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Math Help - Total Differential Problem

  1. #1
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    Total Differential Problem

    Hi

    I am trying to get the total differential for

    y = x1x2/2x1 + x2

    I have been given the answer

    dy = [x2/2x1 + x2]dx1 - [2x1x2/(2x1+x2)^2]dx1 + [x1/2x1 + x2]dx2 - [x1x2/(2x1+x2)^2]dx2

    Is this right?

    It is my understanding that the general formula is F1dx1 + F2dx2, for y=F(x1,x2)

    I don't understand why, in the second and fourth terms of the answer, the WHOLE of the denominator is being squared, rather than just the x1 or x2 that the partial derivative refers to......
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  2. #2
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    Re: Total Differential Problem

    Quote Originally Posted by econolondon View Post
    Hi

    I am trying to get the total differential for

    y = x1x2/2x1 + x2

    I have been given the answer

    dy = [x2/2x1 + x2]dx1 - [2x1x2/(2x1+x2)^2]dx1 + [x1/2x1 + x2]dx2 - [x1x2/(2x1+x2)^2]dx2

    Is this right?

    It is my understanding that the general formula is F1dx1 + F2dx2, for y=F(x1,x2)

    I don't understand why, in the second and fourth terms of the answer, the WHOLE of the denominator is being squared, rather than just the x1 or x2 that the partial derivative refers to......
    1. I assume that your function reads: \displaystyle{y = \frac{x_1 \cdot x_2}{2x_1+x_2}}

    2. You have to use the quotient rule: f = \frac uv~\implies~ f'=\frac{v \cdot u' - u \cdot v'}{v^2}

    3. I'll show you the F_1dx_1 part:

    The variable is x_1. x_2 is considered to be constant.

    F_1dx_1 = \frac{(2x_1+x_2) \cdot x_2 - x_1 \cdot x_2 \cdot 2}{(2x_1+x_2)^2}

    4. The F_2dx_2 part has to be done in excatly the same way.
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  3. #3
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    Re: Total Differential Problem

    Thanks, that's very helpful.
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