1. #1

    What is the First Derivative of this:

    34x^12 - 3x^5 + 23xy^3 - y^2 - xy

    If you can answer it (correctly), you get a cookie!
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  2. #2
    Total or partial and wrt to x or y ?

  3. #3
    Bloodsail Admiral Wass's Avatar
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    I have no idea if I understand your question right since you got two unkown variables in the same function, and we don't really deal with that (so far) at the uni. But sure, I'll have a shot at that;

    D(34x^12) is obv 34*12x^11 = 408x^11

    D(3x^5) is also obv 3*5x^4 = 15x^4

    23xy^3, here is a little trickier one (albeit still easy for me if I understand your question right). if we are supposed to derive both x and y (which I don't really know if we are, your first question doesn't tell me as much) you got a specific rule as to how that works, I have no idea what it's called in English though. But basically f(x)*f(y) (or normally f(x)*g(x)) would have the derivative f'(x)*f(y) + f(x)*f'(y). This would result in 23*y^3 + 23x*3y^2, which is equal to 23y^2(y + 3x).

    D(y^2) = 2y

    D(xy) = y + x (using the same rule as mentioned above.)

    Now you got:

    408x^11 - 15x^4 + 23y^2(y + 3x) - 2y - (y + x) = 408x^11 - 15x^4 + 23y^2(y + 3x) - 2y - y - x

    Bunched together:

    408x^11 - 15x^4 + 23y^2(y + 3x) - 3y - x, which could be written a bit differently but not necessarily easier. Your teacher should be able to accept that solution, if that is what it's all about.

    Note, I did not use any program like mathlab or wolframalpha for this. Pure head calculus.

    However I'm still not sure if I understood your question correctly. Might be that this is meant to be some kind of special case with two unknown variables, and in that case I'm probably not gonna be of much use since I haven't gotten there yet.
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    I don't do calculus for anything less than double chocolate chip.

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  5. #5
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    Quote Originally Posted by schwarzkopf View Post
    Total or partial and wrt to x or y ?
    This is the correct response to the OP.
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  6. #6
    (d)/(dx)(-3 x^5+34 x^12-x y-y^2+23 x y^3) = 408 x^11-15 x^4+23 y^3-y ?

  7. #7
    Quote Originally Posted by Wass View Post
    if we are supposed to derive both x and y (which I don't really know if we are, your first question doesn't tell me as much) you got a specific rule as to how that works, I have no idea what it's called in English though. But basically f(x)*f(y) (or normally f(x)*g(x)) would have the derivative f'(x)*f(y) + f(x)*f'(y).
    This is called the product rule.

    In English we use boring names for them. We have the product rule, the quotient rule, the chain rule, and the power rule. There's probably more, but those are the ones I know.

  8. #8
    assuming it's a function f with x and y being variables:
    d(f)/d(x)=408x^11-15x^4+23y^3-y
    d(f)/d(y)=69xy^2-2y-x

  9. #9
    Quote Originally Posted by Jaylock View Post
    34x^12 - 3x^5 + 23xy^3 - y^2 - xy

    If you can answer it (correctly), you get a cookie!
    Assuming

    y = 34x^12 - 3x^5 + 23xy^3 - y^2 - xy

    then assuming you're taking dy/dx (means taking derivative of function y with respect to x, and because of y being in the equation, means it's a two unknown variable equation).

    So.

    dy/dx = 408x^11 - 15x^4 + 23y^3 + 69xy^2dy/dx - 2ydy/dx - y -xdy/dx

    Need to have the dy/dx variables on one side, and other variables on another side.

    so.

    subtract 69xy^2dy/dx to the other side, and add 2ydy/dx to the other side, and add xdy/dx to the other side.

    you get.

    dy/dx - 69xy^2dy/dx + 2ydy/dx + xdy/dx = 408x^11 - 15x^4 + 23y^3 - y

    factor dy/dx out of left side

    dy/dx(1 - 69xy^2 + 2y + x) = 408x^11 - 15x^4 + 23y^3 - y

    divide (1 - 69xy^2 + 2y + x) to the other side.

    And finally you end up with:

    dy/dx = (408x^11 - 15x^4 + 23y^3 - y) / (1 - 69xy^2 + 2y + x)

    Meh, with two unknown variables, should be the way to do it.



    Of course, I could be completely wrong, seeing as the op didn't have y = that function, so it could be a differential equation needing to be set to zero, and then variables switched to the appropriate sides, then finally be able to derive it....buuuuut, I don't feel like doing that. Or wass had the correct answer.

    Calculus!
    Last edited by Hastings; 2013-02-23 at 04:49 AM.
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