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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4.2 #4
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![**Problem Statement:**
Use the product rule to find the derivative of the following.
\[ y = (x + 5)\left(5\sqrt{x} + 9\right) \]
**Solution:**
To find the derivative \( y' \), we will apply the product rule. The product rule states that if \( y = u(x) \cdot v(x) \), then the derivative \( y' = u'(x) \cdot v(x) + u(x) \cdot v'(x) \).
Let:
- \( u(x) = x + 5 \)
- \( v(x) = 5\sqrt{x} + 9 \)
**Steps:**
1. **Find \( u'(x) \):**
- \( u(x) = x + 5 \)
- Derivative: \( u'(x) = 1 \)
2. **Find \( v'(x) \):**
- \( v(x) = 5\sqrt{x} + 9 = 5x^{1/2} + 9 \)
- Derivative: \( v'(x) = \frac{5}{2}x^{-1/2} = \frac{5}{2\sqrt{x}} \)
3. **Apply the product rule:**
- \( y' = u'(x)v(x) + u(x)v'(x) \)
- Substitute the values:
\[
y' = 1 \cdot (5\sqrt{x} + 9) + (x + 5) \cdot \frac{5}{2\sqrt{x}}
\]
- Simplifying further gives:
\[
y' = 5\sqrt{x} + 9 + \frac{5(x + 5)}{2\sqrt{x}}
\]
**Conclusion:**
The derivative \( y' \) of the function \( y = (x + 5)(5\sqrt{x} + 9) \) using the product rule is:
\[ y' = 5\sqrt{x} + 9 + \frac{5(x + 5)}{2\sqrt{x}} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F63591bc8-75d6-42dc-963a-0ef4cf7b290a%2F924b9f6c-3b22-495d-b9d0-01ab428f6031%2F5ef6zzb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Use the product rule to find the derivative of the following.
\[ y = (x + 5)\left(5\sqrt{x} + 9\right) \]
**Solution:**
To find the derivative \( y' \), we will apply the product rule. The product rule states that if \( y = u(x) \cdot v(x) \), then the derivative \( y' = u'(x) \cdot v(x) + u(x) \cdot v'(x) \).
Let:
- \( u(x) = x + 5 \)
- \( v(x) = 5\sqrt{x} + 9 \)
**Steps:**
1. **Find \( u'(x) \):**
- \( u(x) = x + 5 \)
- Derivative: \( u'(x) = 1 \)
2. **Find \( v'(x) \):**
- \( v(x) = 5\sqrt{x} + 9 = 5x^{1/2} + 9 \)
- Derivative: \( v'(x) = \frac{5}{2}x^{-1/2} = \frac{5}{2\sqrt{x}} \)
3. **Apply the product rule:**
- \( y' = u'(x)v(x) + u(x)v'(x) \)
- Substitute the values:
\[
y' = 1 \cdot (5\sqrt{x} + 9) + (x + 5) \cdot \frac{5}{2\sqrt{x}}
\]
- Simplifying further gives:
\[
y' = 5\sqrt{x} + 9 + \frac{5(x + 5)}{2\sqrt{x}}
\]
**Conclusion:**
The derivative \( y' \) of the function \( y = (x + 5)(5\sqrt{x} + 9) \) using the product rule is:
\[ y' = 5\sqrt{x} + 9 + \frac{5(x + 5)}{2\sqrt{x}} \]
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