Explanation Given information: The child’s dosage is represents as, D ( t ) = a t t + 12 (1) In the above equation a is the adult dosage in milligrams and t is the child’s age in years and the adult dosage of a substance is 500 mg that mean a = 500 . Formula used : The quotient rule of differentiation is, f ′ ( x ) = d d x [ f ( x ) g ( x ) ] = g ( x ) ⋅ d d x f ( x ) − f ( x ) ⋅ d d x g ( x ) [ g ( x ) ] 2 The above formula is defined when the denominator g ( x ) ≠ 0 . Calculation: Section1: Substitute 500 for a in the equation (1) to get D ( t ) in terms of t , D ( t ) = 500 t t + 12 (2) Use the quotient rule of differentiation for the equation (2) to get the rate of change of child’s dosage with respect to child’s age that is D ′ ( t ) . D ′ ( t ) = d d t [ 500 t t + 12 ] = ( t + 12 ) ⋅ d d t ( 500 t ) − ( 500 t ) ⋅ d d t ( t + 12 ) ( t + 12 ) 2 = ( t + 12 ) ⋅ ( 500 ) − ( 500 t ) ⋅ ( 1 ) ( t + 12 ) 2 = 500 t + 6000 − 500 t ( t + 12 ) 2 ( Apply distributive property ) Further solve the above equation for D ′ ( t )

Applied Calculus for the Managerial, Life, and Social Sciences (MindTap Course List)

10th Edition

ISBN: 9781305657861

Author: Soo T. Tan

Publisher: Cengage Learning

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Chapter 3.2, Problem 54E

To determine

To find: The rate of change of child’s dosage with respect to child’s age and the rate of change of child’s dosage for a 6 year-old and 10 year-old child.

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Chapter 3 Solutions

Applied Calculus for the Managerial, Life, and Social Sciences (MindTap Course List)

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The rate of change of child’s dosage with respect to child’s age and the rate of change of child’s dosage for a 6 year-old and 10 year-old child.

Solution Summary: The author explains the rate of change of child's dosage with respect to child’s age, and the quotient rule of differentiation.

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To find: The rate of change of child’s dosage with respect to child’s age and the rate of change of child’s dosage for a 6 year-old and 10 year-old child.

Chapter 3 Solutions