下册 7.4 隐函数的微分 第17题

\section*{解题过程：} （1）关于 $x$ 求导得 $$ \left\{\begin{array}{l} f_{x}+f_{y} y_{x}+f_{z} z_{x}=0 \\ z_{x}=g_{x}+g_{y} y_{x} \end{array}\right. $$ 解之得 $\displaystyle \frac{\mathrm{d} y}{\mathrm{~d} x}=-\frac{f_{x}+f_{z} g_{x}}{f_{y}+f_{z} g_{z}}, \frac{\mathrm{~d} z}{\mathrm{~d} x}=\frac{g_{x} f_{y}-f_{z} g_{z}}{f_{y}+f_{z} g_{y}}$ ． （2）方程组求微分得 $$ \left\{\begin{array}{l} F_{1} \mathrm{~d} x+F_{2}\left(f_{y} \mathrm{~d} y+f_{u} \mathrm{~d} u\right)=0 \\ G_{1} \mathrm{~d} y+G_{2}\left(g_{x} \mathrm{~d} x+g_{v} \mathrm{~d} v\right)=0 \end{array}\right. $$ 解之得 $\displaystyle \mathrm{d} x=-\frac{F_{2} G_{1} f_{u}}{F_{1} G_{1}-F_{2} G_{2} g_{x} f_{y}} \mathrm{~d} u+\frac{F_{2} G_{2} g_{v} f_{y}}{F_{1} G_{1}-F_{2} G_{2} g_{x} f_{y}} \mathrm{~d} v$ 。于是 $$ x_{u}=-\frac{F_{2} G_{1} f_{u}}{F_{1} G_{1}-F_{2} G_{2} g_{x} f_{y}}, \quad x_{v}=\frac{F_{2} G_{2} g_{v} f_{y}}{F_{1} G_{1}-F_{2} G_{2} g_{x} f_{y}} $$ （3）当 $g, h$ 对各变元有连续的偏导数，且 $\displaystyle \frac{\partial(g, h)}{\partial(z, t)} \neq 0$ 时，方程组 $\left\{\begin{array}{l}g(y, z, t)=0 \\ h(z, t)=0\end{array}\right.$ 可确定函数组 $\left\{\begin{array}{l}z=z(y), \\ t=t(y) .\end{array}\right.$ 解方程组 $\left\{\begin{array}{l}g_{y} \mathrm{~d} y+g_{z} \mathrm{~d} z+g_{t} \mathrm{~d} t=0, \\ h_{z} \mathrm{~d} z+h_{t} \mathrm{~d} t=0,\end{array}\right.$ 得 $\displaystyle \mathrm{d} z=\frac{-g_{y} h_{t}}{J} \mathrm{~d} y, \mathrm{~d} t=\frac{g_{y} h_{z}}{J} \mathrm{~d} y$ ，其中 $\displaystyle J=\frac{\partial(g, h)}{\partial(z, t)}$ ． 将函数组 $\left\{\begin{array}{l}z=z(y) \\ t=t(y)\end{array}\right.$ 代人 $u=f(x, y, z, t)$ ，即得 $u$ 是 $x, y$ 的函数 $u=f(x, y, z(y), t(y))$ ．于是 $$ \mathrm{d} u=f_{x} \mathrm{~d} x+f_{y} \mathrm{~d} y+f_{z} \frac{-g_{y} h_{t} \mathrm{~d} y}{J}+f_{t} \frac{g_{y} h_{z} \mathrm{~d} y}{J}=f_{x} \mathrm{~d} x+\left(f_{y}+\frac{g_{y}}{J} \frac{\partial(h, f)}{\partial(z, t)}\right) \mathrm{d} y . $$ 从而 $\displaystyle u_{x}=f_{x}, u_{y}=f_{y}+\frac{g_{y}}{J} \frac{\partial(h, f)}{\partial(z, t)}$ ．