验证二元函数的可微性(B012)

问题

若有二元函数 $z$ $=$ $f(x, y)$, 且 $\phi$ $=$ $\sqrt{(\Delta x)^{2}+(\Delta y)^{2}}$, 则,如何验证该二元函数的可微性?

选项

[A].   $\lim _{\rho \rightarrow 0}$ $\frac{\Delta z-f_x^{\prime}(x, y) \Delta x-f_y^{\prime}(x, y) \Delta y}{\rho}$ $=$ $0$ $\Rightarrow$ $f(x, y)$ 可微

[B].   $\lim _{\rho \rightarrow 0}$ $\frac{\Delta z-f_x^{\prime}(x, y) \Delta x-f_y^{\prime}(x, y) \Delta y}{\rho^{2}}$ $=$ $0$ $\Rightarrow$ $f(x, y)$ 可微

[C].   $\lim _{\rho \rightarrow 0}$ $\frac{\Delta z+f_x^{\prime}(x, y) \Delta x-f_y^{\prime}(x, y) \Delta y}{\rho}$ $=$ $0$ $\Rightarrow$ $f(x, y)$ 可微

[D].   $\lim _{\rho \rightarrow 0}$ $\frac{\Delta z-f_x^{\prime}(x, y) \Delta x+f_y^{\prime}(x, y) \Delta y}{\rho}$ $=$ $0$ $\Rightarrow$ $f(x, y)$ 可微


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若 $\lim _{\rho \rightarrow 0}$ $\frac{\Delta z-f_x^{\prime}(x, y) \Delta x-f_y^{\prime}(x, y) \Delta y}{\rho}$ $=$ $0$,

或者:

$\lim _{\rho \rightarrow 0}$ $\frac{\Delta z – \frac{\partial z}{\partial x} \Delta x – \frac{\partial z}{\partial y} \Delta y}{\rho}$ $=$ $0$,

则函数 $z$ $=$ $f(x, y)$ 可微.