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Differnentiating e^x

gimme5minutes · 2009-05-12 13:45:25

Ok so if you differentiate e^x you get e^x. And if you differentiate e^kx, you get ke^kx. So what do you get if you differentiate e^(x-1)? It looks like (x-1)e^(x-1), but that seems almost too simple...but is it correct?

12-05-2009 02:45PM

#1

gimme5minutes

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Ok so if you differentiate e^x you get e^x. And if you differentiate e^kx, you get ke^kx. So what do you get if you differentiate e^(x-1)?

It looks like (x-1)e^(x-1), but that seems almost too simple...but is it correct?

0

Comments

#2 12-05-2009 03:03PM

Thomas_S_Hunterson

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Chain rule

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#3 12-05-2009 03:16PM

gimme5minutes

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So e(x-1) would be (x-1)e(x-2).e(x-1)? Does that look alright?

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#4 12-05-2009 04:03PM

Grudaire

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gimme5minutes wrote: »

So e(x-1) would be (x-1)e(x-2).e(x-1)? Does that look alright?

No, you want to differentiate e^u, where u = x-1
u = x - 1 => du = dx

So integral{e^(x-1)dx} = integral{e^u du} = e^u + c = e^(x - 1) + c

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