Theorem: Quotient Rule If f(x) = T(x) B(x) is the quotient of differentiable functions, then. f'(x) = B'(x)T'(x) – T'(x)B′(x) - [B'(x)]2 T'(x) f'(x) = B'(x) f'(x) = f'(x) = B(x)T'(x)+T(x)B′(x) [B(x)]2 - B(x)T'(x) T(x)B'(x) [B(x)]2 B(x)T'(x) – T(x)B′(x) - f'(x) = [B'(x)]² B'(x)T(x) T'(x)B(x) f'(x) = [B(x)]2
Theorem: Quotient Rule If f(x) = T(x) B(x) is the quotient of differentiable functions, then. f'(x) = B'(x)T'(x) – T'(x)B′(x) - [B'(x)]2 T'(x) f'(x) = B'(x) f'(x) = f'(x) = B(x)T'(x)+T(x)B′(x) [B(x)]2 - B(x)T'(x) T(x)B'(x) [B(x)]2 B(x)T'(x) – T(x)B′(x) - f'(x) = [B'(x)]² B'(x)T(x) T'(x)B(x) f'(x) = [B(x)]2
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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Question
![Theorem: Quotient Rule
If f(x)
=
T(x)
B(x)
is the quotient of differentiable functions, then
B'(x)T'(x) T'(x)B'(x)
-
f'(x) =
[B'(x)]2
T'(x)
f'(x) =
B'(x)
f'(x)
=
B(x)T'(x) + T(x)B′(x)
[B(x)]2
-
B(x)T'(x) T(x)B'(x)
f'(x) =
=
[B(x)]²
B(x)T'(x) – T(x)B′(x)
-
f'(x) =
[B'(x)]²
B'(x)T(x) T'(x)B(x)
f'(x) =
[B(x)]²](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3e723b24-d59d-4231-b966-9a7027336941%2Fd3a9ffcc-99fe-4957-a5e7-14b8df9fb486%2Frbqo8a8_processed.png&w=3840&q=75)
Transcribed Image Text:Theorem: Quotient Rule
If f(x)
=
T(x)
B(x)
is the quotient of differentiable functions, then
B'(x)T'(x) T'(x)B'(x)
-
f'(x) =
[B'(x)]2
T'(x)
f'(x) =
B'(x)
f'(x)
=
B(x)T'(x) + T(x)B′(x)
[B(x)]2
-
B(x)T'(x) T(x)B'(x)
f'(x) =
=
[B(x)]²
B(x)T'(x) – T(x)B′(x)
-
f'(x) =
[B'(x)]²
B'(x)T(x) T'(x)B(x)
f'(x) =
[B(x)]²
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