Articles
Register
Sign In
Search
LikeADemonicMind
@LikeADemonicMind
May 2019
1
137
Report
Bonjour! Est-ce que quelqu'un pourrait m'aider à dériver cette fonction en m'expliquant la méthode s'il-vous-plaît?
f(t) =
Merci d'avance!
Please enter comments
Please enter your name.
Please enter the correct email address.
Agree to
terms and service
You must agree before submitting.
Send
Lista de comentários
scoladan
Verified answer
Bonjour,
de manière générale :
[fog(x)]' = [f(g(x))]' = g'(x) x f'(g(x))
Avec f(x) = cos(x) et g(x) = √(αt + β) :
soit f'(x) = -sin(x)
[cos(√(αt + β))]' = [√(αt + β)]' x -sin(√(αt + β))
De même, si on pose u(x) = √x et v(x) = αt + β :
u'(x) = 1/2√x et v'(x) = α
⇒ [√(αt + β)]' = α x 1/2√(αt + β)
Au final :
[cos(√(αt + β))]'
= -α/2√(αt + β) x sin(√(αt + β))
0 votes
Thanks 0
More Questions From This User
See All
LikeADemonicMind
January 2021 | 0 Respostas
Responda
LikeADemonicMind
January 2021 | 0 Respostas
Responda
LikeADemonicMind
January 2021 | 0 Respostas
Responda
LikeADemonicMind
January 2021 | 0 Respostas
Responda
LikeADemonicMind
May 2019 | 0 Respostas
Responda
LikeADemonicMind
May 2019 | 0 Respostas
Responda
LikeADemonicMind
May 2019 | 0 Respostas
Responda
×
Report "Bonjour! Est-ce que quelqu'un pourrait m'aider à dériver cette fonction en m'expliquant la méthode s.... Pergunta de ideia de LikeADemonicMind"
Your name
Email
Reason
-Select Reason-
Pornographic
Defamatory
Illegal/Unlawful
Spam
Other Terms Of Service Violation
File a copyright complaint
Description
Helpful Links
Sobre nós
Política de Privacidade
Termos e Condições
direito autoral
Contate-Nos
Helpful Social
Get monthly updates
Submit
Copyright © 2024 ELIBRARY.TIPS - All rights reserved.
Lista de comentários
Verified answer
Bonjour,de manière générale :
[fog(x)]' = [f(g(x))]' = g'(x) x f'(g(x))
Avec f(x) = cos(x) et g(x) = √(αt + β) :
soit f'(x) = -sin(x)
[cos(√(αt + β))]' = [√(αt + β)]' x -sin(√(αt + β))
De même, si on pose u(x) = √x et v(x) = αt + β :
u'(x) = 1/2√x et v'(x) = α
⇒ [√(αt + β)]' = α x 1/2√(αt + β)
Au final :
[cos(√(αt + β))]'
= -α/2√(αt + β) x sin(√(αt + β))