a. Explanation Proof: The quotient rule for differentiation is given below, d d x [ f ( x ) g ( x ) ] = g ( x ) f ′ ( x ) − f ( x ) g ′ ( x ) [ g ( x ) ] 2 Let the value of function k ( x ) is, k ( x ) = f ( x ) g ( x ) (1) Substitute x + h for x in above equation. k ( x + h ) = f ( x + h ) g ( x + h ) (2) Subtract equation (1) from equation (2). k ( x + h ) − k ( x ) = f ( x + h ) g ( x + h ) − b. To determine To show: The equation k ( x + h ) − k ( x ) h = 1 g ( x + h ) g ( x ) [ ( f ( x + h ) − f ( x ) h ) g ( x ) − ( g ( x + h ) − g ( x ) h ) f ( x ) ] . c. To determine To prove: The derivative k ′ ( x ) = g ( x ) f ′ ( x ) − f ( x ) g ′ ( x ) [ g ( x ) ] 2 .

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ISBN: 9781337825597

Author: Tan

Publisher: INTER CENG

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Question

Chapter 3.2, Problem 74E

To determine

To prove: The result k(x+h)−k(x)h=f(x+h)g(x)−f(x)g(x+h)hg(x+h)g(x) .

To determine

To show: The equation k(x+h)−k(x)h=1g(x+h)g(x)[(f(x+h)−f(x)h)g(x)−(g(x+h)−g(x)h)f(x)] .

To determine

To prove: The derivative k′(x)=g(x)f′(x)−f(x)g′(x)[g(x)]2 .

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Chapter 3.2, Problem 73E

Chapter 3.2, Problem 1TE

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Chapter 3 Solutions

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To prove: The result k ( x + h ) − k ( x ) h = f ( x + h ) g ( x ) − f ( x ) g ( x + h ) h g ( x + h ) g ( x ) .

Solution Summary: The author explains that the quotient rule for differentiation is given below.

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To prove: The result k(x+h)−k(x)h=f(x+h)g(x)−f(x)g(x+h)hg(x+h)g(x) .

To show: The equation k(x+h)−k(x)h=1g(x+h)g(x)[(f(x+h)−f(x)h)g(x)−(g(x+h)−g(x)h)f(x)] .

To prove: The derivative k′(x)=g(x)f′(x)−f(x)g′(x)[g(x)]2 .

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Chapter 3 Solutions