### Calculus Problem: DifferentiationIn 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 $.

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Answered: If f(z) = JE sin I, find f'(x).

Math

Calculus

If f(z) = JE sin I, find f'(x).

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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Question

### 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 $.

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