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Theorem: Product Rule If f(x) = F(x)S(x) is the product of differentiable functions, then f'(x) = S(x)F'(x) F(x)S' (x) [S(x)]² O f'(x) = F(x)S' (x) + S(x)F' (x) O f'(x) = F'(x)S' (x) + S(x)F(x) O f'(x) = F(x)S' (x) – S(x)F" (x) O f'(x) = F'[S(x)]S'(x) O f'(x) = F'(x)S" (x)

Theorem: Product Rule If f(x) = F(x)S(x) is the product of differentiable functions, then f'(x) = S(x)F'(x) F(x)S' (x) [S(x)]² O f'(x) = F(x)S' (x) + S(x)F' (x) O f'(x) = F'(x)S' (x) + S(x)F(x) O f'(x) = F(x)S' (x) – S(x)F" (x) O f'(x) = F'[S(x)]S'(x) O f'(x) = F'(x)S" (x)

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter1: Fundamental Concepts Of Algebra
Section1.3: Algebraic Expressions
Problem 20E
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Theorem: Product Rule If f(x) = F(x)S(x) is the product of differentiable functions, then f'(x) = S(x)F'(x) F(x)S' (x) [S(x)]² O f'(x) = F(x)S' (x) + S(x)F' (x) O f'(x) = F'(x)S' (x) + S(x)F(x) ○ f'(x) = F(x)S" (x) – S(x)F' (x) - ○ f'(x) = F'[S(x)]S′(x) O f'(x) = F'(x)S" (x)
Transcribed Image Text:Theorem: Product Rule If f(x) = F(x)S(x) is the product of differentiable functions, then f'(x) = S(x)F'(x) F(x)S' (x) [S(x)]² O f'(x) = F(x)S' (x) + S(x)F' (x) O f'(x) = F'(x)S' (x) + S(x)F(x) ○ f'(x) = F(x)S" (x) – S(x)F' (x) - ○ f'(x) = F'[S(x)]S′(x) O f'(x) = F'(x)S" (x)
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