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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![The image shows a mathematical expression that poses a calculus problem. The text reads:
"If y = e^(x^2 - 3x), then y' ="
To solve for y', the derivative of y with respect to x, you will need to use the chain rule. The function y is of the form e^u, where u is a function of x, specifically u = x^2 - 3x.
Using the chain rule:
1. Differentiate y = e^u with respect to u to get dy/du = e^u.
2. Then differentiate u = x^2 - 3x with respect to x to get du/dx = 2x - 3.
3. Combine these results to find dy/dx = dy/du * du/dx = e^(x^2 - 3x) * (2x - 3).
Thus, the derivative y' is:
\[ y' = e^{x^2 - 3x} \cdot (2x - 3) \]
This problem introduces the concept of differentiation of composite functions and demonstrates the application of the chain rule in calculus.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9973d75d-32e3-4ebe-9432-aeaa4634e542%2Fa05bfd4f-7073-4f65-8fe6-3411d617c793%2Fcds19l_processed.png&w=3840&q=75)
Transcribed Image Text:The image shows a mathematical expression that poses a calculus problem. The text reads:
"If y = e^(x^2 - 3x), then y' ="
To solve for y', the derivative of y with respect to x, you will need to use the chain rule. The function y is of the form e^u, where u is a function of x, specifically u = x^2 - 3x.
Using the chain rule:
1. Differentiate y = e^u with respect to u to get dy/du = e^u.
2. Then differentiate u = x^2 - 3x with respect to x to get du/dx = 2x - 3.
3. Combine these results to find dy/dx = dy/du * du/dx = e^(x^2 - 3x) * (2x - 3).
Thus, the derivative y' is:
\[ y' = e^{x^2 - 3x} \cdot (2x - 3) \]
This problem introduces the concept of differentiation of composite functions and demonstrates the application of the chain rule in calculus.
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