考研数学 - 第94页共147页

空间曲线的法平面方程：基于参数方程（B013）

问题

若已知空间曲线 $\Gamma$ 的参数方程为 $\left\{\begin{array}{l}x=x(t), \\ y=y(t) \\ z=z(t)\end{array}\right.$, 则曲线 $\Gamma$ 在点 $(x_{0}, y_{0}, z_{0})$（对应参数 $t$ $=$ $t_{0}$）处的法平面方程为多少？

选项

[A]. $x^{\prime}\left(t_{0} \right)$ $\left(x-x_{0} \right)$ $\times$ $y^{\prime}\left(t_{0} \right)$ $\left(y-y_{0} \right)$ $\times$ $z^{\prime}\left(t_{0} \right)$ $\left(z-z_{0} \right)$ $=$ $1$

[B]. $x \left(t_{0} \right)$ $\left(x-x_{0} \right)$ $+$ $y \left(t_{0} \right)$ $\left(y-y_{0} \right)$ $+$ $z \left(t_{0} \right)$ $\left(z-z_{0} \right)$ $=$ $0$

[C]. $x^{\prime}\left(t_{0} \right)$ $\left(x+x_{0} \right)$ $-$ $y^{\prime}\left(t_{0} \right)$ $\left(y+y_{0} \right)$ $-$ $z^{\prime}\left(t_{0} \right)$ $\left(z+z_{0} \right)$ $=$ $0$

[D]. $x^{\prime}\left(t_{0} \right)$ $\left(x-x_{0} \right)$ $+$ $y^{\prime}\left(t_{0} \right)$ $\left(y-y_{0} \right)$ $+$ $z^{\prime}\left(t_{0} \right)$ $\left(z-z_{0} \right)$ $=$ $0$

答案

$x^{\prime}\left(t_{0} \right)$ $\left(x-x_{0} \right)$ $+$ $y^{\prime}\left(t_{0} \right)$ $\left(y-y_{0} \right)$ $+$ $z^{\prime}\left(t_{0} \right)$ $\left(z-z_{0} \right)$ $=$ $0$

空间曲线的切线方程：基于参数方程（B013）

问题

若已知空间曲线 $\Gamma$ 的参数方程为 $\left\{\begin{array}{l}x=x(t), \\ y=y(t) \\ z=z(t)\end{array}\right.$, 则曲线 $\Gamma$ 在点 $(x_{0}, y_{0}, z_{0})$（对应参数 $t$ $=$ $t_{0}$）处的切线方程为多少？

选项

[A]. $\frac{x-x_{0}}{x \left(t_{0} \right)}$ $=$ $\frac{y-y_{0}}{y \left(t_{0} \right)}$ $=$ $\frac{z-z_{0}}{z \left(t_{0} \right)}$

[B]. $\frac{x+x_{0}}{x^{\prime}\left(t_{0} \right)}$ $=$ $\frac{y+y_{0}}{y^{\prime}\left(t_{0} \right)}$ $=$ $\frac{z+z_{0}}{z^{\prime}\left(t_{0} \right)}$

[C]. $\frac{x-x_{0}}{x^{\prime}\left(t_{0} \right)}$ $=$ $\frac{y-y_{0}}{y^{\prime}\left(t_{0} \right)}$ $=$ $\frac{z-z_{0}}{z^{\prime}\left(t_{0} \right)}$

[D]. $\frac{x-x_{0}}{x^{\prime \prime}\left(t_{0} \right)}$ $=$ $\frac{y-y_{0}}{y^{\prime \prime}\left(t_{0} \right)}$ $=$ $\frac{z-z_{0}}{z^{\prime \prime}\left(t_{0} \right)}$

答案

$\frac{x-x_{0}}{x^{\prime}\left(t_{0} \right)}$ $=$ $\frac{y-y_{0}}{y^{\prime}\left(t_{0} \right)}$ $=$ $\frac{z-z_{0}}{z^{\prime}\left(t_{0} \right)}$

三元函数求单条件极值：拉格朗日函数的使用（B013）

问题

若要求函数 $u$ $=$ $f(x, y, z)$ 在 $\varphi(x, y, z)$ $=$ $0$ 条件下的极值，且已经构造出了如下的拉格朗日函数：

$F(x, y, z)$ $=$ $f(x, y, z)$ $+$ $\lambda$ $\varphi(x, y, z)$

则，根据拉格朗日乘数法，还需要构造以下哪个选项中的方程组并计算才可能得出与极值对应的驻点 $(x_{0}, y_{0}, z_{0})$ ?

