Why does
d/dx e^x = e^x
instead of
xe^(x-1)
Why does
d/dx e^x = e^x
instead of
xe^(x-1)
Because you can't use the power rule with functions raised to a variable.
d/dx k^x = k^(bx)*ln k*b
Away.
you mean constants raised to a variable?
Either one.Originally posted by DavidP
you mean constants raised to a variable?
Instead of using a shortcut rule, actually put it through the long definition of a derivative and work it out.
-Govtcheez
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for me I did not understand it until I found a proof, so unless your going to look at a proof take it as is.
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just take it as it is. but if you want to know why, as cheez said, put it through the long definition of the derivative.
http://archives.math.utk.edu/visual....definition.12/ The definition of a derivative can be found there if you want to have a go at it
Away.
so is this setup what cheez and you guys meant:
(e^(x+h) - e^(x)) / (h)
Lim h->0
I saw what confuted put for the answer but I'm not 100% sure the steps you have to take to get there.
EDIT: does b stand for base above? (change of base)
Last edited by Silvercord; 11-03-2003 at 04:38 PM.
b and k were arbitrary constants.
(e^(x+h) - e^(x)) / (h)
Lim h->0
(e^((x+h)/x)) / (h)
Lim h->0
You'll have to take an ln() in there to get that to something you can work with.
Away.
Use the Maclaurin series for e^x, and it becomes really obvious:
e^x = sum(k=0, infinity) [ x^k / k! ]
So for each term,
d[x^k / k!]/dx = k*x^(k-1) / k! = x^(k-1) / (k-1)
That is, when you take the derivative, each term becomes its predecessor, so the series remains unchanged.
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