Hey Pit, I was just wondering who could answer a quick AP Calculus problem.

It has to do with Derivatives of Exponential functions; I know that the derivative of (e^x)=(e^x) , but I don't know what to do with the coefficients and such, when dealing with more complex problems.

Basically, I'd like to know how to go about doing a problem like:

Find an algebraic expression for f'(x) given f(x)=(e^5)-4e^x

Any help is appreciated.
(by the way, it's hard to see the f prime symbol with this font, but the problem is finding the derivative of the function).

Sorry, mistyped the first time.
Last edited by manmanman133 at Nov 9, 2008,
There's no x. It's a constant. Should be 0.
there is no x in that expression so the derrivative is zero
Quote by magic medicine
there is no x in that expression so the derrivative is zero

Sorry, a typo, it's corrected now.
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The new one's derivative is -4e^x.
derivitave will just be -4e^x , it doesnt change
e^5 is a constant, so it turns into 0-int(4e^x)
4 is constant, so you get -4int(e^x)
which is -4e^x
Thanks all, I've got it now.
He's not integrating, but it's right nonetheless.
-4x(e^[x-1])
No no no no and no to the user above me...
This is how you do it...

f(x)=e^u
f '(x)=u'e^u