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Math Help - Total Differentiation; total mess.

  1. #1
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    Total Differentiation; total mess.

    Hello all,

    A question about total differentiation is doing my head in. It is:

    u = x^5 - 10x^3y^2 + 5xy^4 and y = 1 + x^2

    It wants me to show that du/dx is some expression in x involving x's up to the power 8. But my technique gives rise to x's up to the power 12. I've been using du = (partial du/dx)dx + (partial du/dy)dy, and dividing through by dx, leaving du/dx on the left and then multiples of dx/dx (=1) and dy/dx (=2x) on the right.

    Am I doing this wrong? Any help appreciated.
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  2. #2
    Super Member Random Variable's Avatar
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    Are you sure you copied the right problem?

     \frac{du}{dx} = \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} \frac{dy}{dx}

     = 5x^{4} -30x^{2}y^{2}+5y^{4} + (-20x^{3}y + 20xy^{3})(2x)

     = 5x^{4} -30x^{2}y^{2} + 5y^{4} -40x^{4}y + 40 x^{2}y^{3}
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  3. #3
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    You seem to have arrived at the same thing as me - the last term would appear to yield a term in x^12 when the substitution for y is used. This is a copy paste from the actual question on the website (i'd link you but you need a university account to get in):

    Given that u = x5 -10x3y2 + 5xy4, and y = 1 + x2, show that
    du
    dx
    = 5 + 30x2 + 55x4 + 70x6 + 45x8

    The formatting isn't perfect but you can see what they think the answer is.

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  4. #4
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    Never mind - I think every time I was taught about indices must have been a friday.

    But anyway, assuming it does all work out, am I using the correct method?
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  5. #5
    Super Member Random Variable's Avatar
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    Yeah, I forgot to substitute for y.


    5x^{4} -30x^{2}y^{2} + 5y^{4} -40x^{4}y + 40 x^{2}y^{3}

     = 5x^{4} -30x^{2} (1+2x^{2}+x^{4}) + 5 (1+4x^{2}+6x^{4} +40x^{6} +x^{8}) -4x^{4}  (1+x^{2}) + 40x^{2}(1+3x^{2}+3x^{4}+x^{6})

     = 5 + 30x^{2} + +55x^{4} +70x^{6}+  45x^{8}



    In general, if  f(t) = f\big(x_{1}(t),x_{2}(t),...,x_{n}(t)\big)

    then  \frac{df}{dt} = \frac{\partial f}{\partial x_{1}} \frac{dx_{1}}{dt}+  \frac{\partial f}{\partial x_{2}} \frac{dx_{2}}{dt} + ... +  \frac{\partial f}{\partial x_{n}} \frac{dx_{n}}{dt}


    In this particular problem we have  u(x) = u\big(x,y(x)\big)

    so  \frac{du}{dx} = \frac{\partial u}{\partial x} \frac{dx}{dx}+  \frac{\partial u}{\partial y} \frac{dy}{dx} = \frac{\partial u}{\partial x} +  \frac{\partial u}{\partial y} \frac{dy}{dx}
    Last edited by Random Variable; May 13th 2010 at 01:33 PM.
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