Explanation d d θ ( sin 2 ( θ ) ) + d d θ ( sin ( θ ) ) d d θ ( sin 2 ( θ ) ) Apply the chain rule : d f ( u ) d x = d f d u · d u d x f = u 2 , u = sin ( θ ) = d d u ( u 2 ) d d θ ( sin ( θ ) ) = d d u ( u 2 ) Apply the Power Rule : d d x ( x a ) = a · x a − 1 = 2 u 2 − 1 =2u d d θ ( sin ( θ ) ) Apply the common derivative : d d θ ( sin ( θ ) ) = cos ( θ ) = cos ( θ ) = 2 u cos ( θ ) = 2 sin ( &

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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The extreme points of a function are the maximum and the minimum points of the function. A maximum is attained when the function takes the maximum value and a minimum is attained when the function takes the minimum value.

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In calculus, concavity is a descriptor of mathematics that tells about the shape of the graph. It is the parameter that helps to estimate the maximum and minimum value of any of the functions and the concave nature using the graphical method. We use the …

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Chapter 4.3, Problem 47E

To determine

Find the critical point and endpoint at the minimum and maximum value of the function of y=sin2 θ+sin θ.

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Chapter 4.3, Problem 46E

Chapter 4.3, Problem 48E

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Chapter 4 Solutions

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Ch. 4.1 - Prob. 1PQ Ch. 4.1 - Prob. 2PQ Ch. 4.1 - Prob. 3PQ Ch. 4.1 - Prob. 4PQ Ch. 4.1 - Prob. 1E Ch. 4.1 - Prob. 2E Ch. 4.1 - Prob. 3E Ch. 4.1 - Prob. 4E Ch. 4.1 - Prob. 5E Ch. 4.1 - Prob. 6E

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Find the critical point and endpoint at the minimum and maximum value of the function of y = sin 2 θ + sin θ .

Solution Summary: The author explains how to determine the critical point and endpoint at the minimum and maximum value of the function of y=sin2.

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Find the critical point and endpoint at the minimum and maximum value of the function of y=sin2 θ+sin θ.

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Chapter 4 Solutions