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 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 T'(x) f'(x) = B'(x) О 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

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
icon
Related questions
Question

quesion is attached in ss

thnaks for hle

paephaehp

aeohkpe

hoek

tkw

tptk

jow

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
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
T'(x)
f'(x) =
B'(x)
О
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
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 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 T'(x) f'(x) = B'(x) О 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
Expert Solution
steps

Step by step

Solved in 3 steps with 1 images

Blurred answer
Recommended textbooks for you
Calculus For The Life Sciences
Calculus For The Life Sciences
Calculus
ISBN:
9780321964038
Author:
GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:
Pearson Addison Wesley,
Trigonometry (MindTap Course List)
Trigonometry (MindTap Course List)
Trigonometry
ISBN:
9781337278461
Author:
Ron Larson
Publisher:
Cengage Learning
Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:
9781133382119
Author:
Swokowski
Publisher:
Cengage