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Step 2 Now write the derivative in terms of x by recalling that u = x³ = x + 8. [eu] = (eu) du dx [ex³-x+8] = (x³ -x + 8) (

Step 2 Now write the derivative in terms of x by recalling that u = x³ = x + 8. [eu] = (eu) du dx [ex³-x+8] = (x³ -x + 8) (

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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Step 2 Now write the derivative in terms of \( x \) by recalling that \( u = x^3 - x + 8 \). \[ \frac{d}{dx} \left[ e^u \right] = \left( e^u \right) \frac{du}{dx} \] \[ \frac{d}{dx} \left[ e^{x^3 - x + 8} \right] = \left( e^{x^3 - x + 8} \right) \frac{d}{dx} \left( \underline{\phantom{u}} \right) \]
Transcribed Image Text:Step 2 Now write the derivative in terms of \( x \) by recalling that \( u = x^3 - x + 8 \). \[ \frac{d}{dx} \left[ e^u \right] = \left( e^u \right) \frac{du}{dx} \] \[ \frac{d}{dx} \left[ e^{x^3 - x + 8} \right] = \left( e^{x^3 - x + 8} \right) \frac{d}{dx} \left( \underline{\phantom{u}} \right) \]
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