Explanation Given information: The given function is g ( t ) = ( 2 t + 3 ) 2 ( 3 t 2 − 1 ) − 3 . Formula used: (1) Product rule: The derivative of the function h ( x ) = [ f ( x ) g ( x ) ] is, h ′ ( x ) = f ( x ) d d x g ( x ) + g ( x ) d d x f ( x ) (2) Chain rule for power function: The derivative of h ( x ) = [ f ( x ) ] n where n is a real number is, h ′ ( x ) = n [ f ( x ) ] n − 1 f ′ ( x ) Calculation: The derivative of the given function is, g ′ ( t ) = d d t ( 2 t + 3 ) 2 ( 3 t 2 − 1 ) − 3 g ′ ( t ) = ( 2 t + 3 ) 2 ⋅ d d t ( 3 t 2 − 1 ) − 3 + ( 3 t 2 − 1 ) − 3 ⋅ d d t ( 2 t + 3 ) 2 ( Use product rule ) Use chain rule for the power function to simplify it. g ′ ( t ) = ( 2 t + 3 ) 2 ⋅ [ − 3 ( 3 t 2 − 1 ) − 3 − 1 d d t ( 3 t 2 − 1 ) ] + ( 3 t 2 − 1 ) − 3 [ 2 ( 2 t + 3 ) 2 − 1 d d t ( 2 t + 3 ) ] = ( 2 t + 3 ) 2 [ − 3 ( 3 t 2 − 1 ) − 4 ( 6 t ) ] + ( 3 t 2 − 1 )

Applied Calculus for the Managerial, Life, and Social Sciences

10th Edition

ISBN: 9781305887831

Author: Tan

Publisher: CENGAGE CO

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Chapter 3.3, Problem 48E

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The derivative of the given function.

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Chapter 3.3, Problem 47E

Chapter 3.3, Problem 49E

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Chapter 3 Solutions

Applied Calculus for the Managerial, Life, and Social Sciences

Ch. 3.1 - State the following rules of differentiation in...Ch. 3.1 - Prob. 2CQ Ch. 3.1 - Prob. 3CQ Ch. 3.1 - Prob. 4CQ Ch. 3.1 - Prob. 1E Ch. 3.1 - Prob. 2E Ch. 3.1 - Prob. 3E Ch. 3.1 - Prob. 4E Ch. 3.1 - Prob. 5E Ch. 3.1 - Prob. 6E

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