Explanation Given: The function, y = e α x sin β x . 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 ( cos x ) = − sin x d d x ( sin x ) = cos x Derivative of any constant C is zero, that is, d d x ( C ) = 0 Calculation: In order to find the derivatives of the function, use chain rule andproduct rule of derivatives and simplify further to get y ' = d d x ( e α x sin β x ) = e α x ⋅ d d x ( sin β x ) + sin β x ⋅ d d x ( e α x ) = e α x ⋅ ( cos β x ⋅ d d x ( β x ) ) + sin β x ⋅ ( e α x ⋅ d d x ( α x ) ) = e α x ⋅ ( cos β x ⋅ ( β ) ) + sin β x ⋅ ( e α x ⋅ ( α ) ) = e α x β cos β x + e α x α sin β x = e α x ( β cos β x + α sin β x ) So, the first derivative is y ' = e α x ( β cos β x + α sin β x )

15th Edition

ISBN: 9781305177291

Author: James Stewart, Troy Day

Publisher: Cengage Learning

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Question

Chapter 3.5, Problem 39E

To determine

The firstderivative y' and second derivative y'' of the function, y=eαxsinβx .

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Chapter 3.5, Problem 38E

Chapter 3.5, Problem 40E

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Chapter 3 Solutions

Biocalculus: Calculus For Life Sciences

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The firstderivative y ' and second derivative y ' ' of the function, y = e α x sin β x .

Solution Summary: The author explains the first and second derivatives of the function, y=ealpha xmathrmsinbeta.

The firstderivative y' and second derivative y'' of the function, y=eαxsinβx .

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Chapter 3 Solutions