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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![### Calculus Problem: Differentiation
In this section, we will focus on calculating the derivative of a given function. Below, you will see a boxed input area where you can enter your answer for the first derivative of the function \( f(x) \).
#### Example Function for Differentiation:
\[ f'(x) = \]
**Instructions:**
1. Carefully read the given function in your problem statement.
2. Use appropriate differentiation rules (product rule, quotient rule, chain rule, etc.) to find \( f'(x) \).
3. Enter your result in the input box provided.
To the right of the input box, there is a button with a grid icon. Clicking this button might open a mathematical keyboard, aiding in inputting mathematical symbols and expressions more easily.
Below the input box, there are two buttons:
- **Previous Question (Button not fully shown)**
- **Next Question (Button not fully shown)**
These buttons can be used to navigate between different questions in your problem set.
Always double-check your entered derivative for accuracy.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5698be52-85e5-4fb1-8c09-8b1bee5b6fa9%2Fc67cdd11-b77e-41b6-9afd-89d4eaff23e2%2Fm7y4pse_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Calculus Problem: Differentiation
In this section, we will focus on calculating the derivative of a given function. Below, you will see a boxed input area where you can enter your answer for the first derivative of the function \( f(x) \).
#### Example Function for Differentiation:
\[ f'(x) = \]
**Instructions:**
1. Carefully read the given function in your problem statement.
2. Use appropriate differentiation rules (product rule, quotient rule, chain rule, etc.) to find \( f'(x) \).
3. Enter your result in the input box provided.
To the right of the input box, there is a button with a grid icon. Clicking this button might open a mathematical keyboard, aiding in inputting mathematical symbols and expressions more easily.
Below the input box, there are two buttons:
- **Previous Question (Button not fully shown)**
- **Next Question (Button not fully shown)**
These buttons can be used to navigate between different questions in your problem set.
Always double-check your entered derivative for accuracy.
![**Mathematics: Calculus - Differentiation**
**Problem Statement:**
Given the function \( f(x) = \sqrt{x} \sin x \), find the derivative \( f'(x) \).
**Detailed Explanation:**
In this task, we aim to find the derivative of the given function \( f(x) = \sqrt{x} \sin x \). To accomplish this, we apply the product rule of differentiation, which is used when differentiating products of two or more functions. The product rule states that if \( u(x) \) and \( v(x) \) are functions of \( x \), then:
\[ \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \]
Here, let:
- \( u(x) = \sqrt{x} \)
- \( v(x) = \sin x \)
First, we find the derivatives of \( u(x) \) and \( v(x) \):
1. For \( u(x) = \sqrt{x} \):
\[ u(x) = x^{1/2} \]
\[ u'(x) = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}} \]
2. For \( v(x) = \sin x \):
\[ v'(x) = \cos x \]
Now, apply the product rule:
\[ f'(x) = u'(x)v(x) + u(x)v'(x) \]
\[ f'(x) = \left( \frac{1}{2\sqrt{x}} \right) \sin x + \sqrt{x} (\cos x) \]
\[ f'(x) = \frac{\sin x}{2\sqrt{x}} + \sqrt{x} \cos x \]
Thus, the derivative of \( f(x) = \sqrt{x} \sin x \) is:
\[ f'(x) = \frac{\sin x}{2\sqrt{x}} + \sqrt{x} \cos x \]
This expression gives the rate of change of the function \( f(x) \) with respect to \( x \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5698be52-85e5-4fb1-8c09-8b1bee5b6fa9%2Fc67cdd11-b77e-41b6-9afd-89d4eaff23e2%2Fzj4evk5_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Mathematics: Calculus - Differentiation**
**Problem Statement:**
Given the function \( f(x) = \sqrt{x} \sin x \), find the derivative \( f'(x) \).
**Detailed Explanation:**
In this task, we aim to find the derivative of the given function \( f(x) = \sqrt{x} \sin x \). To accomplish this, we apply the product rule of differentiation, which is used when differentiating products of two or more functions. The product rule states that if \( u(x) \) and \( v(x) \) are functions of \( x \), then:
\[ \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \]
Here, let:
- \( u(x) = \sqrt{x} \)
- \( v(x) = \sin x \)
First, we find the derivatives of \( u(x) \) and \( v(x) \):
1. For \( u(x) = \sqrt{x} \):
\[ u(x) = x^{1/2} \]
\[ u'(x) = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}} \]
2. For \( v(x) = \sin x \):
\[ v'(x) = \cos x \]
Now, apply the product rule:
\[ f'(x) = u'(x)v(x) + u(x)v'(x) \]
\[ f'(x) = \left( \frac{1}{2\sqrt{x}} \right) \sin x + \sqrt{x} (\cos x) \]
\[ f'(x) = \frac{\sin x}{2\sqrt{x}} + \sqrt{x} \cos x \]
Thus, the derivative of \( f(x) = \sqrt{x} \sin x \) is:
\[ f'(x) = \frac{\sin x}{2\sqrt{x}} + \sqrt{x} \cos x \]
This expression gives the rate of change of the function \( f(x) \) with respect to \( x \).
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