多元函数的微分学（一）

1多元函数的极限与连续性

1.1 平面点集

1. 1.

   点$M_0$的$\delta$邻域（开区域）：$\bar{U}(M_0,\delta )\stackrel{\Delta }{=}\{M|\left|M{M}_0\right|<\delta \}$.

2. 2.

   点集$E$的内点$M_0$：$\exists \delta >0$，使得$\bar{U}(M_0,\delta )\subset E$.

3. 3.

   开集：全由内点¹组成的集合.

4. 4.

   （开）区域$D$：$D$中任意两点均可用曲线连接（连通性）.

5. 5.

   闭区域（闭域）$\bar{D}=D+\partial D$，其中$\partial D$为$D$的边界.

6. 6.

   ${\mathbb{R}}^2,\varnothing$既为开集，又为闭集.

1.2 二元函数的极限与连续性

1. 1.

   极限：设点$M(x,y),M_0(x_0,y_0)$，则若对$\forall \epsilon >0$，总$\exists \delta (\epsilon ,M_0)>0$，当$\left|M{M}_0\right|<\delta$时，满足$\left|f(M)-A\right|<\epsilon$，则称当$M\to M_0$时，$f(M)$以$A$为极限，记作$\underset{M\to M_0}{\lim }f(M)=A$.

2. 2.

   连续：特别地，若$A=f(M_0)$，即$\underset{(x,y)\to (x_0,y_0)}{\lim }f(x,y)=f(x_0,y_0)$，则称$f(x,y)$在点$M_0$处连续. 若$f(x,y)$在点集$E$内中每一点都连续，则称$f(x,y)$在$E$上连续.

3. 3.

   二元函数极限的常用（通用）求解方法：作极坐标变换$\{\begin{array}{l}x=x_0+\rho \cos \theta ,\\ y=y_0+\rho \sin \theta ,\end{array}$代入表达式中得到$g(\rho ,\theta )=f(x(\rho ,\theta ),y(\rho ,\theta ))$，再求$\rho \to 0$的极限. 若要求沿某一轨迹的极限，则写出轨迹的极坐标方程，代入$\theta =\theta (\rho )$关系再求解.

1.3 有界闭区域上连续函数的性质

1. 1.

   有界性.

2. 2.

   最值性.

3. 3.

   介值性.

4. 4.

   零值性.

5. 5.

   一致连续性：对$\forall \epsilon >0$，总$\exists \delta (\epsilon )>0$，使得当$\left|M{M}_0\right|<\delta (\epsilon )$时，满足$\left|f(M)-f(M_0)\right|<\epsilon$.

1.4 例题

1. 1.

   判断下列极限是否存在，若有极限，求出其极限； 若无极限，说明理由.

   1. (1)

      $\underset{(x,y)\to (0,0)}{\lim }\frac{x^2{y}^2}{x^4+y^2}$.

   2. (2)

      $\underset{(x,y)\to (0,0)}{\lim }\frac{x^{2}y}{x^4+y^2}$.

   3. (3)

      $\underset{(x,y)\to (0,0)}{\lim }\frac{\sqrt{xy+1}-1}{x+y}$.

   4. (4)

      $\underset{(x,y)\to (+\infty ,+\infty )}{\lim }(x^2+y^2){\mathrm{e}}^{-(x+y)}$.

   5. (5)

      $\underset{(x,y)\to (0,0)}{\lim }\frac{x^2+y^2}{|x|+|y|}$.

   6. (6)

      $\underset{(x,y)\to (0,0)}{\lim }\frac{x^3+y^3}{x^2+y^2}.$

2. 2.

   研究$f(x,y)=\{\begin{array}{ll}\frac{xy}{x-y},& x\ne y,\\ 0,& x=y\end{array}$的连续性.

2偏导数与全微分

2.1 偏导数的定义

设$z=f(x,y)$在$\bar{U}(M_0,\delta )$内有定义，且$M_1(x_0+\Delta x,y_0),M_2(x_0,y_0+\Delta y)\subset \bar{U}(M_0,\delta )$，则

上述两式分别称为$z=f(x,y)$关于$x,y$的偏导数.

