DavidP

11-02-2003, 11:27 PM

Why does

d/dx e^x = e^x

instead of

xe^(x-1)

d/dx e^x = e^x

instead of

xe^(x-1)

View Full Version : d/dx e^x

DavidP

11-02-2003, 11:27 PM

Why does

d/dx e^x = e^x

instead of

xe^(x-1)

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

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.

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)

(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.

(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.

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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