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Math Help - Youngs theorum.

  1. #1
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    Youngs theorum.

    Says  f_{xy}=f_{yx}

    The original equation is
     f(x,y)=x^3\ln(1-xy)

    So for the first derivative with respect to x ,I used the product rule on first part then the quotient rule on second part to get the derivative below.

    f_x= 3x^2\ln(1-xy)-\frac{x^3y}{1-xy}

    Then for this one I just had to differentiate with respect to y which made it this.
    f_y=\frac{x^4}{1-xy}

    However i want to find  f_{xy}
    But i dont quite understand what it means , and the answers i get do not match each other according to youngs theorum.


    Any suggestions?
    Youngs theorum =  f_{xy}=f_{yx}
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  2. #2
    MHF Contributor
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    Quote Originally Posted by el123 View Post
    Says  f_{xy}=f_{yx}

    The original equation is
     f(x,y)=x^3\ln(1-xy)

    So for the first derivative with respect to x ,I used the product rule on first part then the quotient rule on second part to get the derivative below.

    f_x= 3x^2\ln(1-xy)-\frac{x^3y}{1-xy} Correct

    Then for this one I just had to differentiate with respect to y which made it this.
    f_y=\frac{x^4}{1-xy} Correct if you make it negative

    However i want to find  f_{xy}
    But i dont quite understand what it means , and the answers i get do not match each other according to youngs theorum.

    Any suggestions?
    Youngs theorum =  f_{xy}=f_{yx}
    OK, so what have you got so far for...

    differentiating your f_x with respect to y (holding x constant) i.e. f_{yx}

    ... and...

    differentiating your f_y with respect to x (holding y constant) i.e. f_{xy}

    ...?



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