2.2 全微分的定义

设$z=f(x,y)$在$\bar{U}(M_0,\delta )$内有定义，且$M(x_0+\Delta x,y_0+\Delta y)\subset \bar{U}(M_0,\delta )$，则若全增量$\Delta z=f(M)-f(M_0)$可写为如下形式：

其中$\rho =\sqrt{(\Delta x{)}^2+(\Delta y{)}^2}$，则称$z=f(x,y)$在点$M_0(x_0,y_0)$处可微，$A\Delta x+B\Delta y$称为其在$M_0$处的全微分.

2.3 三个定理

1. 1.

   可微$⟹$可偏导，且$A=f_x^{\prime }(M_0),B=f_y^{\prime }(M_0)$（反之未必）.

2. 2.

   可微$⟹$连续（反之未必）.

3. 3.

   $z=f(x,y)$在$M_0(x_0,y_0)$处可微$⟺\underset{(\Delta x,\Delta y)\to (0,0)}{\lim }\frac{\Delta z-f_x^{\prime }(M_0)\Delta x-f_y^{\prime }(M_0)\Delta y}{\rho }$，其中$\rho =\sqrt{(\Delta x{)}^2+(\Delta y{)}^2}$.

2.4 例题

1. 1.

   证明：$f(x,y)=\sqrt{x^2+y^2}$在$(0,0)$处连续但不可偏导.

2. 2.

   证明：$f(x,y)=\{\begin{array}{ll}\frac{x^{2}y}{x^4+y^2},& x^2+y^2\ne 0,\\ 0,& x^2+y^2=0\end{array}$在$(0,0)$处可偏导但不连续，从而不可微，并求$f_x^{\prime }(1,1),f_y^{\prime }(2,3)$.

3. 3.

   设$x,y,z>0$，$a$为正的常数，$u(x,y,z)=x^{y^z}+x^{a^y}+a^{y^x}+y^{z^a}$，求$\frac{\partial u}{\partial x},\frac{\partial u}{\partial y},\frac{\partial u}{\partial z}$及${\mathrm{d}u|}_{(1,1,1)}$.

4. 4.

   设$f(x,y)={\int }_1^{x^{2}y}\frac{\sin t}{t}\mathrm{d}t$，求$\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}$.

5. 5.

   设变换 $\{\begin{array}{l}u=x-2y\\ v=x+ay\end{array}$可把方程$6\frac{{\partial }^{2}z}{\partial x^2}+\frac{{\partial }^{2}z}{\partial x\partial y}-\frac{{\partial }^{2}z}{\partial y^2}=0$简化为$\frac{{\partial }^{2}z}{\partial u\partial v}=0$，求常数$a$（其中二阶偏导数均连续）.

3高阶偏导数与多元复合函数微分法

3.1 多元函数微分学概念关系

1. 1.

   可微$⟹$可偏导、连续.

2. 2.

   存在且连续的$f_x^{\prime }(M_0),f_y^{\prime }(M_0)$$⟹$可微.

3. 3.

   存在且有界的$f_x^{\prime }(M_0),f_y^{\prime }(M_0)$$⟹$连续.

3.2 高阶偏导数及其特性

1. 1.

   若$z=f(x,y)$在区域$D$上连续，则$\frac{{\partial }^{2}f}{\partial x\partial y}=\frac{{\partial }^{2}f}{\partial y\partial x},\forall (x,y)\in D$.

2. 2.

   若$z=f(u,v),u=\phi (x,y),v=\psi (x,y)$皆可微，且复合有意义，则有链式法则

   $$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x},$$

   $$\frac{\partial f}{\partial y}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial y}.$$

3.3 例题

1. 1.

   设$u=f(x+y+z,x^2+y^2+z^2)$连续，求$\frac{\partial u}{\partial x},\frac{{\partial }^{2}u}{\partial x^2},\frac{{\partial }^{3}u}{{\partial }^{2}x\partial y}$.

2. 2.

