Explanation y = 1 x 2 + 1 1 x 2 + 1 : x = 0 Critical points are points where the function is defined and derivative is zero undefined Find where f ’ ( x ) equal to zero or undefined f ' ( x ) = − 2 x ( x 2 + 1 ) 2 d d x ( 1 x 2 + 1 ) Apply exponent rule 1 a = a − 1 = d d x ( ( x 2 + 1 ) − 1 ) Apply the chin rule d f ( u ) d x = d f d u ⋅ d u d x f = u − 1 , u = x 2 + 1 = d d u ( u − 1 ) d d x ( x 2 + 1 ) d d u ( u − 1 ) = − 1 u 2 d d u ( u − 1 ) Apply the power rule: d d x ( x a ) = a ⋅ x a − 1 = − 1 ⋅ u − 1 − 1 Simplify − 1 ⋅ u − 1 − 1 : − 1 u 2 − 1 ⋅ u − 1 − 1 = − 1 ⋅ 1 u 2 Multiply : 1 ⋅ 1 u 2 = 1 u 2 = − 1 u 2 d d x ( x 2 + 1 ) = 2 x d d x ( x 2 + 1 ) = d d x ( x 2 ) + d d x ( 1 ) d d x ( x 2 ) = 2 x d d x ( x 2 ) Apply the Power Rule : d d x ( x a ) = a ⋅ x a − 1 = 2 x 2 − 1 Simplify =2x d d x ( 1 ) = 0 d d x ( 1 ) Derivative of a constant :

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 4.3, Problem 41E

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y=1x2+1

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Chapter 4.3, Problem 40E

Chapter 4.3, Problem 42E

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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y = 1 x 2 + 1

Solution Summary: The author explains that critical points are points where the function is defined and derivative is zero undefined.

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