选项

[A]. $\left\{\begin{array}{l}f_{x}^{\prime}(x, y, z)+\lambda \varphi(x, y, z)=0, \\ f_{y}^{\prime}(x, y, z)+\lambda \varphi(x, y, z)=0, \\ f_{x}^{\prime}(x, y, z)+\lambda \varphi(x, y, z)=0, \\ \varphi(x, y, z)=0. \end{array}\right.$

[B]. $\left\{\begin{array}{l}f(x, y, z)+\lambda \varphi_{x}^{\prime}(x, y, z)=0, \\ f(x, y, z)+\lambda \varphi_{y}^{\prime}(x, y, z)=0, \\ f(x, y, z)+\lambda \varphi_{z}^{\prime}(x, y, z)=0, \\ \varphi(x, y, z)=0. \end{array}\right.$

[C]. $\left\{\begin{array}{l}f_{x}^{\prime}(x, y, z)+\lambda \varphi_{x}^{\prime}(x, y, z)=1, \\ f_{y}^{\prime}(x, y, z)+\lambda \varphi_{y}^{\prime}(x, y, z)=1, \\ f_{x}^{\prime}(x, y, z)+\lambda \varphi_{z}^{\prime}(x, y, z)=1, \\ \varphi(x, y, z)=1. \end{array}\right.$

[D]. $\left\{\begin{array}{l}f_{x}^{\prime}(x, y, z)+\lambda \varphi_{x}^{\prime}(x, y, z)=0, \\ f_{y}^{\prime}(x, y, z)+\lambda \varphi_{y}^{\prime}(x, y, z)=0, \\ f_{x}^{\prime}(x, y, z)+\lambda \varphi_{z}^{\prime}(x, y, z)=0, \\ \varphi(x, y, z)=0. \end{array}\right.$

答案

$\left\{\begin{array}{l}f_{x}^{\prime}(x, y, z)+\lambda \varphi_{x}^{\prime}(x, y, z)=0, \\ f_{y}^{\prime}(x, y, z)+\lambda \varphi_{y}^{\prime}(x, y, z)=0, \\ f_{x}^{\prime}(x, y, z)+\lambda \varphi_{z}^{\prime}(x, y, z)=0, \\ \varphi(x, y, z)=0. \end{array}\right.$

二元函数求单条件极值：拉格朗日函数的使用（B013）

问题

若要求函数 $z$ $=$ $f(x, y)$ 在 $\varphi(x, y)$ $=$ $0$ 条件下的极值，且已经构造出了如下的拉格朗日函数：

$F(x, y)$ $=$ $f(x, y)$ $+$ $\lambda$ $\varphi(x, y)$

则，根据拉格朗日乘数法，还需要构造以下哪个选项中的方程组并计算才可能得出与极值对应的驻点 $(x_{0}, y_{0})$ ?

选项

[A]. $\left\{\begin{array}{l}f_{x}^{\prime}(x, y)+\lambda \varphi_{x y}^{\prime \prime}(x, y)=0, \\ f_{y}^{\prime}(x, y)+\lambda \varphi_{y x}^{\prime \prime}(x, y)=1, \\ \varphi(x, y)=0.\end{array}\right.$

[B]. $\left\{\begin{array}{l}f_{x}^{\prime}(x, y)-\lambda \varphi_{x}^{\prime}(x, y)=0, \\ f_{y}^{\prime}(x, y)-\lambda \varphi_{y}^{\prime}(x, y)=0, \\ \varphi(x, y)=1. \end{array}\right.$

[C]. $\left\{\begin{array}{l}f(x, y)+\lambda \varphi_{x}^{\prime}(x, y)=0, \\ f(x, y)+\lambda \varphi_{y}^{\prime}(x, y)=0, \\ \varphi(x, y)=0. \end{array}\right.$

