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01-21-2010   #1 (permalink)
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Default 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)
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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.
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01-21-2010   #3 (permalink)
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That's what I did at first, but it said I was wrong @,@
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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)
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Okay, I finally got it.
F_x = xe^(-x^2).
Thanks SilentSaber and Toasty!
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