   证明：$u(t,x)=\phi (x-at)+\psi (x+at)$是波动方程$\frac{{\partial }^{2}u}{\partial t^2}=a^2\frac{{\partial }^{2}u}{\partial x^2}(a,t>0,-\infty <x<+\infty )$的通解，其中$\phi ,\psi$连续.

3. 3.

   证明：$u(t,x)=\frac{1}{2a\sqrt{\pi t}}{\mathrm{e}}^{-x^2/4a^{2}t}$满足热传导方程$\frac{\partial u}{\partial t}=a^2\frac{{\partial }^{2}u}{\partial x^2}$.

4复合函数与隐函数微分法

4.1 隐函数存在定理及隐函数微分法

设$F(x,y)$连续，$F(M_0)=F(x_0,y_0)=0$，且$F_x^{\prime }(M_0)$或$F_y^{\prime }(M_0)$非零，则有

1. 1.

   $\exists \delta >0$，使得方程$F(x,y)=0$在$I=(x_0-\delta ,x_0+\delta )$中可唯一确定函数$y=\phi (x)$（隐函数），且$y_0=\phi (x_0)$.

2. 2.

   ${\phi }^{\prime }(x)=\frac{\mathrm{d}y}{\mathrm{d}x}$在$I$中连续，且${\phi }^{\prime }(x)=-\frac{F_x^{\prime }(x,y)}{F_y^{\prime }(x,y)}$.

计算隐函数的技巧：结合链式法则，从外层逐层剥离所需的复合函数关于某变量的偏导数，再进行回代.

4.2 例题

1. 1.

   设$y=f(x,t)$，而$t$是由方程$y+g(x,t)=0$所确定的$x,y$的函数，求$\frac{\mathrm{d}y}{\mathrm{d}x}$.

2. 2.

   设$x=x(z),y=y(z)$是方程组 $\{\begin{array}{l}x+y+z=0,\\ x^2+y^2+z^2=1\end{array}$所确定的隐函数组，求$\frac{\mathrm{d}x}{\mathrm{d}z},\frac{\mathrm{d}y}{\mathrm{d}z}$.

3. 3.

   设$u=u(x,y),v=v(x,y)$是由下列方程所确定的隐函数组，求$\frac{\partial (u,v)}{\partial (x,y)}$.

   1. (1)

      $\{\begin{array}{l}{u}^2+v^2+x^2+y^2=1\\ u+v+x+y=0\end{array}$;

   2. (2)

      $\{\begin{array}{l}xu-yv=0\\ yu+xv=1\end{array}$;

   3. (3)

      $\{\begin{array}{l}u=f(ux,v+y)\\ v=g(u-x,v^{2}y)\end{array}$

4. 4.

   求下列函数组所确定的反函数组的偏导数$\frac{\partial u}{\partial x},\frac{\partial u}{\partial y},\frac{\partial v}{\partial x},\frac{\partial v}{\partial y}$.

   1. (1)

      $\{\begin{array}{l}x=f(u,v)\\ y=g(u,v)\end{array}$

   2. (2)

      $\{\begin{array}{l}x={\mathrm{e}}^u+u\sin v\\ y={\mathrm{e}}^u-u\cos v\end{array}$

5补充习题

1. 1.

   证明：在极坐标变换$\{\begin{array}{l}x=\rho \cos \theta ,\\ y=\rho \sin \theta \end{array}$下，二维Laplace方程${\Delta }_{2}u=0$即$\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}=0$可写为如下形式：

   $$\frac{1}{\rho }\frac{\partial u}{\partial \rho }+\frac{{\partial }^{2}u}{\partial {\rho }^2}+\frac{1}{{\rho }^2}\frac{{\partial }^{2}u}{\partial {\theta }^2}=0\text{或}\frac{1}{\rho }\frac{\partial }{\partial \rho }\left(\rho \frac{\partial u}{\partial \rho }\right)+\frac{1}{{\rho }^2}\frac{{\partial }^{2}u}{\partial {\theta }^2}=0.$$
   证

