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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
Transcribed Image Text:### Calculus Problem: Differentiation
**Problem: 1. Differentiate \( y = \sqrt{3x - 7} \)**
#### Solution Steps:
1. **Identify the function to differentiate:**
\( y = \sqrt{3x - 7} \)
2. **Rewrite the function in exponential form:**
\( y = (3x - 7)^{1/2} \)
3. **Apply the chain rule for differentiation:**
- The chain rule states that the derivative of \( y \) with respect to \( x \) is \( dy/dx = (dy/du) \cdot (du/dx) \), where \( u = 3x - 7 \).
4. **Differentiate the outer function (power rule):**
- Let \( u = 3x - 7 \)
- Then, \( y = u^{1/2} \)
- By the power rule, \( dy/du = \frac{1}{2}u^{-1/2} \)
5. **Differentiate the inner function:**
- \( du/dx = 3 \)
6. **Combine the derivatives using the chain rule:**
- \( dy/dx = (dy/du) \cdot (du/dx) \)
- \( dy/dx = \frac{1}{2}u^{-1/2} \cdot 3 \)
- Substitute back \( u = 3x - 7 \)
- \( dy/dx = \frac{3}{2}(3x - 7)^{-1/2} \)
7. **Simplify the expression:**
- \( dy/dx = \frac{3}{2\sqrt{3x - 7}} \)
Thus, the derivative of \( y = \sqrt{3x - 7} \) with respect to \( x \) is \( \frac{3}{2\sqrt{3x - 7}} \).
### Conclusion:
The differentiation problem demonstrates the application of the chain rule in finding the derivative of a composite function. This rule is essential for handling more complex differentiation tasks in calculus. Always ensure to express the final solution in its simplest form.
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