Why does
d/dx e^x = e^x
instead of
xe^(x-1)
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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
you mean constants raised to a variable?
Either one.Quote:
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.
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.
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
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)
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.
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.