   由题意，

   $${\partial }_{\rho }u=u_x\cos \theta +u_y\sin \theta ,$$

   $${\partial }_{\rho }(\rho {\partial }_{\rho }u)={\partial }_{\rho }u+\rho (u_{xx}{\cos }^2\theta +2u_{xy}\sin \theta \cos \theta +u_{yy}{\sin }^2\theta ),$$

   $${\partial }_{\theta }u=u_x\cdot (-\rho \sin \theta )+u_y\cdot \rho \cos \theta ,$$

   $${\partial }_{\theta }^{2}u={\rho }^2(u_{xx}{\sin }^2\theta -2u_{xy}\sin \theta \cos \theta +u_{yy}{\sin }^2\theta )-\rho (u_x\cos \theta +u_y\sin \theta ),$$

   因此

   $$0=\frac{1}{\rho }\frac{\partial }{\partial \rho }\left(\rho \frac{\partial u}{\partial \rho }\right)+\frac{1}{{\rho }^2}\frac{{\partial }^{2}u}{\partial {\theta }^2}$$

   $$=u_{xx}({\cos }^2\theta +{\sin }^2\theta )+0+u_{yy}({\sin }^2\theta +{\cos }^2\theta )+\frac{0}{\rho }=\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}.$$
2. 2.

   证明：在球坐标变换$\{\begin{array}{l}x=r\sin \theta \cos \phi ,\\ y=r\sin \theta \sin \phi ,\\ z=r\cos \theta \end{array}$下，三维Laplace方程${\Delta }_{3}u=0$即$\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}+\frac{{\partial }^{2}u}{\partial z^2}=0$可写为如下形式：

   $$\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial u}{\partial r}\right)+\frac{1}{r^2\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \theta \frac{\partial u}{\partial \theta }\right)+\frac{1}{r^2{\sin }^2\theta }\frac{{\partial }^{2}u}{\partial {\phi }^2}=0.$$
   证

   由题意，

   $${\partial }_{r}u=u_x\sin \theta \cos \phi +u_y\sin \theta \sin \phi +u_z\cos \theta ,$$

   $${\partial }_r(r^2{\partial }_{r}u)=2r{\partial }_{r}u+r^2(u_{xx}{\sin }^2\theta {\cos }^2\phi +u_{yy}{\sin }^2\theta {\sin }^2\phi +u_{zz}{\cos }^2\theta$$

   $$+2u_{xy}{\sin }^2\theta \sin \phi \cos \phi +2u_{yz}\sin \theta \cos \theta \sin \phi +2u_{zx}\sin \theta \cos \theta \cos \phi ),$$

   $${\partial }_{\theta }u=u_x\cdot r\cos \theta \cos \phi +u_y\cdot r\cos \theta \sin \phi +u_z\cdot (-r\sin \theta ),$$

   $${\partial }_{\theta }(\sin \theta {\partial }_{\theta }u)=\cos \theta {\partial }_{\theta }u+r\sin \theta (-u_x\sin \theta \cos \phi -u_y\sin \theta \sin \phi -u_z\cos \theta )$$

   $$+r^2\sin \theta (u_{xx}{\cos }^2\theta {\cos }^2\phi +u_{yy}{\cos }^2\theta {\sin }^2\phi +u_{zz}{\sin }^2\theta$$

   $$+2u_{xy}{\cos }^2\theta \sin \phi \cos \phi -2u_{yz}\sin \theta \cos \theta \sin \phi -2u_{zx}\sin \theta \cos \theta \cos \phi ),$$

   $${\partial }_{\phi }u=-u_x\cdot r\sin \theta \sin \phi +u_y\cdot r\sin \theta \cos \phi ,$$

   $${\partial }_{\phi }^{2}u=-r\sin \theta (u_x\cos \phi +\sin \phi )+r^2{\sin }^2\theta (u_{xx}{\sin }^2\phi +u_{yy}{\cos }^2\phi -2u_{xy}\sin \phi \cos \phi ),$$

   代入，按$u$的各阶导数合并项可证.

3. 3.

