Finding $\frac{\partial ^8 f}{\partial x^4\partial y^4}$ > Given the function $f(x,y)=\frac{1}{1-xy}$ find the value of$\frac{\partial ^8 f}{\partial x^4\partial y^4}(0,0)$. First I developed the function into a taylor series using geometric series around $(0,0)$: $$f(x,y)=\frac{1}{1-xy}=\displaystyle{\sum_{n=0}^\infty(xy)^n}$$. The taylor approximation of order 8 is given by $$f(x,y)\thickapprox\displaystyle{\sum_{n=0}^4(xy)^n}=1+xy+(xy)^2+(xy)^3+(xy)^4$$, where the degree of the last element is 8 so indeed we got the approximation. It's the taylor series by its singularity. looking at the element in original taylor series deriving 4 times by x and 4 by y is the only term which is with coefficient 1 (all other derivatives exist twice since around $(0,0)$ f is continuous and so its derivatives). So prima facie $$\frac{\partial ^8 f}{\partial x^4\partial y^4}=8!\cdot 1=40320$$ but mypad claims that $\frac{\partial ^8 f}{\partial x^4\partial y^4}=576$. Where am I mistaken?

$$\frac{\partial^4 x^4y^4}{\partial x^4}=y^4\frac{\partial^4 x^4}{\partial x^4}=24y^4$$ $$\frac{\partial^4 24y^4}{\partial y^4}=24\times 24 = 576$$