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: Chain Rule Application
**Problem Statement:**
Calculate \(\left. \frac{df}{dx} \right|_{x=2}\) if \( f(u) = 6u^2 \), \( u(2) = -5 \), and \( u'(2) = -5 \).
*(Give your answer as a whole or exact number.)*
**Solution:**
Consider the given functions and derivatives:
1. \( f(u) = 6u^2 \)
2. \( u(2) = -5 \)
3. \( u'(2) = -5 \)
We need to find the derivative \(\left. \frac{df}{dx} \right|_{x=2}\).
Using the chain rule:
\[
\frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx}
\]
First, compute \(\frac{df}{du}\):
\[
f(u) = 6u^2 \implies \frac{df}{du} = 12u
\]
Next, evaluate \(\frac{df}{du}\) at \( u = u(2) = -5 \):
\[
\left. \frac{df}{du} \right|_{u = -5} = 12(-5) = -60
\]
We also know that \( \left. \frac{du}{dx} \right|_{x=2} = u'(2) = -5 \).
Now, multiply these values together:
\[
\left. \frac{df}{dx} \right|_{x=2} = \left( \left. \frac{df}{du} \right|_{u=-5} \right) \cdot \left( \left. \frac{du}{dx} \right|_{x=2} \right)
= (-60) \cdot (-5) = 300
\]
Therefore,
\[
\left. \frac{df}{dx} \right|_{x=2} = 300
\]
**Answer:**
\[
\left. \frac{df}{dx} \right|_{x=2} = 300
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F575d721b-26eb-4db6-af93-50c207cd3fab%2Ff6f49230-eb3f-448f-9f75-47cdbcd8649a%2Ftlj85tl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Calculus Problem: Chain Rule Application
**Problem Statement:**
Calculate \(\left. \frac{df}{dx} \right|_{x=2}\) if \( f(u) = 6u^2 \), \( u(2) = -5 \), and \( u'(2) = -5 \).
*(Give your answer as a whole or exact number.)*
**Solution:**
Consider the given functions and derivatives:
1. \( f(u) = 6u^2 \)
2. \( u(2) = -5 \)
3. \( u'(2) = -5 \)
We need to find the derivative \(\left. \frac{df}{dx} \right|_{x=2}\).
Using the chain rule:
\[
\frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx}
\]
First, compute \(\frac{df}{du}\):
\[
f(u) = 6u^2 \implies \frac{df}{du} = 12u
\]
Next, evaluate \(\frac{df}{du}\) at \( u = u(2) = -5 \):
\[
\left. \frac{df}{du} \right|_{u = -5} = 12(-5) = -60
\]
We also know that \( \left. \frac{du}{dx} \right|_{x=2} = u'(2) = -5 \).
Now, multiply these values together:
\[
\left. \frac{df}{dx} \right|_{x=2} = \left( \left. \frac{df}{du} \right|_{u=-5} \right) \cdot \left( \left. \frac{du}{dx} \right|_{x=2} \right)
= (-60) \cdot (-5) = 300
\]
Therefore,
\[
\left. \frac{df}{dx} \right|_{x=2} = 300
\]
**Answer:**
\[
\left. \frac{df}{dx} \right|_{x=2} = 300
\]
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