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Thread: another partial derivative

  1. #1
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    another partial derivative

    $\displaystyle 3x^2\ln(1-xy)$

    i want to differentiate with respect to y.

    So.
    $\displaystyle [(0)(\ln(1-xy)]+[(3x^2)(-x)]$

    which is $\displaystyle -3x^3$


    Is that correct?
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  2. #2
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    again you're running into this same problem where you want to do a partial derivative in terms of y, but you're not holding x as a constant.

    what type of function of y are you needing to differentiate (ie. polynomial, quotient etc.)
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by el123 View Post
    $\displaystyle 3x^2\ln(1-xy)$

    i want to differentiate with respect to y.

    So.
    $\displaystyle [(0)(\ln(1-xy)]+[(3x^2)(-x)]$

    which is $\displaystyle -3x^3$


    Is that correct?
    Just treat $\displaystyle x$ like a number. $\displaystyle 3x^2$ and $\displaystyle x$ are just constants.

    If you were derivating $\displaystyle c_1\ln(1-c_2 y)$, what would you get? You'd get $\displaystyle \frac{c_1}{1-c_2 y}\cdot -c_2$. Just extend that notion to this problem.
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  4. #4
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    so its $\displaystyle \frac{3x^3}{1-xy}$

    Is that what your saying?
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  5. #5
    Super Member redsoxfan325's Avatar
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    You have to multiply that by $\displaystyle \frac{d}{dx}[1-xy]=-x$, to complete the chain rule, so the answer is $\displaystyle \frac{-3x^3}{1-xy}$.
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