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)
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter9: Multivariable Calculus
Section9.CR: Chapter 9 Review
Problem 54CR
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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)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2c634e35-7043-49c5-9dee-7b7e699d3d45%2F1e7b337f-6285-454d-9757-17f00cc1160e%2Fy3dcg2_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
-
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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