# d/dx e^x

• 11-02-2003
DavidP
d/dx e^x
Why does

d/dx e^x = e^x

xe^(x-1)
• 11-02-2003
confuted
Because you can't use the power rule with functions raised to a variable.

d/dx k^x = k^(bx)*ln k*b
• 11-02-2003
DavidP
you mean constants raised to a variable?
• 11-03-2003
Govtcheez
Quote:

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.
• 11-03-2003
whistlenm1
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.
• 11-03-2003
alpha
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.
• 11-03-2003
confuted
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
• 11-03-2003
Silvercord
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)
• 11-03-2003
confuted
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.
• 11-04-2003
Zach L.
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.