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Thread: partial derivative

  1. #1
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    partial derivative

    I have

    $\displaystyle x^2\ln(xy+4)$


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

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

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


    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.

    $\displaystyle x^2\ln(xy+4)$

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

    $\displaystyle a^2\ln(ay+4)$

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

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


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

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

    subsitute x back in for a:

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

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

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

    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

    $\displaystyle (x^3*\frac{1}{1-xy})*\frac{\partial}{\partial y}
    (1-xy)$

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

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

    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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