Explanation y = 2 x + 1 x 2 + 1 x = − 1 + 5 2 , x = 5 − 1 2 f ' ( x ) = − 2 x 2 − 2 x + 2 ( x 2 + 1 ) 2 d d x ( 2 x + 1 x 2 + 1 ) Apply the Quotient Rule : ( f g ) ' = f ' . g − g ' . f g = d d x ( 2 x + 1 ) ( x 2 + 1 ) − d d x ( x 2 + 1 ) ( 2 x + 1 ) ( x 2 + 1 ) 2 d d x ( 2 x + 1 ) = 2 d d x ( 2 x + 1 ) Apply the Sum/Difference Rule : ( f ± g ) ' = f ' ± g ' = d d x ( 2 x ) + d d x ( 1 ) d d x ( 2 x ) = 2 Apply the common derivative : d d x ( x ) = 1 = 2 ⋅ 1 Simplify =2 d d x ( 1 ) = 0 =2+0 =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 =2x+0 =2x = 2 ( x 2 + 1 ) − 2 x ( 2 x + 1 ) ( x 2 + 1 ) 2 Expand 2 ( x 2 + 1 ) − 2 x ( 2 x + 1 ) : − 2 x 2 − 2 x + 2 2 ( x 2 + 1 ) − 2 x ( 2 x + 1 ) Expand 2 ( x 2 + 1 ) : 2 x 2 + 2 Expand 2 ( x 2 + 1 ) : 2 x 2 + 2 2 ( x 2 + 1 ) Distribute parentheses using : a ( b + c ) = a b + a c a = 2 , b = x 2 , c = 1 = 2 ⋅ x 2 + 2 ⋅ 1 <

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 4.3, Problem 42E

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y=2x+1x2+1

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Chapter 4.3, Problem 41E

Chapter 4.3, Problem 43E

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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 = 2 x + 1 x 2 + 1

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