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Re: Simply derivative question, Math 5.
*To*: mathgroup at smc.vnet.net
*Subject*: [mg50257] Re: [mg50240] Simply derivative question, Math 5.
*From*: Murray Eisenberg <murray at math.umass.edu>
*Date*: Mon, 23 Aug 2004 06:34:25 -0400 (EDT)
*Organization*: Mathematics & Statistics, Univ. of Mass./Amherst
*References*: <200408220419.AAA10290@smc.vnet.net>
*Reply-to*: murray at math.umass.edu
*Sender*: owner-wri-mathgroup at wolfram.com
As the Mathematica Book (printed or electronic) explains, ALL built-in
objects in Mathematica have names that begin with an upper-case letter.
So the Napierian "e" is represented by E, not e. (Or, if you want it
to look more like conventional mathematical notation, type Esc e Esc
to obtain the Mathematica stylized e; in fact, you'll see that stylized
e in the result.)
So Mathematica was treating e as an unknown constant and gave the
correct answer.
When e = E, then Log[e] = 1, so no "extra" factor:
D[(1 + c E^t)/(1 - c E^t), t] // Simplify
By the way, you don't need an explicit multiplication operator after the
c here -- just a space (so Mathematica can see you have two
single-letter names rather than a two-letter name cE).
It's REALLY worth a couple hours of your time to read the early material
in the Mathematica Book! It will save you a lot of time avoiding such
basic difficulties.
Ted Kahn wrote:
> Hello- I am trying to take the derivative of the following function:
>
> (1 + c*e^t)/(1 - c*e^t)
>
> with respect to t.
>
> ================
> \!\(Simplify[D[\(1 + c\ e\^t\)\/\(1 - c\ e\^t\), t]]\)
> ============
>
> The answer I get includes an "extra" Log[e] in the numerator. Am I not
> using the program correctly or am I not understanding the answer
> correctly? Other?
>
> thanks, -ted
--
Murray Eisenberg murray at math.umass.edu
Mathematics & Statistics Dept.
Lederle Graduate Research Tower phone 413 549-1020 (H)
University of Massachusetts 413 545-2859 (W)
710 North Pleasant Street fax 413 545-1801
Amherst, MA 01003-9305
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