Explanation Given information : f g ( x ) = f ( x ) g ( x ) is a function where g ( x ) ≠ 0 . Formula used : f ′ ( x ) = lim Δ x → 0 f ( x + Δ x ) − f ( x ) Δ x Calculation : f ′ ( x ) = lim Δ x → 0 f ( x + Δ x ) − f ( x ) Δ x Replace f by f g in above equation, ( f g ) ′ ( x ) = lim Δ x → 0 f g ( x + Δ x ) − f g ( x ) Δ x Simplifying right-hand expression, f g ( x + Δ x ) − f g ( x ) Δ x = f ( x + Δ x ) g ( x + Δ x ) − f ( x ) g ( x ) Δ x = f ( x + Δ x ) g ( x ) − g ( x + Δ x ) f ( x ) Δ x g ( x + Δ x ) g ( x ) = f ( x + Δ x ) g ( x ) − g ( x + Δ x ) f ( x ) Δ x Δ x g ( x + Δ x ) g ( x ) Δ x [ Divide by <

7th Edition

ISBN: 9781524916817

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Chapter 3.2, Problem 56PS

To determine

To prove : (fg)′=gf′−fg′g2

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Chapter 3.2, Problem 55PS

Chapter 3.2, Problem 57PS

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Chapter 3 Solutions

Calculus

Ch. 3.1 - Prob. 1PS Ch. 3.1 - Prob. 2PS Ch. 3.1 - Prob. 3PS Ch. 3.1 - Prob. 4PS Ch. 3.1 - Prob. 5PS Ch. 3.1 - Prob. 6PS Ch. 3.1 - Prob. 7PS Ch. 3.1 - Prob. 8PS Ch. 3.1 - Prob. 9PS Ch. 3.1 - Prob. 10PS

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To prove : ( f g ) ′ = g f ′ − f g ′ g 2

Solution Summary: The author explains how to prove that (fg)prime is a function.

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To prove : (fg)′=gf′−fg′g2

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Chapter 3 Solutions