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Who's good at calculus?

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jambone
Posts: 934 - Karma: 11
My calc professor offered 20 points extra credit on the final to anyone that could differentiate a chain rule problem, using the derivative.  (His example was xe^x).  I consider myself pretty good at math, but after an hour of trying I just can't get it.

So could anyone show the steps to differentiate xe^x using the  definition of the derivative (f(x+h)-f(x))/h
Dec 6 2007 6:58PM
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Pman13a
Posts: 3800 - Karma: 452
i can tell you its x(e^x)^2
Dec 6 2007 7:10PM
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Seths22
Posts: 272 - Karma: -3
oh christ.  the power rule makes everything so much easier


Dec 6 2007 7:13PM
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Jhorski
Posts: 648 - Karma: 45
No its x(e^x)+(e^x)
Dec 6 2007 7:16PM
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rhcp
Posts: 2598 - Karma: 18
uhh no im pretty sure its derivative is e^x + xe^x
Dec 6 2007 7:17PM
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rhcp
Posts: 2598 - Karma: 18
for a second i was an idiot and thought it was logarithmic differentiation and was like "damn, you have hard high school teachers"
Dec 6 2007 7:18PM
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almostaskier
Posts: 13809 - Karma: 93
I'm at lim h->o (xe^x+h)-(e^x)/h + e^x+h

The right side is what you want... But i don't know how to cancel your - e^x and h... ill keep trying.
Dec 6 2007 7:28PM
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spliff.Life
Posts: 7847 - Karma: 1,549
im confused
Dec 6 2007 7:30PM
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4-Front
Posts: 5306 - Karma: 199
Well since you have to do it by the definition the teacher is really just testing your algebra skills.   I am assuming that you know what the definiton of the derivative is.  If you do just solve the limit.
Dec 6 2007 7:31PM
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jason...
Posts: 9006 - Karma: 780
theres a grade 10 in my drama class thats doing college calculus and getting 97% he might be able to help u
Dec 6 2007 7:53PM
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hurrisun
Posts: 6284 - Karma: 49
fuck calculus haha
Dec 6 2007 8:00PM
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TheQuailman
Posts: 7382 - Karma: 147

I would ask another teacher at your school.  Or at least try yahoo answers

Dec 6 2007 8:05PM
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deezy.
Posts: 2538 - Karma: 26
yeah ask asian allen or something
Dec 6 2007 8:06PM
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attitashfreestyle29
Posts: 3070 - Karma: 209
This kids right. To do the derivative of xe^x you use the product rule which is the derivative of u times v plus the derivative of v times u. u is x and v is e^x. So the derivative of x is 1 so the first term is e^x then the derivative of e^x is just e^x. so the second term is xe^x. so you get e^x + xe^x. And this isn't a chain rule problem it's a product rule problem. You could probably find that using the definition of the derivative, but that's why more work and impossible to explain online
Dec 6 2007 8:40PM
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Tom.
Posts: 1954 - Karma: 12
ummm... use the product rule?
Dec 6 2007 8:43PM
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rhcp
Posts: 2598 - Karma: 18
sorry i read it wrong.

alright, so you want to find  (fg)', and this can be done by taking the two limits of f and g and combining them.  so

f'(x) = lim   f(x+h)g(x+h) - f(x)g(x+h)+f(x)+g(x+h)-f(x)g(x)
         h->0 ---------------------------------------------------------
                                                h

which simplifies to


f'(x) = lim [f(x+h)-f(x)]g(x+h) + f(x)[g(x+h) - g(x)]
          h->0----------------------------------------------
                                          h

which simplifies to something you can handle

f'(x) = lim   f(x+h) - f(x)  *   g(x+h)   +   f(x)   *  g(x+h)-g(x)
          h->0-------------                                     -----------------
                      h                                                        h  

and i trust you can figure that out by yourself.
Dec 6 2007 8:55PM
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rhcp
Posts: 2598 - Karma: 18
Nope, not impossible.
Dec 6 2007 9:38PM
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freeski07
Posts: 696 - Karma: 47
isnt it xe^(x-1)

product rule is bring down the power then subtract one from the power on top
i dont see where u gotta use the chain rule there.....
Dec 6 2007 9:58PM
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Chummer3
Posts: 599 - Karma: 9
x*e^x does not require the chain rule, you use the product rule. you end up getting (x+1)e^x
Dec 6 2007 10:11PM
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SirFryanator
Posts: 11149 - Karma: 1,415
You can definitely solve it but it would take about a page and a half and it's not really explainable online.
Dec 6 2007 10:29PM
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rhcp
Posts: 2598 - Karma: 18
why are you people so mystified?  Its really not that bad.
Dec 6 2007 10:53PM
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Chummer3
Posts: 599 - Karma: 9
ya i dont know what the big deal is 20 extra credit points for that wow i wish i had your teacher. Okay so product rule says:

x*(e^x)' + x' * (e^x), with ' meaning derivative

that gives

x*e^x + 1* (e^x), or
(1+x)e^x

someone else explained it right in this thread also, i forget who
Dec 6 2007 10:58PM
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SirFryanator
Posts: 11149 - Karma: 1,415
no, it's not bad. But it takes a lot of paper to do it and I really don't like to type out math equations online.
Dec 6 2007 11:18PM
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