[D]. $\left\{\begin{array}{l}f_{x}^{\prime}(x, y)+\lambda \varphi_{x}^{\prime}(x, y)=0, \\ f_{y}^{\prime}(x, y)+\lambda \varphi_{y}^{\prime}(x, y)=0, \\ \varphi(x, y)=0. \end{array}\right.$

答案

$\left\{\begin{array}{l}f_{x}^{\prime}(x, y)+\lambda \varphi_{x}^{\prime}(x, y)=0, \\ f_{y}^{\prime}(x, y)+\lambda \varphi_{y}^{\prime}(x, y)=0, \\ \varphi(x, y)=0. \end{array}\right.$

三元函数求双条件极值：拉格朗日函数的构造（B013）

问题

根据拉格朗日乘数法，若要求函数 $u$ $=$ $f(x, y, z)$ 在 $\varphi_{1}(x, y, z)$ $=$ $0$ 和 $\varphi_{2}(x, y, z)$ $=$ $0$ 两个条件约束下的极值，如何构造拉格朗日函数 $F(x, y, z)$ ?

选项

[A]. $F(x, y, z)$ $=$ $f(x, y, z)$ $-$ $\lambda_{1}$ $\varphi_{1}(x, y, z)$ $-$ $\lambda_{2}$ $\varphi_{2}(x, y, z)$

[B]. $F(x, y, z)$ $=$ $f(x, y, z)$ $+$ $\lambda_{1}$ $\varphi_{1}(x, y, z)$ $+$ $\lambda_{2}$ $\varphi_{2}(x, y, z)$

[C]. $F(x, y, z)$ $=$ $f(x, y, z)$ $+$ $\frac{1}{\lambda_{1}}$ $\varphi_{1}(x, y, z)$ $+$ $\frac{1}{\lambda_{2}}$ $\varphi_{2}(x, y, z)$

[D]. $F(x, y, z)$ $=$ $\lambda$ $f(x, y, z)$ $+$ $\lambda_{1}$ $\varphi_{1}(x, y, z)$ $+$ $\lambda_{2}$ $\varphi_{2}(x, y, z)$

答案

$F(x, y, z)$ $=$ $f(x, y, z)$ $+$ $\lambda_{1}$ $\varphi_{1}(x, y, z)$ $+$ $\lambda_{2}$ $\varphi_{2}(x, y, z)$

三元函数求单条件极值：拉格朗日函数的构造（B013）

问题

根据拉格朗日乘数法，若要求函数 $u$ $=$ $f(x, y, z)$ 在 $\varphi(x, y, z)$ $=$ $0$ 条件约束下的极值，如何构造拉格朗日函数 $F(x, y, z)$ ?

选项

[A]. $F(x, y, z)$ $=$ $\lambda$ $f(x, y, z)$ $+$ $\varphi(x, y, z)$

[B]. $F(x, y, z)$ $=$ $f(x, y, z)$ $-$ $\lambda$ $\varphi(x, y, z)$

[C]. $F(x, y, z)$ $=$ $f(x, y, z)$ $+$ $\varphi(x, y, z)$

[D]. $F(x, y, z)$ $=$ $f(x, y, z)$ $+$ $\lambda$ $\varphi(x, y, z)$

答案

$F(x, y, z)$ $=$ $f(x, y, z)$ $+$ $\lambda$ $\varphi(x, y, z)$

二元函数求单条件极值：拉格朗日函数的构造（B013）

问题

根据拉格朗日乘数法，若要求函数 $z$ $=$ $f(x, y)$ 在 $\varphi(x, y)$ $=$ $0$ 条件约束下的极值，如何构造拉格朗日函数 $F(x, y)$ ?

