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DavidP
11-02-2003, 11:27 PM
Why does

d/dx e^x = e^x

instead of

xe^(x-1)

confuted
11-02-2003, 11:35 PM
Because you can't use the power rule with functions raised to a variable.

d/dx k^x = k^(bx)*ln k*b

DavidP
11-02-2003, 11:42 PM
you mean constants raised to a variable?

Govtcheez
11-03-2003, 06:11 AM
Originally posted by DavidP
you mean constants raised to a variable? Either one.

Instead of using a shortcut rule, actually put it through the long definition of a derivative and work it out.

whistlenm1
11-03-2003, 02:11 PM
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.

alpha
11-03-2003, 03:13 PM
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.

confuted
11-03-2003, 03:51 PM
http://archives.math.utk.edu/visual.calculus/2/definition.12/ The definition of a derivative can be found there if you want to have a go at it

Silvercord
11-03-2003, 04:34 PM
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)

confuted
11-03-2003, 05:34 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.

Zach L.
11-04-2003, 06:09 PM
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.