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Mathematics 19 Online
OpenStudy (anonymous):

How do I differentiatie e to the power minus x? So let's say: g(x) = e^-x. In steps please, I'm somewhat confused...

myininaya (myininaya):

u=u(x) \[(e^u)'=u'e^u \]

myininaya (myininaya):

So we have u=-x => u'=-1 just plug in

OpenStudy (anonymous):

let -x =y, hence dy/dx = -1, hence (d/dx)e^-x = (d/dy)e^y.(dy/dx) = e^y.(-1) = -e^y = -e^-x

OpenStudy (anonymous):

Merci!

myininaya (myininaya):

au revoir ( i don't know how to say your welcome in french :()

OpenStudy (anonymous):

Haha, me neither :)

myininaya (myininaya):

lol

OpenStudy (anonymous):

I got it, Bienvenue!!!

OpenStudy (anonymous):

Can I always "plug" in x = -1?

OpenStudy (anonymous):

no you need to check your substitution and use the du/dx as myininaya showed

OpenStudy (anonymous):

Hmm, okay.

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