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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![### Calculating the Derivative of a Given Function
Given the function:
\[ f(x) = (3x^2 - 7)^3 \cdot (-7x^2 + 8)^{12} \]
We are required to determine the first derivative, \( f'(x) \).
Use the product rule for differentiation which states that if \( u(x) \) and \( v(x) \) are differentiable functions of \( x \), then the derivative of their product \( u(x) \cdot v(x) \) is given by:
\[ \frac{d}{dx} \left[ u(x)v(x) \right] = u'(x)v(x) + u(x)v'(x) \]
In this scenario:
- Let \( u(x) = (3x^2 - 7)^3 \)
- Let \( v(x) = (-7x^2 + 8)^{12} \)
First, we need to find the derivatives \( u'(x) \) and \( v'(x) \).
#### Finding \( u'(x) \)
\[ u(x) = (3x^2 - 7)^3 \]
Using the chain rule,
\[ u'(x) = 3 \cdot (3x^2 - 7)^2 \cdot \frac{d}{dx} (3x^2 - 7) \]
\[ = 3 \cdot (3x^2 - 7)^2 \cdot 6x \]
\[ = 18x(3x^2 - 7)^2 \]
#### Finding \( v'(x) \)
\[ v(x) = (-7x^2 + 8)^{12} \]
Using the chain rule,
\[ v'(x) = 12 \cdot (-7x^2 + 8)^{11} \cdot \frac{d}{dx} (-7x^2 + 8) \]
\[ = 12 \cdot (-7x^2 + 8)^{11} \cdot (-14x) \]
\[ = -168x(-7x^2 + 8)^{11} \]
#### Applying the Product Rule
Now, substitute \( u(x) \), \( u'(x) \), \( v(x) \), and \( v'(x) \) back into the product rule formula](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F19a44d99-9126-45f2-b575-2ff09f428fba%2Fca4d5ca7-6201-44f6-9a4e-f627f0ae43d7%2Fza84dfo.png&w=3840&q=75)
Transcribed Image Text:### Calculating the Derivative of a Given Function
Given the function:
\[ f(x) = (3x^2 - 7)^3 \cdot (-7x^2 + 8)^{12} \]
We are required to determine the first derivative, \( f'(x) \).
Use the product rule for differentiation which states that if \( u(x) \) and \( v(x) \) are differentiable functions of \( x \), then the derivative of their product \( u(x) \cdot v(x) \) is given by:
\[ \frac{d}{dx} \left[ u(x)v(x) \right] = u'(x)v(x) + u(x)v'(x) \]
In this scenario:
- Let \( u(x) = (3x^2 - 7)^3 \)
- Let \( v(x) = (-7x^2 + 8)^{12} \)
First, we need to find the derivatives \( u'(x) \) and \( v'(x) \).
#### Finding \( u'(x) \)
\[ u(x) = (3x^2 - 7)^3 \]
Using the chain rule,
\[ u'(x) = 3 \cdot (3x^2 - 7)^2 \cdot \frac{d}{dx} (3x^2 - 7) \]
\[ = 3 \cdot (3x^2 - 7)^2 \cdot 6x \]
\[ = 18x(3x^2 - 7)^2 \]
#### Finding \( v'(x) \)
\[ v(x) = (-7x^2 + 8)^{12} \]
Using the chain rule,
\[ v'(x) = 12 \cdot (-7x^2 + 8)^{11} \cdot \frac{d}{dx} (-7x^2 + 8) \]
\[ = 12 \cdot (-7x^2 + 8)^{11} \cdot (-14x) \]
\[ = -168x(-7x^2 + 8)^{11} \]
#### Applying the Product Rule
Now, substitute \( u(x) \), \( u'(x) \), \( v(x) \), and \( v'(x) \) back into the product rule formula
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