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:**
Find \(\frac{dy}{dx}\) for \(y = 7x^2 \sin x + 14x \cos x - 14 \sin x\).
**Solution:**
To find \(\frac{dy}{dx}\), we need to differentiate each term of the function with respect to \(x\).
1. **Differentiate \(7x^2 \sin x\):**
- Use the product rule: \((uv)' = u'v + uv'\)
- Let \(u = 7x^2\) and \(v = \sin x\).
- \(u' = 14x\)
- \(v' = \cos x\)
- The derivative is:
\[
(uv)' = (14x)(\sin x) + (7x^2)(\cos x)
\]
2. **Differentiate \(14x \cos x\):**
- Use the product rule again.
- Let \(u = 14x\) and \(v = \cos x\).
- \(u' = 14\)
- \(v' = -\sin x\)
- The derivative is:
\[
(uv)' = (14)(\cos x) + (14x)(-\sin x)
\]
3. **Differentiate \(-14 \sin x\):**
- The derivative of \(\sin x\) is \(\cos x\).
- So, \(-14 \sin x\) becomes \(-14 \cos x\).
**Combine all parts:**
\[
\frac{dy}{dx} = (14x)(\sin x) + (7x^2)(\cos x) + (14)(\cos x) - (14x)(\sin x) - 14 \cos x
\]
Simplifying, we cancel out \((14x)(\sin x)\) terms:
\[
\frac{dy}{dx} = (7x^2)(\cos x)
\]
Therefore, the derivative is:
\[
\frac{dy}{dx} = 7x^2 \cos x
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1a0dff6b-75e5-4857-9f6c-486756f4401c%2F72b403b5-867e-4822-b07d-394338125465%2Fz5i0mv_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem:**
Find \(\frac{dy}{dx}\) for \(y = 7x^2 \sin x + 14x \cos x - 14 \sin x\).
**Solution:**
To find \(\frac{dy}{dx}\), we need to differentiate each term of the function with respect to \(x\).
1. **Differentiate \(7x^2 \sin x\):**
- Use the product rule: \((uv)' = u'v + uv'\)
- Let \(u = 7x^2\) and \(v = \sin x\).
- \(u' = 14x\)
- \(v' = \cos x\)
- The derivative is:
\[
(uv)' = (14x)(\sin x) + (7x^2)(\cos x)
\]
2. **Differentiate \(14x \cos x\):**
- Use the product rule again.
- Let \(u = 14x\) and \(v = \cos x\).
- \(u' = 14\)
- \(v' = -\sin x\)
- The derivative is:
\[
(uv)' = (14)(\cos x) + (14x)(-\sin x)
\]
3. **Differentiate \(-14 \sin x\):**
- The derivative of \(\sin x\) is \(\cos x\).
- So, \(-14 \sin x\) becomes \(-14 \cos x\).
**Combine all parts:**
\[
\frac{dy}{dx} = (14x)(\sin x) + (7x^2)(\cos x) + (14)(\cos x) - (14x)(\sin x) - 14 \cos x
\]
Simplifying, we cancel out \((14x)(\sin x)\) terms:
\[
\frac{dy}{dx} = (7x^2)(\cos x)
\]
Therefore, the derivative is:
\[
\frac{dy}{dx} = 7x^2 \cos x
\]
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