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:**
Find \( f''(x) \) for the function \( f(x) = x \cos x \).
**Solution:**
To find the second derivative \( f''(x) \), we first need to find the first derivative \( f'(x) \).
The function \( f(x) = x \cos x \) is a product of two functions, \( u(x) = x \) and \( v(x) = \cos x \). We can use the product rule to find \( f'(x) \):
\[ f'(x) = u'(x)v(x) + u(x)v'(x) \]
Calculating each part:
- \( u(x) = x \) so \( u'(x) = 1 \)
- \( v(x) = \cos x \) so \( v'(x) = -\sin x \)
Substitute these derivatives into the product rule formula:
\[ f'(x) = (1)(\cos x) + (x)(-\sin x) \]
\[ f'(x) = \cos x - x \sin x \]
Next, we find the second derivative \( f''(x) \). We differentiate \( f'(x) = \cos x - x \sin x \):
\[ f''(x) = (\cos x)' - (x \sin x)' \]
Calculating each derivative:
- The derivative of \( \cos x \) is \( -\sin x \).
- We again use the product rule for \( (x \sin x)' \), where \( u = x \) and \( v = \sin x \).
- \( u'(x) = 1 \)
- \( v'(x) = \cos x \)
\[ (x \sin x)' = (1)(\sin x) + (x)(\cos x) \]
\[ (x \sin x)' = \sin x + x \cos x \]
Plugging these into the equation for \( f''(x) \):
\[ f''(x) = -\sin x - (\sin x + x \cos x) \]
\[ f''(x) = -\sin x - \sin x - x \cos x \]
\[ f''(x) = -2\sin x - x \cos x \]
So, the second derivative is:
\[ f''(](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F53d9a1e2-5877-4e8e-930c-9c60a45f5a51%2F48ae7ab8-d518-4dd5-b08c-ce028d6b3e52%2Fh8tq94o_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find \( f''(x) \) for the function \( f(x) = x \cos x \).
**Solution:**
To find the second derivative \( f''(x) \), we first need to find the first derivative \( f'(x) \).
The function \( f(x) = x \cos x \) is a product of two functions, \( u(x) = x \) and \( v(x) = \cos x \). We can use the product rule to find \( f'(x) \):
\[ f'(x) = u'(x)v(x) + u(x)v'(x) \]
Calculating each part:
- \( u(x) = x \) so \( u'(x) = 1 \)
- \( v(x) = \cos x \) so \( v'(x) = -\sin x \)
Substitute these derivatives into the product rule formula:
\[ f'(x) = (1)(\cos x) + (x)(-\sin x) \]
\[ f'(x) = \cos x - x \sin x \]
Next, we find the second derivative \( f''(x) \). We differentiate \( f'(x) = \cos x - x \sin x \):
\[ f''(x) = (\cos x)' - (x \sin x)' \]
Calculating each derivative:
- The derivative of \( \cos x \) is \( -\sin x \).
- We again use the product rule for \( (x \sin x)' \), where \( u = x \) and \( v = \sin x \).
- \( u'(x) = 1 \)
- \( v'(x) = \cos x \)
\[ (x \sin x)' = (1)(\sin x) + (x)(\cos x) \]
\[ (x \sin x)' = \sin x + x \cos x \]
Plugging these into the equation for \( f''(x) \):
\[ f''(x) = -\sin x - (\sin x + x \cos x) \]
\[ f''(x) = -\sin x - \sin x - x \cos x \]
\[ f''(x) = -2\sin x - x \cos x \]
So, the second derivative is:
\[ f''(
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