Explanation y = x 3 x 2 + 1 f ' ( x ) = x 4 + 3 x 2 ( x 2 + 1 ) 2 d d x ( x 3 x 2 + 1 ) Apply the Quotient Rule : ( f g ) = f ' g − g ' . f g = d d x ( x 3 ) ( x 2 + 1 ) − d d x ( x 2 + 1 ) x 3 ( x 2 + 1 ) 2 d d x ( x 3 ) = 3 x 2 d d x ( x 3 ) Apply the Power Rule : d d x ( x a ) = a ⋅ x a − 1 = 3 x 3 − 1 = 3 x 2 d d x ( x 2 + 1 ) = 2 x d d x ( x 2 + 1 ) Apply the Sum/Difference Rule : ( f ± g ) ' = f ' ± g ' = 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 =2x d d x ( 1 ) = 0 d d x ( 1 ) Derivative of a constant : d d x ( a ) = 0 =0 =2x+0 =2x Simplify 3 x 2 ( x 2 + 1 ) − 2 x x 3 ( x 2 + 1 ) 2 : x 4 + 3 x 2 ( x 2 + 1 ) 2 3 x 2 ( x 2 + 1 ) − 2 x x 3 ( x 2 + 1 ) 2 3 x 2 ( x 2 + 1 ) − 2 x x 3 = 3 x 2 ( x 2 + 1 ) − 2 x 4 3 x 2 ( x 2 + 1 ) − 2 x x 3

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 4.3, Problem 43E

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y=x3x2+1

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Chapter 4.3, Problem 42E

Chapter 4.3, Problem 44E

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Chapter 4 Solutions

Calculus - Standalone book

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 = x 3 x 2 + 1

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