Let $U\subset R^k, V\subset R^l$ be open. Consider smooth maps $f:U\rightarrow R^K, g: V\rightarrow R^L$ and the map $f\times g:U\times V \rightarrow R^{K+L}$, which is also smooth since $f$ and $g$ are. I'm trying to show that $$d(f\times g)_{(x,y)}=df_x\times dg_y.$$
My attempt is as follows. Let $A=d(f\times g)_{(x,y)}$. By definition, $A$ is the (unique) linear map such that $$\lim_{(u,v)\rightarrow (0,0)}\frac{||(f\times g)(x+u,y+v)-(f\times g)(x,y)-A(u,v)||}{||(u,v)||}=0,$$ where $u\in R^k,v\in R^l$, and $(u,v)$ means the vector in $R^{k+l}$ obtained from $u$ and $v$ by appending coordinates in the "natural order" (you know what I mean). If we substitute $df_x\times dg_y$ for $A$ in the expression of the limit, we get $$\lim_{(u,v)\rightarrow (0,0)}\frac{||(f(x+u)-f(x)-df_x(u), g(y+v)-g(y)-dg_y(v))||}{||(u,v)||}.$$
If we manage to show that this limit is zero, then $d(f\times g)_{(x,y)}=df_x\times dg_y$ will follow from the uniqueness of $A$. But I'm not sure how exactly to show that the $\lim$ is zero. I do realize that $$\lim_{u\rightarrow 0}\frac{||f(x+u)-f(x)-df_x(u)||}{||u||}=0$$ and similarly for $v$, but how to connect it explicitly with the big limit above?