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Math Help - How to find the partial derivative of this using the definition?

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    How to find the partial derivative of this using the definition?

    f(x,y) = 2(x^2)y - 4x/y

    Prove that f'y (with respect to y) = 2x^2 + 4x/(y^2) using the definition of a partial derivative (limit)
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    Quote Originally Posted by AlphaRock View Post
    f(x,y) = 2(x^2)y - 4x/y

    Prove that f'y (with respect to y) = 2x^2 + 4x/(y^2) using the definition of a partial derivative (limit)
    To take the derivative with respect to y, you treat x as a constant and use the standard derivative rules.


    f(x, y) = 2x^2y - \frac{4x}{y}

     = 2x^2y^1 - 4x\,y^{-1}.


    \frac{\partial f}{\partial y} = 1\cdot 2x^2 y^{1-1} - 1(-4x)y^{-1-1}

     = 2x^2y^0 + 4x\,y^{-2}

     = 2x + \frac{4x}{y^2}.
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    Quote Originally Posted by Prove It View Post
    To take the derivative with respect to y, you treat x as a constant and use the standard derivative rules.


    f(x, y) = 2x^2y - \frac{4x}{y}

     = 2x^2y^1 - 4x\,y^{-1}.


    \frac{\partial f}{\partial y} = 1\cdot 2x^2 y^{1-1} - 1(-4x)y^{-1-1}

     = 2x^2y^0 + 4x\,y^{-2}

     = 2x + \frac{4x}{y^2}.
    Thanks for your help, Prove it.

    How would we find f'y using limits?
    The definition of this partial derivative would be:

    lim (f(x,y+k) - f(x,y))/k
    k->0
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    Quote Originally Posted by AlphaRock View Post
    Thanks for your help, Prove it.

    How would we find f'y using limits?
    The definition of this partial derivative would be:

    lim (f(x,y+k) - f(x,y))/k
    k->0
    The calculation is tedious but straightforward.

    \frac{\partial f}{\partial y} = \lim_{h\to 0}\frac{f(x,y+h)-f(x,y)}{h}

    =\lim_{h\to  0}\frac{2x^2(y+h)-\frac{4x}{y+h}-(2x^2y-\frac{4x}{y})}{h}

    =\lim_{h\to   0}\frac{2hx^2-\frac{4x}{y+h}+\frac{4x}{y}}{h}

    =\lim_{h\to   0}\bigg(2x^2+\frac{1}{h}\bigg(\frac{-4x(y)}{y(y+h)}+\frac{4x(y+h)}{y(y+h)}\bigg)\bigg)

    =\lim_{h\to   0}\bigg(2x^2+\frac{-4xy+4xy+4xh}{h(y)(y+h)}\bigg)

    =\lim_{h\to   0}\bigg(2x^2+\frac{4x}{(y)(y+h)}\bigg)

    =2x^2+\frac{4x}{(y)(y+0)}

    =2x^2+\frac{4x}{y^2}
    Last edited by undefined; April 6th 2010 at 09:28 PM. Reason: (lots of) typos, formatting
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    \frac{f(x, y + h) - f(x, y)}{h}

     = \frac{2x^2(y + h) - \frac{4x}{y + h} - \left(2x^2y - \frac{4x}{y}\right)}{h}

     = \frac{2x^2h + \frac{4xh}{y(y + h)}}{h}

     = \frac{\frac{2x^2h\,y(y + h) + 4h}{y(y + h)}}{h}

     = \frac{2x^2h\,y(y + h) + 4xh}{h\,y(y + h)}

     = \frac{h[2x^2y(y + h) + 4x]}{h\,y(y + h)}

     = \frac{2x^2y(y + h) + 4x}{y(y + h)}.


    Now making h \to 0 we get

    \frac{2x^2y^2 + 4x}{y^2}

     = 2x^2 + \frac{4x}{y^2}.
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