Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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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.
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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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