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 01-21-2010 #1 (permalink) and more came out Games   Join Date: Oct 2008 Location: On my toilet atm @_@ Posts: 902 Blog Entries: 6 iTrader: (0) IGN: BANNED! Multivariable Partial Derivatives I'm sorta lost on what to do for F_x. For F_y, I guessed 0, and it was correct. __________________
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 01-21-2010 #2 (permalink) ggFTW Stalker   Join Date: Oct 2008 Location: 自分の幻想 Posts: 1,425 iTrader: (0) Class: MME Level: -2147483647 Guild: YetAnotherUC well, I'll guess you just take the integral. So it would be: ∫(4,x)[te^(-t^2)dt] =-1/2(e^(-t^2))|(4, x) =-1/2(e^(-x^2))+1/2(e^(-16)) Or something like that. __________________ Human history is the story of complaisance. While disaster is fresh in our memory, we take precautions. But as the memory of disaster recedes, we start to take risks.
 01-21-2010 #3 (permalink) and more came out Games   Join Date: Oct 2008 Location: On my toilet atm @_@ Posts: 902 Blog Entries: 6 iTrader: (0) IGN: BANNED! That's what I did at first, but it said I was wrong @,@ __________________
 01-21-2010 #4 (permalink) ggFTW Lurker   Join Date: Jul 2009 Posts: 18 iTrader: (0) Thinking about it with one variable, suppose you have G(x) = ∫(4, x) [g(t) dt]. What is G'(x)? The same principle will apply for f_x in your question, especially since f(x, y) is really just a single-variable function (there is no y in your expression).
 01-21-2010 #5 (permalink) and more came out Games   Join Date: Oct 2008 Location: On my toilet atm @_@ Posts: 902 Blog Entries: 6 iTrader: (0) IGN: BANNED! Okay, I finally got it. F_x = xe^(-x^2). Thanks SilentSaber and Toasty! __________________

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