选项

[A]. $F(x, y)$ $=$ $\lambda$ $f(x, y)$ $+$ $\varphi(x, y)$

[B]. $F(x, y)$ $=$ $f(x, y)$ $-$ $\lambda$ $\varphi(x, y)$

[C]. $F(x, y)$ $=$ $f(x, y)$ $+$ $\lambda$ $\varphi(x, y)$

[D]. $F(x, y)$ $=$ $f(x, y)$ $+$ $\frac{1}{\lambda}$ $\varphi(x, y)$

答案

$F(x, y)$ $=$ $f(x, y)$ $+$ $\lambda$ $\varphi(x, y)$

极值存在的充分条件：判别公式中的 $A$, $B$, $C$ 都是多少？（B013）

问题

若已知函数 $z$ $=$ $f(x, y)$ 在点 $\left(x_{0}, y_{0} \right)$ 的某邻域内有连续的二阶偏导数,且 $f_{x}^{\prime}\left(x_{0}, y_{0} \right)$ $=$ $0$, $f_{y}^{\prime}\left(x_{0}, y_{0} \right)$ $=$ $0$; 则极值判别公式 $AC$ $-$ $B^{2}$ 中的 $A$, $B$ 和 $C$ 各等于多少？

选项

[A]. $\begin{cases} & A = f_{x}^{\prime}\left(x_{0}, y_{0}\right) \\ & B= f_{x y}^{\prime \prime}\left(x_{0} \right., \left.y_{0} \right)\\ & C = f_{y}^{\prime}\left(x_{0}, y_{0} \right) \end{cases}$

[B]. $\begin{cases} & A = f_{y x}^{\prime \prime}\left(x_{0}, y_{0}\right) \\ & B= f_{x y}^{\prime \prime}\left(x_{0} \right., \left.y_{0} \right)\\ & C = f_{y y}^{\prime \prime}\left(x_{0}, y_{0} \right) \end{cases}$

[C]. $\begin{cases} & A = f_{y y}^{\prime \prime}\left(x_{0}, y_{0}\right) \\ & B= f_{x y}^{\prime \prime}\left(x_{0} \right., \left.y_{0} \right)\\ & C = f_{x x}^{\prime \prime}\left(x_{0}, y_{0} \right) \end{cases}$

[D]. $\begin{cases} & A = f_{x x}^{\prime \prime}\left(x_{0}, y_{0}\right) \\ & B= f_{x y}^{\prime \prime}\left(x_{0} \right., \left.y_{0} \right)\\ & C = f_{y y}^{\prime \prime}\left(x_{0}, y_{0} \right) \end{cases}$

答案

$\begin{cases} & A = f_{x x}^{\prime \prime}\left(x_{0}, y_{0}\right) \\ & B= f_{x y}^{\prime \prime}\left(x_{0} \right., \left.y_{0} \right)\\ & C = f_{y y}^{\prime \prime}\left(x_{0}, y_{0} \right) \end{cases}$

极值存在的充分条件：判断是极大值点还是极小值点（B013）

问题

若已知函数 $z$ $=$ $f(x, y)$ 在点 $\left(x_{0}, y_{0} \right)$ 的某邻域内有连续的二阶偏导数,且 $f_{x}^{\prime}\left(x_{0}, y_{0} \right)$ $=$ $0$, $f_{y}^{\prime}\left(x_{0}, y_{0} \right)$ $=$ $0$; $A$ $=$ $f_{x x}^{\prime \prime}\left(x_{0}, y_{0}\right)$, $B$ $=$ $f_{x y}^{\prime \prime}\left(x_{0} \right.$, $\left.y_{0} \right)$, $C$ $=$ $f_{y y}^{\prime \prime}\left(x_{0}, y_{0} \right)$.

则以下哪个选项可以说明点 $\left(x_{0}, y_{0} \right)$ 为函数 $z$ $=$ $f(x, y)$ 的一个极值大点或极小值点？

选项

[A]. $A C$ $-$ $B^{2}$ $>$ $0$ $\Rightarrow$ $\begin{cases} & A<0 \Rightarrow 极小值点 \\ & A>0 \Rightarrow 极大值点 \end{cases}$

[B]. $A C$ $-$ $B^{2}$ $>$ $0$ $\Rightarrow$ $\begin{cases} & A>0 \Rightarrow 极小值点 \\ & A<0 \Rightarrow 极大值点 \end{cases}$

[C]. $A C$ $-$ $B^{2}$ $<$ $0$ $\Rightarrow$ $\begin{cases} & A>0 \Rightarrow 极小值点 \\ & A<0 \Rightarrow 极大值点 \end{cases}$

