Theorem: Product Rule ||f 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) 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 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) 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
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
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)
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)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2c634e35-7043-49c5-9dee-7b7e699d3d45%2Fc58024c6-cb6c-4f7d-b3fa-ed40344e8da7%2Fqiybau_processed.png&w=3840&q=75)
Transcribed Image Text:Theorem: Product Rule
||f 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)
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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