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Math Help - Partial derivative

  1. #1
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    Partial derivative

     3x^2 \ln(1-xy)

    the partial derivatvie with respect to y is

     \frac{x}{1-xy}

    is that right?
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  2. #2
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    Quote Originally Posted by el123 View Post
     3x^2 \ln(1-xy)

    the partial derivatvie with respect to y is

     \frac{x}{1-xy}

    is that right?
    Not quite.

    It's actually

    3x^2\left(\frac{-x}{1 - xy}\right)

     = -\frac{3x^3}{1 - xy}.
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  3. #3
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    but i thought you treat x as a constant so its derivative is 0?
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    Quote Originally Posted by el123 View Post
    but i thought you treat x as a constant so its derivative is 0?
    Yes, it is treated like a constant.

    What do you do to differentiate a constant times a function?

    It's the constant times the derivative of that function.

    i.e. \frac{d}{dx}\left[c\cdot f(x)\right] = c\cdot f'(x).


    So in your case, you have 3x^2 \ln{(1 - xy)}

    The 3x^2 is your constant.

    So the partial derivative will be

    3x^2 \frac{\partial}{\partial y}\left[\ln{(1 - xy)}\right].
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  5. #5
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    true to that , thank you very much !
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