   设方程组$\{\begin{array}{l}F(x,y,u,v)=0,\\ G(x,y,u,v)=0\end{array}$确定了隐函数组$\{\begin{array}{l}u=u(x,y),\\ v=v(x,y)\end{array}$及反函数组$\{\begin{array}{l}x=x(u,v),\\ y=y(u,v).\end{array}$

   1. (1)

      求$\frac{\partial u}{\partial x},\frac{\partial u}{\partial y},\frac{\partial v}{\partial x},\frac{\partial v}{\partial y},\mathrm{d}u,\mathrm{d}v$.

   2. (2)

      证明²：$\frac{\partial (u,v)}{\partial (x,y)}\cdot \frac{\partial (x,y)}{\partial (u,v)}=1$.

      解

      1. i.

         将方程组两端取微分得

         $$\{\begin{array}{l}{F}_x^{\prime }\mathrm{d}x+F_y^{\prime }\mathrm{d}y+F_u^{\prime }\mathrm{d}u+F_v^{\prime }\mathrm{d}v=0,\\ G_x^{\prime }\mathrm{d}x+G_y^{\prime }\mathrm{d}y+G_u^{\prime }\mathrm{d}u+G_v^{\prime }\mathrm{d}v=0,\end{array}⟹\{\begin{array}{l}\mathrm{d}u=\frac{(F_v^{\prime }{G}_x^{\prime }-F_x^{\prime }{G}_v^{\prime })\mathrm{d}x+(F_v^{\prime }{G}_y^{\prime }-F_y^{\prime }{G}_v^{\prime })\mathrm{d}y}{F_u^{\prime }{G}_v^{\prime }-F_v^{\prime }{G}_u^{\prime }},\\ \mathrm{d}v=\frac{(F_u^{\prime }{G}_x^{\prime }-F_x^{\prime }{G}_u^{\prime })\mathrm{d}x+(F_u^{\prime }{G}_y^{\prime }-F_y^{\prime }{G}_u^{\prime })\mathrm{d}y}{F_v^{\prime }{G}_u^{\prime }-F_u^{\prime }{G}_v^{\prime }},\end{array}$$

      利用$\mathrm{d}u,\mathrm{d}v$的表达式，可求出$u,v$对$x,y$的偏导数. 不想算辣！

      注

      此结论可推广至$n$元情形：$\frac{\partial (u_1,u_2,\cdots ,u_n)}{\partial (x_1,x_2,\cdots ,x_n)}\cdot \frac{\partial (x_1,x_2,\cdots ,x_n)}{\partial (u_1,u_2,\cdots ,u_n)}=1$，其中 $\frac{\partial (u_1,u_2,\cdots ,u_n)}{\partial (x_1,x_2,\cdots ,x_n)}$为行列式，其第$(i,j)(i,j=1,2,\cdots ,n)$元为$\frac{\partial u_j}{\partial x_i}$（简记为${\partial }_i{u}_j$）.

4. 4.

   证明：方程$\frac{{\partial }^{2}u}{\partial x^2}+2\frac{{\partial }^{2}u}{\partial x\partial y}-3\frac{{\partial }^{2}u}{\partial y^2}+2\frac{\partial u}{\partial x}+6\frac{\partial u}{\partial y}=0$经变换$\xi =x+y,\eta =3x-y$后变成$\frac{{\partial }^{2}u}{\partial \xi \partial \eta }+\frac{1}{2}\frac{\partial u}{\partial \xi }=0$（其中二阶导数均连续），并求解此方程. 将$\xi (x,y),\eta (x,y)$反解得到$x=(\xi +\eta )/4,y=(3\xi -\eta )/4$，因此

   $${\partial }_{\xi }u=\frac{1}{4}{u}_x+\frac{3}{4}{u}_y,{\partial }_{\xi \eta }u=\frac{1}{16}{u}_{xx}+\frac{1}{8}{u}_{xy}-\frac{3}{16}{u}_{yy},$$

