Explanation Given: Function h ( θ ) = θ ( θ − 1 2 − θ − 2 ) Formula Used: Formulas of derivatives are as follows: d d x ( c ) = 0 , where c is a constant. d d x ( x n ) = n x n − 1 d d x { c . f ( x ) } = c . d d x f ( x ) Calculation: Function is given as, h ( θ ) = θ ( θ − 1 2 − θ − 2 ) h ( θ ) = ( θ 1 − 1 2 − θ 1 − 2 ) h ( θ ) = θ 1 2 − θ − 1 Derivative of the above function is calculated as: d d θ h ( θ ) = d d θ ( θ 1 2 − θ − 1

Applied Calculus, 6e WileyPLUS + Loose-leaf

6th Edition

ISBN: 9781119408901

Author: Deborah Hughes-Hallett, Andrew M. Gleason, Patti Frazer Lock, Daniel E. Flath

Publisher: Wiley (WileyPLUS Products)

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Chapter 3.1, Problem 29P

To determine

To find derivative of the below function:

h(θ)=θ(θ−12−θ−2)

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Chapter 3.1, Problem 28P

Chapter 3.1, Problem 30P

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Applied Calculus, 6e WileyPLUS + Loose-leaf

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To find derivative of the below function: h ( θ ) = θ ( θ − 1 2 − θ − 2 )

Solution Summary: The author explains the formula used to find derivatives of the above function.

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To find derivative of the below function:

h(θ)=θ(θ−12−θ−2)

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Chapter 3 Solutions

To find derivative of the below function: h ( θ ) = θ ( θ − 1 2 − θ − 2 )

Solution Summary: The author explains the formula used to find derivatives of the above function.

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To find derivative of the below function: h(θ)=θ(θ−12−θ−2)

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Chapter 3 Solutions

To find derivative of the below function:

h(θ)=θ(θ−12−θ−2)