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Quick help for ridiculously easy question...

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    What is \frac{\partial}{\partial y}\left{(x)}\right?

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    The differential of x with respect to y.
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    Man that is ridiculously easy.
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    (Original post by 117r)
    Man that is ridiculously easy.
    Just pull the trigger and fire the slug through my brain so that I may be enlightened in my moment of death as to what the answer to this 'mickey mouse' question is.

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    I don't think you understand what a differential is.
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    (Original post by electriic_ink)
    I don't think you understand what a differential is.
    It's 0, isn't it? Since x is treated as a constant.

    Just tell me the answer you patronising piece of cabbage!
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    Well, it's just the rate of change of x with respect to y...
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    Let me break it down for you eggheads,

    s=x. I am trying to find \frac{\partial{s}}{\partial{y}}

    (Original post by electriic_ink)
    I don't think you understand what a differential is.
    I don't think you know what a differential is. Because if you did, you would have provided me with some insight rather than write out that God forsaken useless 'I don't think you understand...' reply!

    (Original post by 117r)
    Well, it's just the rate of change of x with respect to y...
    Okay, I know that. Care to answer my question?
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    FFS, why doesn't someone just answer his question?

    You can't say what it is without further information.

    eg if x=y^2 then it's 2y. If x and y are independent variables, then it's 0.
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    (Original post by Dirac Delta Function)
    If x and y are independent variables, then it's 0.
    Thank you sir! :hat:
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    0 is answer
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    (Original post by GreenLantern1)
    0 is answer
    Thank you sir! :hat:
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    i know it sounds stupid but can someone explain this to me?
    i thought the answer was 1 :/
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    (Original post by HuyTaa)
    i know it sounds stupid but can someone explain this to me?
    i thought the answer was 1 :/
    f(x, y) = x. Then \frac{\partial f}{\partial y} = \displaystyle \lim _{h\to 0} \frac{f(x, y+h) - f(x, y)}{h} = 0 since f(x, y+h) = f(x, y) = x.

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