Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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# Calculating Derivatives: A Practice Problem

## Problem Statement

Given the function:

\[ f(x) = (x^2 + 3x + 4)^3 \]

Find:

1. The first derivative of \( f(x) \), denoted as \( f'(x) \).
2. The value of the derivative at \( x =3 \), denoted as \( f'(3) \).

### Instructions

1. For the first part, apply the chain rule to find the first derivative \( f'(x) \).
2. Once you have \( f'(x) \), substitute \( x=3 \) to find \( f'(3) \).

### Guided Solution

To find \( f'(x) \), use the chain rule, which in general form for a function \( g(h(x)) \) is:

\[ \frac{d}{dx} [g(h(x))] = g'(h(x)) \cdot h'(x) \]

For our specific function \( f(x) = (x^2 + 3x + 4)^3 \):

1. Let \( g(u) = u^3 \) and \( h(x) = x^2 + 3x + 4 \).

Therefore, \( f(x) = g(h(x)) \).

2. Differentiate \( g(u) = u^3 \) with respect to \( u \):

\[ g'(u) = 3u^2 \]

3. Differentiate \( h(x) = x^2 + 3x + 4 \) with respect to \( x \):

\[ h'(x) = 2x + 3 \]

4. Apply the chain rule:

\[ f'(x) = g'(h(x)) \cdot h'(x) = 3(h(x))^2 \cdot h'(x) \]

\[ f'(x) = 3(x^2 + 3x + 4)^2 \cdot (2x + 3) \]

5. To find \( f'(3) \):

\[ f'(3) = 3(3^2 + 3 \cdot 3 + 4)^2 \cdot (2 \cdot 3 + 3) \]

\[ = 3(9 + 9 + 4)^2 \cdot (6 + 3) \]

\[ =
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Transcribed Image Text:# Calculating Derivatives: A Practice Problem ## Problem Statement Given the function: \[ f(x) = (x^2 + 3x + 4)^3 \] Find: 1. The first derivative of \( f(x) \), denoted as \( f'(x) \). 2. The value of the derivative at \( x =3 \), denoted as \( f'(3) \). ### Instructions 1. For the first part, apply the chain rule to find the first derivative \( f'(x) \). 2. Once you have \( f'(x) \), substitute \( x=3 \) to find \( f'(3) \). ### Guided Solution To find \( f'(x) \), use the chain rule, which in general form for a function \( g(h(x)) \) is: \[ \frac{d}{dx} [g(h(x))] = g'(h(x)) \cdot h'(x) \] For our specific function \( f(x) = (x^2 + 3x + 4)^3 \): 1. Let \( g(u) = u^3 \) and \( h(x) = x^2 + 3x + 4 \). Therefore, \( f(x) = g(h(x)) \). 2. Differentiate \( g(u) = u^3 \) with respect to \( u \): \[ g'(u) = 3u^2 \] 3. Differentiate \( h(x) = x^2 + 3x + 4 \) with respect to \( x \): \[ h'(x) = 2x + 3 \] 4. Apply the chain rule: \[ f'(x) = g'(h(x)) \cdot h'(x) = 3(h(x))^2 \cdot h'(x) \] \[ f'(x) = 3(x^2 + 3x + 4)^2 \cdot (2x + 3) \] 5. To find \( f'(3) \): \[ f'(3) = 3(3^2 + 3 \cdot 3 + 4)^2 \cdot (2 \cdot 3 + 3) \] \[ = 3(9 + 9 + 4)^2 \cdot (6 + 3) \] \[ =
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