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Math Help - differentiation question

  1. #1
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    differentiation question

    Could somebody help me 'total differentiate' this question please?

    K = 12x^3y + 10x^2y^2 + 6xy^3 +4y^4

    I don't really know where to begin this is what i have down

    fx = 36x^2 y + 20xy^2 + 6y^3 +4y^4
    fy = 12x^3 + 10x^2(2y) + 6x(3y^2) + 16y^3

    then im lost on the total differential formula
    DK = dk/dx (DK) + dk/dy (DY)

    (where the small 'd' represents the differential symbol

    am i completely lost?
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  2. #2
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    Quote Originally Posted by oligo View Post
    Could somebody help me 'total differentiate' this question please?

    K = 12x^3y + 10x^2y^2 + 6xy^3 +4y^4

    I don't really know where to begin this is what i have down

    fx = 36x^2 y + 20xy^2 + 6y^3 +4y^4
    fy = 12x^3 + 10x^2(2y) + 6x(3y^2) + 16y^3

    then im lost on the total differential formula
    DK = dk/dx (DK) + dk/dy (DY)

    (where the small 'd' represents the differential symbol

    am i completely lost?
    The total differential is gven by

    dK = \frac{\partial K}{\partial x} \,dx +\frac{\partial K}{\partial y} \,dy
    where

    <br />
K_x = 36x^2 y + 20xy^2 + 6y^3
    <br />
K_y = 12x^3 + 10x^2(2y) + 6x(3y^2) + 16y^3<br />
    (note the differences in our derivatives)
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  3. #3
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    If F(x,y) is a function of x and y, and x and y are themselves functions of some variable t, then we could write F(t)= F(x(t),y(t)) as a function of t and, by the chain rule, \frac{dF}{dt}= \frac{\partial F}{\partial x}\frac{dx}{dt}+ \frac{\partial F}{\partial y}\frac{dy}{dt}. Since the differential of a function F(t) is defined as [tex]dF= \frac{dF}{dt}dt[tex], that gives, as the "total differential" dF= \frac{\partial F}{\partial x}dx+ \frac{\partial F}{\partial y}dy

    You almost have the correct partial derivatives
    Kx = 36x^2 y + 20xy^2 + 6y^3 +4y^4
    is not quite right. The derivative of 4y^4 with respect to x is 0, not 4y^4.

    Ky = 12x^3 + 10x^2(2y) + 6x(3y^2) + 16y^3
    is correct- once you have multiplied 10(2)= 20 and 6(3)= 18.

    Then, to find dK by simply writing
    dK= (36x^2 y + 20xy^2 + 6y^3)dx+ (12x^3 + 20x^2y + 18xy^2 + 16y^3)dy
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