Second degree multi variable taylor polynomial Let f (x, y ) = x cos(πy ) − y sin(πx)  point: 1,2

I am following the standard formula, which starts with taking the partial of f with regards to x twice, which gives me:

ysin(πx)π  But plugging in 1,2 gives me 0 because sin(π) = 0. This isn't right, but I am not sure what I am doing wrong.

Example of formula: http://www.maa.org/press/periodicals/loci/resources/calcplot3d-an-exploration-environment-for-multivariable-calculus-taylor-polynomials-of-a-function-of

Thanks.
The Taylor series out to second-order terms for a function of two variables is $$
f(x,y)=f(x_0,y_0)+(x-x_0)f_x(x_0,y_0)+(y-y_0)f_y(x_0,y_0)+\frac12((x-x_0)^2f_{xx}(x_0,y_0)+2(x-x_0)(y-y_0)f_{xy}(x_0,y_0)+(y-y_0)^2f_{yy}(x_0,y_0))+O(\|(x,y)-(x_0,y_0)\|^3).
$$ Just as with the Taylor series for $\cos x$ at $0$, in which every other coefficient is $0$, it’s perfectly fine for some of these partial derivatives to be $0$. In this case, $f_y=-\sin{\pi x}-\pi x\sin{\pi y}=0$ and $f_{xx}=\pi^2y\sin{\pi x}=0$ at $(1,2)$, so the series won’t have a $(y-2)$ or $(x-1)^2$ term.
The Taylor series out to second-order terms for a function of two variables is $$
f(x,y)=f(x_0,y_0)+(x-x_0)f_x(x_0,y_0)+(y-y_0)f_y(x_0,y_0)+\frac12((x-x_0)^2f_{xx}(x_0,y_0)+2(x-x_0)(y-y_0)f_{xy}(x_0,y_0)+(y-y_0)^2f_{yy}(x_0,y_0))+O(\|(x,y)-(x_0,y_0)\|^3).
$$ Just as with the Taylor series for $\cos x$ at $0$, in which every other coefficient is $0$, it’s perfectly fine for some of these partial derivatives to be $0$. In this case, $f_y=-\sin{\pi x}-\pi x\sin{\pi y}=0$ and $f_{xx}=\pi^2y\sin{\pi x}=0$ at $(1,2)$, so the series won’t have a $(y-2)$ or $(x-1)^2$ term.