[D]. $A C$ $-$ $B^{2}$ $>$ $0$ $\Rightarrow$ $\begin{cases} & A>1 \Rightarrow 极小值点 \\ & A<1 \Rightarrow 极大值点 \end{cases}$

答案

$A C$ $-$ $B^{2}$ $>$ $0$ $\Rightarrow$ $\begin{cases} & A>0 \Rightarrow 极小值点 \\ & A<0 \Rightarrow 极大值点 \end{cases}$

极值存在的充分条件：判断是否为极值点（B013）

问题

若已知函数 $z$ $=$ $f(x, y)$ 在点 $\left(x_{0}, y_{0} \right)$ 的某邻域内有连续的二阶偏导数,且 $f_{x}^{\prime}\left(x_{0}, y_{0} \right)$ $=$ $0$, $f_{y}^{\prime}\left(x_{0}, y_{0} \right)$ $=$ $0$; $A$ $=$ $f_{x x}^{\prime \prime}\left(x_{0}, y_{0}\right)$, $B$ $=$ $f_{x y}^{\prime \prime}\left(x_{0} \right.$, $\left.y_{0} \right)$, $C$ $=$ $f_{y y}^{\prime \prime}\left(x_{0}, y_{0} \right)$.

则以下哪个选项可以说明点 $\left(x_{0}, y_{0} \right)$ 为函数 $z$ $=$ $f(x, y)$ 的一个极值点？

选项

[A]. $A C$ $-$ $B^{2}$ $=$ $0$

[B]. $B C$ $-$ $A^{2}$ $<$ $0$

[C]. $A C$ $-$ $B^{2}$ $<$ $0$

[D]. $A B$ $-$ $C^{2}$ $>$ $0$

[E]. $A C$ $-$ $B^{2}$ $>$ $0$

[F]. $A B$ $-$ $C^{2}$ $=$ $0$

答案

$A C$ $-$ $B^{2}$ $>$ $0$ $\Rightarrow$ 点 $\left(x_{0}, y_{0} \right)$ 是极值点.

$A C$ $-$ $B^{2}$ $<$ $0$ $\Rightarrow$ 点 $\left(x_{0}, y_{0} \right)$ 不是极值点.

$A C$ $-$ $B^{2}$ $=$ $0$ $\Rightarrow$ 不确定点 $\left(x_{0}, y_{0} \right)$ 是否是极值点.

极值存在的必要条件（B013）

问题

设 $z=f(x, y)$ 在点 $\left(x_{0}, y_{0}\right)$ 的一阶偏导数存在, 且 $\left(x_{0}, y_{0}\right)$ 是 $z=$ $f(x, y)$ 的极值点, 则可以推出以下哪个选项所示的结论？

选项

[A]. $\left.\frac{\partial z}{\partial x}\right|_{\left(x, y\right)}$ $=$ $0$, $\left.\frac{\partial z}{\partial y}\right|_{\left(x, y \right)}$ $=$ $0$

[B]. $\left.\frac{\partial z}{\partial x}\right|_{\left(x_{0}, y_{0}\right)}$ $\neq$ $\left.\frac{\partial z}{\partial y}\right|_{\left(x_{0}, y_{0}\right)}$

[C]. $\left.\frac{\partial z}{\partial x}\right|_{\left(x_{0}, y_{0}\right)}$ $=$ $1$, $\left.\frac{\partial z}{\partial y}\right|_{\left(x_{0}, y_{0}\right)}$ $=$ $1$

[D]. $\left.\frac{\partial z}{\partial x}\right|_{\left(x_{0}, y_{0}\right)}$ $=$ $0$, $\left.\frac{\partial z}{\partial y}\right|_{\left(x_{0}, y_{0}\right)}$ $=$ $0$

答案

$\left.\frac{\partial z}{\partial x}\right|_{\left(x_{0}, y_{0}\right)}$ $=$ $0$, $\left.\frac{\partial z}{\partial y}\right|_{\left(x_{0}, y_{0}\right)}$ $=$ $0$

