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Trash Talk - Derivatives

Purge - Wed Sep 26, 2007 12:49 am
Post subject: Derivatives
Long time no see. sa_tongue.gif

Anyway, I'm having trouble with one of my Calculus problems (it's not really a problem, just theory).

If any of you can recall derivatives of functions, you probably know that e^x is its own derivative. It is also known that only e^x is its own derivative. However, according to my teacher, there is another function NOT involving any exponential functions that is its own derivative. I just need to know this other function and how it can be proved.

I personally think he's mistaken, but this is an actual assignment... so I'm looking for a logical answer. icon_confused.gif
Blocks - Wed Sep 26, 2007 12:58 am
Post subject: it's
0

As for the proof, iono, use induction or something.
Dr Brain - Wed Sep 26, 2007 8:51 am
Post subject:
f(x) = 0 is it's own derivative, yes.

It can be proved using the linearity property, or just go straight to:

df/dt = (f(x) - f(x+dt))/t

plug in zero, and get out zero.
tcsoccerman - Wed Sep 26, 2007 4:09 pm
Post subject:
simple enough
BDwinsAlt - Wed Sep 26, 2007 6:09 pm
Post subject:
I hate functions. Useless... icon_rolleyes.gif
Samapico - Wed Sep 26, 2007 8:00 pm
Post subject:
......functions........useless????? you're a retard or what?
Purge - Wed Sep 26, 2007 8:16 pm
Post subject:
It's not f(x) = 0, according to my teacher.

He didn't tell the solution since it's due on Friday, so more thinking for me. sa_tongue.gif
CypherJF - Wed Sep 26, 2007 8:35 pm
Post subject:
actually proving calculus functions begins around abstract algebra level; which is way past the point where you run out of alpha-letters and you have to use greek. tongue.gif
Blocks - Thu Sep 27, 2007 2:50 am
Post subject:
I believe the only class of functions that are their on derivatives are of the form f(x) = c*e^x, where c is an arbitrary constant. If you substitute the Taylor series expansion of e^x, you technically not involving exponentials, but if you teacher means that, he is a wonk.
Animate Dreams - Fri Oct 12, 2007 11:37 am
Post subject:
CypherJF wrote:
actually proving calculus functions begins around abstract algebra level; which is way past the point where you run out of alpha-letters and you have to use greek. tongue.gif


Yeah, I suggest skipping your Math classes and just go straight to Ancient Greek. Much easier, and much more fun.
Blocks - Fri Oct 12, 2007 2:17 pm
Post subject:
Hey Purge, how did this all turn out?
SamHughes - Wed Oct 17, 2007 1:40 pm
Post subject:
Blocks wrote:
I believe the only class of functions that are their on derivatives are of the form f(x) = c*e^x, where c is an arbitrary constant.


Yep, you discover that by solving the differential equation f' = f and applying the uniqueness theorem.
Purge - Sun Oct 21, 2007 9:24 pm
Post subject:
Blocks wrote:
Hey Purge, how did this all turn out?


Well, I turned in my paper which was the proof of f(x) = 0 to be its own derivative, and I got full credit for it... and he said that it wasn't f(x) = 0.

Oh well, I'm not complaining. sa_tongue.gif

Thanks for the help, by the way.
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