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LikeADemonicMind
@LikeADemonicMind
May 2019
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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!
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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))
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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))