问题
设曲面 $\Sigma$ 的方程为 $z$ $=$ $f(x, y)$, 则在 $\Sigma$ 上的点 $\left(x_{0}, y_{0}, z_{0}\right)$ 处的切平面方程是多少？选项
[A]. $\left.\frac{\partial z}{\partial x}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}\left(x-x_{0}\right)$ $+$ $\left.\frac{\partial z}{\partial y}\right|_{\left(x_{0}, y_{0}, x_{2}\right)}\left(y-y_{0}\right)$ $-$ $\left(z-z_{0}\right)$ $=$ $0$[B]. $\left.\frac{\partial z}{\partial x}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}\left(x+x_{0}\right)$ $+$ $\left.\frac{\partial z}{\partial y}\right|_{\left(x_{0}, y_{0}, x_{2}\right)}\left(y+y_{0}\right)$ $-$ $\left(z+z_{0}\right)$ $=$ $0$
[C]. $\left.\frac{\partial z}{\partial x}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}\left(x-x_{0}\right)$ $-$ $\left.\frac{\partial z}{\partial y}\right|_{\left(x_{0}, y_{0}, x_{2}\right)}\left(y-y_{0}\right)$ $+$ $\left(z-z_{0}\right)$ $=$ $0$
[D]. $\left.\frac{\partial z}{\partial x}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}\left(x-x_{0}\right)$ $+$ $\left.\frac{\partial z}{\partial y}\right|_{\left(x_{0}, y_{0}, x_{2}\right)}\left(y-y_{0}\right)$ $-$ $\left(z-z_{0}\right)$ $=$ $1$
$\left.\frac{\partial z}{\partial x}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}\left(x-x_{0}\right)$ $+$ $\left.\frac{\partial z}{\partial y}\right|_{\left(x_{0}, y_{0}, x_{2}\right)}\left(y-y_{0}\right)$ $-$ $\left(z-z_{0}\right)$ $=$ $0$