多元微分中值不证了，自己百度给定一二元函数 f(x,y)is\ defined\ in \ the(E \subseteq \mathbb{R^{2}})\\ \wedge D_{1}f\ \ or\ D_{2}f \ or\ D_{21}\ is\ exists \ in \forall(x,y)\in E\\\wedge \exists(a,b)\in E, D_{21}f\
is \ continuous\ in\ it\\ \wedge \exists D_{12}f \ in \ (a,b)\\ Then\ (D_{12}f)(a,b)=(D_{21}f)(a,b) 我们假设 A=(D_{21}f)(x,y) 又有 \forall \varepsilon >0 则对于矩形Q\wedge h,k充分小，其中(a,b)\ and (a+h,a+bk)是 Q 的对顶点则 \forall (x,y)\in Q,\left|A-(D_{21}f)(x,y)\right|<\varepsilon 由中值定理得:\left|A-(D_{21}f)(x,y)\right|=\left|A-\frac{\Delta(f,Q)}{hk}\right|令 h\rightarrow 0\wedge \exists D_{2}f\in E
故\left|\frac{(D_{2}f)(a+h,b)-(D_{2}f)(a,b)}{h}-A\right|\leq\varepsilon
故(D_{12}f)(a,b)=A=(D_{21}f)(a,b) \blacksquare }