1. ## 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 = $\displaystyle f_{xy}=f_{yx}$

2. Originally Posted by el123
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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