多元函数的极值（B013）

问题

以下哪个选项可以说明点 $(x_{0}, y_{0})$ 是二元函数 $z$ $=$ $f(x, y)$ 的极大值点（或极小值点）？

选项

[A]. 在点 $(x_{0}, y_{0})$ 的邻域内有两个异于点 $(x_{0}, y_{0})$ 的点 $(x, y)$ 使得 $f(x, y) < f(x_{0}, y_{0})$（或 $f(x, y)$ $>$ $f(x_{0}, y_{0})$）成立

[B]. 在点 $(x_{0}, y_{0})$ 的邻域内任意一个异于点 $(x_{0}, y_{0})$ 的点 $(x, y)$, 都使得 $f(x, y) < f(x_{0}, y_{0})$（或 $f(x, y)$ $>$ $f(x_{0}, y_{0})$）成立

[C]. 在点 $(x_{0}, y_{0})$ 的邻域内有一个异于点 $(x_{0}, y_{0})$ 的点 $(x, y)$ 使得 $f(x, y) < f(x_{0}, y_{0})$（或 $f(x, y)$ $>$ $f(x_{0}, y_{0})$）成立

[D]. 在点 $(x_{0}, y_{0})$ 的邻域内有多个异于点 $(x_{0}, y_{0})$ 的点 $(x, y)$ 使得 $f(x, y) < f(x_{0}, y_{0})$（或 $f(x, y)$ $>$ $f(x_{0}, y_{0})$）成立

答案

在点 $(x_{0}, y_{0})$ 的邻域内任意一个异于点 $(x_{0}, y_{0})$ 的点 $(x, y)$, 都使得 $f(x, y) < f(x_{0}, y_{0})$（或 $f(x, y)$ $>$ $f(x_{0}, y_{0})$）成立 $\Rightarrow$ $f(x_{0}, y_{0})$ 是函数 $z$ $=$ $f(x, y)$ 的一个极大值（或极小值）

三元隐函数的复合函数求导法则（B012）

问题

设由方程组 $\left\{\begin{array}{l}F(x, y, z)=0 \\ G(x, y, z)=0\end{array}\right.$ 确定的隐函数为 $y$ $=$ $y(x)$ 与 $z$ $=$ $z(x)$, 则 $\frac{\mathrm{d} y}{\mathrm{~d} x}$ 和 $\frac{\mathrm{d} z}{\mathrm{~d} x}$ 可以通过解以下哪个线性方程组求出？

选项

[A]. $\left\{\begin{array}{l} F_{x}^{\prime}+F_{y}^{\prime} \frac{\mathrm{d} y}{\mathrm{~d} x}+F_{z}^{\prime} \frac{\mathrm{d} z}{\mathrm{~d} x}=0 \\ G_{x}^{\prime}+G_{y}^{\prime} \frac{\mathrm{d} y}{\mathrm{~d} x}+G_{z}^{\prime} \frac{\mathrm{d} z}{\mathrm{~d} x}=0 \end{array}\right.$

[B]. $\left\{\begin{array}{l} F_{x}+F_{y}^{\prime} \frac{\mathrm{d} y}{\mathrm{~d} x}+F_{x}^{\prime} \frac{\mathrm{d} z}{\mathrm{~d} x}=0 \\ G_{x}+G_{y}^{\prime} \frac{\mathrm{d} y}{\mathrm{~d} x}+G_{z}^{\prime} \frac{\mathrm{d} z}{\mathrm{~d} x}=0 \end{array}\right.$

[C]. $\left\{\begin{array}{l} F_{z}^{\prime}+F_{x}^{\prime} \frac{\mathrm{d} y}{\mathrm{~d} x}+F_{y}^{\prime} \frac{\mathrm{d} z}{\mathrm{~d} x}=0 \\ G_{z}^{\prime}+G_{x}^{\prime} \frac{\mathrm{d} y}{\mathrm{~d} x}+G_{y}^{\prime} \frac{\mathrm{d} z}{\mathrm{~d} x}=0 \end{array}\right.$