   那么方程可化为

   $$0=\frac{\partial u}{\partial \xi \eta }+\frac{1}{2}\frac{\partial u}{\partial \xi }=\frac{1}{16}\left(\frac{{\partial }^{2}u}{\partial x^2}+2\frac{{\partial }^{2}u}{\partial x\partial y}-3\frac{{\partial }^{2}u}{\partial y^2}+2\frac{\partial u}{\partial x}+6\frac{\partial u}{\partial y}\right).$$

   方程两端同时乘16即可. 现利用化简后的偏微分方程（即$u$关于$\xi ,\eta$的偏导数的方程），记$v=\frac{\partial u}{\partial \xi }$，则方程写为

   $$\frac{\partial v}{\partial \eta }+\frac{1}{2}v=0⟹v(\xi ,\eta )=f^{\prime }(\xi ){\mathrm{e}}^{-\eta /2},f(\xi )\in {\text{C}}^1(\mathbb{R}),$$

   那么

   $$u(x,y)=\int v(\xi ,\eta )\mathrm{d}\xi =f(\xi ){\mathrm{e}}^{-\eta /2}+g(\eta )=f(x+y){\mathrm{e}}^{-(3x-y)/2}+g(3x-y).$$
   注

   关于这类题在数理方程课程中有过讨论，其证明仅利用数学分析的知识，实际上并不困难，但这里只给出结论，有必要的话在后续课程中给出证明. 该偏微分方程对应的特征方程为$(\mathrm{d}y{)}^2-2(\mathrm{d}x)(\mathrm{d}y)-3(\mathrm{d}x{)}^2=0$. 由于$\Delta =(-1{)}^2-1\cdot (-3)=4>0$，故该方程为双曲型的. 解该方程得$\frac{\partial y}{\partial x}=3$和$\frac{\partial y}{\partial x}=-1$，即隐式解

   $$\phi (x,y)=x+y=0,\psi (x,y)=3x-y=0,$$

   因此作变量代换$\xi =\phi (x.y)=x+y,\eta =\psi (x,y)=3x-y$. 由于方程是双曲型的，因此这样的变量代换一定能将方程化为$\frac{{\partial }^{2}u}{\partial \xi \partial \eta }+c_1\frac{\partial u}{\partial \xi }+c_2\frac{\partial u}{\partial \eta }+c_{3}u=0$的形式. 利用此变量代换具体计算后和得系数$c_i$. 当然，因为已知方程形式，一种偷懒的方式是只算各阶偏导数中的一阶导系数，而其余项不管.

5. 5.

   证明：方程$\frac{{\partial }^{2}u}{\partial x^2}+2\frac{{\partial }^{2}u}{\partial x\partial y}\cos x-\frac{{\partial }^{2}u}{\partial y^2}{\sin }^{2}x-\frac{\partial u}{\partial y}\sin x=0$经变换$\xi =x-\sin x+y,\eta =x+\sin x-y$后变成$\frac{{\partial }^{2}u}{\partial \xi \partial \eta }=0$（其中二阶导数均连续），并求解此方程.

   证

   $$u_x={\partial }_{\xi }u(1-\cos x)+{\partial }_{\eta }u(1+\cos x),$$

   $$u_y={\partial }_{\xi }u-{\partial }_{\eta }u,$$

   $$u_{xx}=\sin x({\partial }_{\xi }u-{\partial }_{\eta }u)+{\partial }_{\xi \xi }u(1-\cos x{)}^2+{\partial }_{\eta \eta }(1+\cos x{)}^2+2{\partial }_{\xi \eta }u(1-{\cos }^{2}x),$$

   $$u_{xy}={\partial }_x(u_y)={\partial }_{\xi \xi }u(1-\cos x)+2u_{\xi \eta }\cos x-{\partial }_{\eta \eta }u(1+\cos x),$$

   $$u_{yy}={\partial }_{\xi \xi }u-2{\partial }_{\xi \eta }u+{\partial }_{\eta \eta }u,$$

   代入可证.

6. 6.

   设$u=f(x,y,z),\phi (x^2,{\mathrm{e}}^y,z)=0,y=\sin x$，其中$f,\phi$都具有一阶连续偏导数，且$\frac{\partial \phi }{\partial z}\ne 0$，求$\frac{\mathrm{d}u}{\mathrm{d}x}$.

