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df Calculate if f(u) = 6u², u(2) = −5, and u' (2) = −5. dx x=2 (Give your answer as a whole or exact number.) df dx \x=2

df Calculate if f(u) = 6u², u(2) = −5, and u' (2) = −5. dx x=2 (Give your answer as a whole or exact number.) df dx \x=2

Calculus: Early Transcendentals
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:** \[ \left. \frac{df}{dx} \right|_{x=2} = 300 \]
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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