Explanation Given: The function, y = e − t ( t 2 − 2 t + 2 ) . Concept Used: The product rule of derivatives: d d x [ f ( x ) ⋅ g ( x ) ] = f ( x ) ⋅ d d x g ( x ) + g ( x ) ⋅ d d x f ( x ) Chain rule, d d x ( f ( g ( x ) ) ) = f ' ( g ( x ) ) ⋅ g ' ( x ) Power rule of derivatives, d d x ( x n ) = n x n − 1 d d x ( e x ) = e x Derivative of any constant C is zero, that is, d d x ( C ) = 0 Calculation: In order to find the derivative of the function, use product rule and chain rule of derivatives and simplify further to get y ' = d d t ( e − t ( t 2 − 2 t + 2 ) ) = e − t ⋅ d d t ( t 2 − 2 t + 2 ) + ( t 2 − 2 t + 2 ) ⋅ d d t ( e − t ) = e − t ⋅ ( 2 t 2 − 1 − 2 ⋅ 1 + 0 ) + ( t 2 − 2 t + 2 ) ⋅ ( e − t

15th Edition

ISBN: 9781133109631

Author: Stewart, JAMES, Day, Troy

Publisher: Cengage Learning,

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Chapter 3, Problem 22E

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The derivative y' of the function, y=e−t(t2−2t+2) .

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Chapter 3, Problem 21E

Chapter 3, Problem 23E

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Chapter 3 Solutions

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The derivative y ' of the function, y = e − t ( t 2 − 2 t + 2 ) .

Solution Summary: The author explains how to find the derivative of the function using the product rule and chain rule of derivatives.

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The derivative y' of the function, y=e−t(t2−2t+2) .

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Chapter 3 Solutions