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

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

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Theorem: Product Rule ||f f(x) = F(x)S(x) is the product of differentiable functions, then - S(x)F'(x) F(x)S'(x) f'(x) = = [S(x)]² ○ f'(x) = F(x)S' (x) + S(x)F' (x) O f'(x) = F'(x)S' (x) O f'(x) = F'[S(x)]S" (x) ○ f'(x) = F(x)S' (x) — S(x)F'(x) O f'(x) = F'(x)S' (x) + S(x)F(x)
Transcribed Image Text:Theorem: Product Rule ||f f(x) = F(x)S(x) is the product of differentiable functions, then - S(x)F'(x) F(x)S'(x) f'(x) = = [S(x)]² ○ f'(x) = F(x)S' (x) + S(x)F' (x) O f'(x) = F'(x)S' (x) O f'(x) = F'[S(x)]S" (x) ○ f'(x) = F(x)S' (x) — S(x)F'(x) O f'(x) = F'(x)S' (x) + S(x)F(x)
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