frac{1}{2} \cdot \left( \frac{x}{x+3} \right)^{-1/2} \cdot \frac{3}{(x+3)^2}
   \]

9. **Final Simplified Form:**
   \[
   y' = \frac{1}{2} \cdot \frac{1}{\sqrt{\frac{x}{x+3}}} \cdot \frac{3}{(x+3)^2}
   \]
   \[
   y' = \frac{3}{2} \cdot \frac{1}{\
Transcribed Image Text:The image contains a step-by-step differentiation process using the chain rule and quotient rule. Here's a transcription and explanation: 1. **Starting Expression:** \[ y = \left( \frac{x}{x+3} \right)^{1/2} = u^{1/2} \] 2. **Substitution:** \[ u = \frac{x}{x+3} \] 3. **Functions Definition:** \[ f(u) = u^{1/2} \quad g(x) = \frac{x}{x+3} \] 4. **Expression for y:** \[ y = \left( \frac{x}{x+3} \right)^{1/2} \] 5. **Derivative Start:** \[ y' = \frac{d}{dx} \left( \left( \frac{x}{x+3} \right)^{1/2} \right) \] 6. **Chain Rule Application:** \[ y' = \frac{1}{2} \left( u \right)^{-1/2} \cdot \frac{d}{dx} \left( \frac{x}{x+3} \right) \] - Use the quotient rule for \(\frac{x}{x+3}\). 7. **Quotient Rule Application:** \[ y' = \frac{1}{2} \cdot u^{-1/2} \cdot \frac{(x+3)(1) - (x)(1)}{(x+3)^2} \] 8. **Simplified Derivative:** \[ y' = \frac{1}{2} \cdot \left( \frac{x}{x+3} \right)^{-1/2} \cdot \frac{3}{(x+3)^2} \] 9. **Final Simplified Form:** \[ y' = \frac{1}{2} \cdot \frac{1}{\sqrt{\frac{x}{x+3}}} \cdot \frac{3}{(x+3)^2} \] \[ y' = \frac{3}{2} \cdot \frac{1}{\
**Problem Statement:**

Find the derivative of the function.

\[ y = \sqrt{\frac{x}{x + 3}} \]

**Explanation:**

In this problem, we are asked to find the derivative of the function \( y \), which is defined as the square root of the fraction \(\frac{x}{x+3}\). 

To solve this, we apply the chain rule and quotient rule for derivatives. 

1. **Chain Rule:** 
   - If \( y = \sqrt{u} \), then \( \frac{dy}{du} = \frac{1}{2\sqrt{u}} \).

2. **Quotient Rule**: 
   - For a function \( u = \frac{f(x)}{g(x)} \), the derivative \( \frac{du}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \).

Using these rules will help find the derivative for more complex functions like the given one.
Transcribed Image Text:**Problem Statement:** Find the derivative of the function. \[ y = \sqrt{\frac{x}{x + 3}} \] **Explanation:** In this problem, we are asked to find the derivative of the function \( y \), which is defined as the square root of the fraction \(\frac{x}{x+3}\). To solve this, we apply the chain rule and quotient rule for derivatives. 1. **Chain Rule:** - If \( y = \sqrt{u} \), then \( \frac{dy}{du} = \frac{1}{2\sqrt{u}} \). 2. **Quotient Rule**: - For a function \( u = \frac{f(x)}{g(x)} \), the derivative \( \frac{du}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \). Using these rules will help find the derivative for more complex functions like the given one.
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Step 1: Derivative property used :

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