问题
设曲面 $\Sigma$ 的方程为 $z$ $=$ $f(x, y, z)$, 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]. $\left.F_{x}^{\prime}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}\left(x-x_{0}\right)$ $-$ $\left.F_{y}^{\prime}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}\left(y-y_{0}\right)$ $-$ $\left.F_{x}^{\prime}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}\left(z-z_{0}\right)$ $=$ $0$[B]. $\left.F_{x}^{\prime}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}\left(x-x_{0}\right)$ $+$ $\left.F_{y}^{\prime}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}\left(y-y_{0}\right)$ $+$ $\left.F_{x}^{\prime}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}\left(z-z_{0}\right)$ $=$ $1$
[C]. $\left.F_{x}^{\prime}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}\left(x-x_{0}\right)$ $+$ $\left.F_{y}^{\prime}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}\left(y-y_{0}\right)$ $+$ $\left.F_{x}^{\prime}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}\left(z-z_{0}\right)$ $=$ $0$
[D]. $\frac{x-x_{0}}{\left.F_{x}^{\prime}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}}$ $=$ $\frac{y-y_{0}}{\left.F_{y}^{\prime}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}}$ $=$ $\frac{z-z_{0}}{\left.F_{z}^{\prime}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}}$
$\left.F_{x}^{\prime}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}\left(x-x_{0}\right)$ $+$ $\left.F_{y}^{\prime}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}\left(y-y_{0}\right)$ $+$ $\left.F_{x}^{\prime}\right|_{\left(x_{0}, y_{0}, z_{0}\right)}\left(z-z_{0}\right)$ $=$ $0$