[D]. $\left\{\begin{array}{l} F_{x}^{\prime}+F_{y}^{\prime} \frac{\mathrm{d} z}{\mathrm{~d} x}+F_{x}^{\prime} \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \\ G_{x}^{\prime}+G_{y}^{\prime} \frac{\mathrm{d} z}{\mathrm{~d} x}+G_{z}^{\prime} \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \end{array}\right.$

答案

$\left\{\begin{array}{l} F_{x}^{\prime}+F_{y}^{\prime} \frac{\mathrm{d} y}{\mathrm{~d} x}+F_{z}^{\prime} \frac{\mathrm{d} z}{\mathrm{~d} x}=0 \\ G_{x}^{\prime}+G_{y}^{\prime} \frac{\mathrm{d} y}{\mathrm{~d} x}+G_{z}^{\prime} \frac{\mathrm{d} z}{\mathrm{~d} x}=0 \end{array}\right.$

三元复合函数求导法则（B012）

问题

已知函数 $F(x, y, z)$ $=$ $0$, 若 $F_{z}^{\prime}$ $\neq$ $0$, 则 $\frac{\partial z}{\partial x}$ $=$ $?$, $\frac{\partial z}{\partial y}$ $=$ $?$

选项

[A]. $\frac{\partial z}{\partial x}$ $=$ $-$ $\frac{F_{z}^{\prime}(x, y, z)}{F_{x}^{\prime}(x, y, z)}$, $\frac{\partial z}{\partial y}$ $=$ $-$ $\frac{F_{z}^{\prime}(x, y, z)}{F_{y}^{\prime}(x, y, z)}$

[B]. $\frac{\partial z}{\partial x}$ $=$ $-$ $\frac{F_{x}^{\prime}(x, y, z)}{F_{z}(x, y, z)}$, $\frac{\partial z}{\partial y}$ $=$ $-$ $\frac{F_{y}^{\prime}(x, y, z)}{F_{z}(x, y, z)}$

[C]. $\frac{\partial z}{\partial x}$ $=$ $\frac{F_{x}^{\prime}(x, y, z)}{F_{z}^{\prime}(x, y, z)}$, $\frac{\partial z}{\partial y}$ $=$ $\frac{F_{y}^{\prime}(x, y, z)}{F_{z}^{\prime}(x, y, z)}$

[D]. $\frac{\partial z}{\partial x}$ $=$ $-$ $\frac{F_{x}^{\prime}(x, y, z)}{F_{z}^{\prime}(x, y, z)}$, $\frac{\partial z}{\partial y}$ $=$ $-$ $\frac{F_{y}^{\prime}(x, y, z)}{F_{z}^{\prime}(x, y, z)}$

答案

$\frac{\partial z}{\partial x}$ $=$ $-$ $\frac{F_{x}^{\prime}(x, y, z)}{F_{z}^{\prime}(x, y, z)}$, $\frac{\partial z}{\partial y}$ $=$ $-$ $\frac{F_{y}^{\prime}(x, y, z)}{F_{z}^{\prime}(x, y, z)}$

二元复合函数求导法则（B012）

问题

已知函数 $F(x, y)$ $=$ $0$, 若 $F_y^{\prime}$ $\neq$ $0$, 则 $\frac{\mathrm{d} y}{\mathrm{~d} x}=$ $?$

选项

[A]. $\frac{\mathrm{d} y}{\mathrm{~d} x}=$ $-\frac{F_{y}^{\prime}(x, y)}{F_{x}^{\prime}(x, y)}$

[B]. $\frac{\mathrm{d} y}{\mathrm{~d} x}=$ $\frac{F_{x}^{\prime}(x, y)}{F_{y}^{\prime}(x, y)}$

[C]. $\frac{\mathrm{d} y}{\mathrm{~d} x}=$ $-\frac{F_{x}^{\prime}(x, y)}{F_{y}^{\prime}(x, y)}$

[D]. $\frac{\mathrm{d} y}{\mathrm{~d} x}=$ $\frac{F_{y}^{\prime}(x, y)}{F_{x}^{\prime}(x, y)}$

答案

$\frac{\mathrm{d} y}{\mathrm{~d} x}=$ $-\frac{F_{x}^{\prime}(x, y)}{F_{y}^{\prime}(x, y)}$