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Math Help - partial derivative

  1. #1
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    partial derivative

    I have

     x^2\ln(xy+4)


     f_x = 2x\ln(xy+4)+\frac{x^2y}{xy+4}

    Thats correct i think but this next one im confused about.

     f_y = 2x\ln(xy+4)+\frac{xy^2}{xy+4}


    Is that correct , if not can someone explain to me how to differentiate that function with respect to y?
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  2. #2
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    There's only 1 function in terms of y.

    x^2\ln(xy+4)

    if you view x as a constant let's replace x with a.

    a^2\ln(ay+4)

    The only variable is y, does that help?
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  3. #3
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    so  f_y = \frac{1}{y+4}

    ??
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  4. #4
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    err you dropped some constants.


    f(x)=a^2\ln(ay+4)

    f_y= a^2 \frac{1}{ay+4}*a

    subsitute x back in for a:

    f_y=x^2\frac{1}{xy+4}*x

    =x^3\frac{1}{xy+4}
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  5. #5
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    so if i had  x^3ln(1-xy)

     f_y =\frac{x^4}{1-xy}

    Would that be correct?
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  6. #6
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    um . . . chain rule the constant in front of y is -x

    (x^3*\frac{1}{1-xy})*\frac{\partial}{\partial y}<br />
(1-xy)

    (\frac{x^3}{1-xy})*(-x)

    \frac{-x^4}{1-xy}

    Does that make sense to you?
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  7. #7
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    yeah woops forgot the negative , yip that makes sense , cheers mate.
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