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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![**Mathematics: Differentiation Practice**
**Problem Statement:**
Given the function \( f(x) = \sqrt{x} \sin x \), find the derivative \( f'(x) \).
\[ f'(x) = \]
**Solution:**
To find the derivative \( f'(x) \) of the given function \( f(x) = \sqrt{x} \sin x \), we will apply the product rule of differentiation.
The product rule states that if \( u(x) \) and \( v(x) \) are functions of \( x \), then the derivative of their product is given by:
\[ (uv)' = u'v + uv' \]
Let's set \( u(x) = \sqrt{x} \) and \( v(x) = \sin x \).
1. Compute \( u'(x) \):
\[ u(x) = x^{1/2} \]
\[ u'(x) = \frac{1}{2} x^{-1/2} = \frac{1}{2 \sqrt{x}} \]
2. Compute \( v'(x) \):
\[ v(x) = \sin x \]
\[ v'(x) = \cos x \]
Now apply the product rule:
\[ f'(x) = u'(x)v(x) + u(x)v'(x) \]
Substitute \( u(x) \), \( u'(x) \), \( v(x) \), and \( v'(x) \) into the formula:
\[ f'(x) = \left( \frac{1}{2 \sqrt{x}} \right) (\sin x) + (\sqrt{x}) (\cos x) \]
Simplify the equation:
\[ f'(x) = \frac{\sin x}{2 \sqrt{x}} + \sqrt{x} \cos x \]
Therefore, the derivative \( f'(x) \) is:
\[ f'(x) = \frac{\sin x}{2 \sqrt{x}} + \sqrt{x} \cos x \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffa043fe4-4451-42e8-82d1-21cc59f9ce63%2Ff51758a4-4e47-4557-8648-dfcbdbb63419%2Ftuwqtjc.png&w=3840&q=75)
Transcribed Image Text:**Mathematics: Differentiation Practice**
**Problem Statement:**
Given the function \( f(x) = \sqrt{x} \sin x \), find the derivative \( f'(x) \).
\[ f'(x) = \]
**Solution:**
To find the derivative \( f'(x) \) of the given function \( f(x) = \sqrt{x} \sin x \), we will apply the product rule of differentiation.
The product rule states that if \( u(x) \) and \( v(x) \) are functions of \( x \), then the derivative of their product is given by:
\[ (uv)' = u'v + uv' \]
Let's set \( u(x) = \sqrt{x} \) and \( v(x) = \sin x \).
1. Compute \( u'(x) \):
\[ u(x) = x^{1/2} \]
\[ u'(x) = \frac{1}{2} x^{-1/2} = \frac{1}{2 \sqrt{x}} \]
2. Compute \( v'(x) \):
\[ v(x) = \sin x \]
\[ v'(x) = \cos x \]
Now apply the product rule:
\[ f'(x) = u'(x)v(x) + u(x)v'(x) \]
Substitute \( u(x) \), \( u'(x) \), \( v(x) \), and \( v'(x) \) into the formula:
\[ f'(x) = \left( \frac{1}{2 \sqrt{x}} \right) (\sin x) + (\sqrt{x}) (\cos x) \]
Simplify the equation:
\[ f'(x) = \frac{\sin x}{2 \sqrt{x}} + \sqrt{x} \cos x \]
Therefore, the derivative \( f'(x) \) is:
\[ f'(x) = \frac{\sin x}{2 \sqrt{x}} + \sqrt{x} \cos x \]
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