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 Second Derivative**
To find the second derivative of the function:
\[ f(x) = 5(2 - 7x)^4 \]
Step 1: Find the first derivative \( f'(x) \).
Use the chain rule for differentiation. Let \( u = 2 - 7x \). Then, \( f(x) = 5u^4 \) and
\[ \frac{d}{dx}[5u^4] = 5 \cdot 4u^3 \cdot \frac{du}{dx} \]
First, differentiate \( u = 2 - 7x \). The derivative of \( u \) with respect to \( x \) is:
\[ \frac{du}{dx} = \frac{d}{dx}(2 - 7x) = -7 \]
Hence,
\[ f'(x) = 5 \cdot 4(2 - 7x)^3 \cdot (-7) = -140(2 - 7x)^3 \]
Step 2: Find the second derivative \( f''(x) \).
To find the second derivative, differentiate \( f'(x) \) with respect to \( x \):
\[ f'(x) = -140(2 - 7x)^3 \]
Again, applying the chain rule, let \( u = 2 - 7x \), then:
\[ \frac{d}{dx}[-140u^3] = -140 \cdot 3u^2 \cdot \frac{du}{dx} \]
We already know \( \frac{du}{dx} = -7 \), hence,
\[ f''(x) = -140 \cdot 3(2 - 7x)^2 \cdot (-7) = 2940(2 - 7x)^2 \]
Thus, the second derivative of the function is:
\[ f''(x) = 2940(2 - 7x)^2 \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc8a57943-65a4-4179-bc90-4de17f531e4f%2F267b937c-943f-4dd8-be66-15e04b88c2ee%2Fdyl1c4c_processed.png&w=3840&q=75)
Transcribed Image Text:**Calculating the Second Derivative**
To find the second derivative of the function:
\[ f(x) = 5(2 - 7x)^4 \]
Step 1: Find the first derivative \( f'(x) \).
Use the chain rule for differentiation. Let \( u = 2 - 7x \). Then, \( f(x) = 5u^4 \) and
\[ \frac{d}{dx}[5u^4] = 5 \cdot 4u^3 \cdot \frac{du}{dx} \]
First, differentiate \( u = 2 - 7x \). The derivative of \( u \) with respect to \( x \) is:
\[ \frac{du}{dx} = \frac{d}{dx}(2 - 7x) = -7 \]
Hence,
\[ f'(x) = 5 \cdot 4(2 - 7x)^3 \cdot (-7) = -140(2 - 7x)^3 \]
Step 2: Find the second derivative \( f''(x) \).
To find the second derivative, differentiate \( f'(x) \) with respect to \( x \):
\[ f'(x) = -140(2 - 7x)^3 \]
Again, applying the chain rule, let \( u = 2 - 7x \), then:
\[ \frac{d}{dx}[-140u^3] = -140 \cdot 3u^2 \cdot \frac{du}{dx} \]
We already know \( \frac{du}{dx} = -7 \), hence,
\[ f''(x) = -140 \cdot 3(2 - 7x)^2 \cdot (-7) = 2940(2 - 7x)^2 \]
Thus, the second derivative of the function is:
\[ f''(x) = 2940(2 - 7x)^2 \]
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