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Thread: Youngs theorum.

  1. #1
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    Youngs theorum.

    Says $\displaystyle f_{xy}=f_{yx}$

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

    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.

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

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

    However i want to find $\displaystyle f_{xy}$
    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 = $\displaystyle f_{xy}=f_{yx}$
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  2. #2
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    Quote Originally Posted by el123 View Post
    Says $\displaystyle f_{xy}=f_{yx}$

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

    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.

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

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

    However i want to find $\displaystyle f_{xy}$
    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 = $\displaystyle f_{xy}=f_{yx}$
    OK, so what have you got so far for...

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

    ... and...

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

    ...?



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