   解

   $y=\sin x$确定了$y(x)$，那么代入$y(x)$后，$\phi (x,{\mathrm{e}}^y,z)=0$确定了$z(x)$. 由隐函数求辅导法则，

   $$z_x^{\prime }=-\frac{{\phi }_1^{\prime }+{\phi }_2^{\prime }\cdot {\mathrm{e}}^y{y}_x^{\prime }}{{\phi }_3^{\prime }}=-\frac{{\phi }_1^{\prime }+{\phi }_2^{\prime }{\mathrm{e}}^y\cos x}{{\phi }_3^{\prime }}.$$

   因此

   $$\frac{\mathrm{d}u}{\mathrm{d}x}=f_x^{\prime }+f_y^{\prime }{y}_x^{\prime }+f_z^{\prime }{z}_x^{\prime }=f_x^{\prime }+f_y^{\prime }\cos x-f_z^{\prime }\frac{{\phi }_1^{\prime }+{\phi }_2^{\prime }{\mathrm{e}}^y\cos x}{{\phi }_3^{\prime }}.$$
7. 7.

   设$y=y(x),z=z(x)$是由方程$z=xf(x+y)$和$F(x,y,z)=0$所确定的函数，其中$f$和$F$分别具有一阶连续导数和一阶连续偏导数，求$\frac{\mathrm{d}z}{\mathrm{d}x}$.

   解

   对两方程求全微分得

   $$\{\begin{array}{l}\mathrm{d}z=[f(x+y)+x{f}^{\prime }(x+y)]\mathrm{d}x+x{f}^{\prime }(x+y)\mathrm{d}y,\\ F_x^{\prime }\mathrm{d}x+F_y^{\prime }\mathrm{d}y+F_z^{\prime }\mathrm{d}z=0,\end{array}$$

   从中消去$\mathrm{d}y$项，最终可得

   $$\frac{\mathrm{d}z}{\mathrm{d}x}=\frac{F_y^{\prime }f(x+y)+x{f}^{\prime }(x+y)(F_y^{\prime }-F_x^{\prime })}{F_y^{\prime }+x{f}^{\prime }(x+y)F_z^{\prime }}.$$
8. 8.

   函数$u=u(x,y)$由方程组$u=f(x,y,z,t),g(y,z,t)=0,h(z,t)=0$定义，求$\frac{\partial u}{\partial x},\frac{\partial u}{\partial y}$.

   解

   先从外层剥离分析：由于需要求$u$关于$x,y$的偏导数，因此结合第一式，希望得到$z(x,y)$和$t(x,y)$。 由最后两式可知，$z,t$均只依赖于$y$，而与$x$无关，因此$\frac{\partial u}{\partial x}=f_x^{\prime }$. 将后两式取全微分得³：

   $$\{\begin{array}{l}{g}_y^{\prime }\mathrm{d}y+g_z^{\prime }\mathrm{d}z+g_t^{\prime }\mathrm{d}t=0,\\ h_z^{\prime }\mathrm{d}z+h_t^{\prime }\mathrm{d}t=0,\end{array}$$

   从中解得

   $$z_y^{\prime }=\frac{-h_t^{\prime }{g}_y^{\prime }}{h_t^{\prime }{g}_z^{\prime }-g_t^{\prime }{h}_z^{\prime }},t_y^{\prime }=\frac{h_z^{\prime }{g}_y^{\prime }}{h_t^{\prime }{g}_z^{\prime }-g_t^{\prime }{h}_z^{\prime }},$$

   因此

   $$\frac{\partial u}{\partial y}=f_y^{\prime }+f_z^{\prime }{z}_y^{\prime }+f_t^{\prime }{t}_y^{\prime }=f_y^{\prime }-g_y^{\prime }\frac{h_t^{\prime }{f}_z^{\prime }-f_t^{\prime }{h}_z^{\prime }}{h_t^{\prime }{g}_z^{\prime }-g_t^{\prime }{h}_z^{\prime }},\frac{\partial u}{\partial x}=f_x^{\prime }.$$