1. 中间变量为一元函数#

定理： 若 $u=\phi(t)$、$v=\psi(t)$ 在 $t$ 处可导，且 $z=f(u,v)$ 在 $(u,v)$ 处偏导连续，则 $z=f[\phi(t),\psi(t)]$ 在 $t$ 可导，且

$z=f(u,v,w)$，$u=u(t)$，$v=v(t)$，$w=w(t)$ 在 $t$ 处可导，偏导连续，有

2. 中间变量为多元函数#

若 $z=f(u,v)$ 在 $(u,v)$ 处偏导连续，且 $u=\phi(x,y)$、$v=\psi(x,y)$ 在 $(x,y)$ 偏导存在，则 $z=f[\phi(x,y),\psi(x,y)]$ 在 $(x,y)$ 偏导存在，且

$w=f(u)$，$u=u(x,y,z)$，有

3. 中间变量既有一元函数，又有二元函数#

$z=f[\phi(x,y),\psi(y)]$ 在 $(x,y)$ 偏导存在，有

4. 中间变量也是自变量#

若 $u=\phi(x,y)$ 在 $(x,y)$ 偏导存在，且 $z=f(u,x,y)$ 在 $(u,x,y)$ 偏导连续，则 $z=f[\phi(x,y),x,y]$ 在 $(x,y)$ 偏导存在

例题#

1. $z=x^2e^y$，$x=2\cos t$，$y=t+\sin t$，求 $\dfrac{\mathrm dz}{\mathrm dt}$

   解:

   • 法一：

     $z=(2\cos t)^2e^{t+\sin t}=4\cos^2te^{t+\sin t}$

     $$\begin{align*} \dfrac{\mathrm dz}{\mathrm dt}&=4\cdot[(-2\sin t\cos t)e^{t+\sin t}+\cos ^2t(1+\cos t)e^{t+\sin t}]\\ &=-4\sin 2te^{t+\sin t}+4\cos^2t(1+\cos t)e^{t+\sin t} \end{align*}$$

   • 法二：

     $$\begin{align*} \dfrac{\mathrm dz}{\mathrm dt}&=\dfrac{\partial f}{\partial x}\cdot \dfrac{\mathrm dx}{\mathrm dt}+\dfrac{\partial f}{\partial y}\cdot \dfrac{\mathrm dy}{\mathrm dt}\\ &=2xe^y\cdot(-2\sin t)+x^2e^y(1+\cos t)\\ &=2\cdot 2\cos te^{t+\sin t}(-2\sin t)+4\cos^2te^{t+\sin t}(1+\cos t)\\ &=-4\sin 2te^{t+\sin t}+(4\cos^2t+4\cos^3t)e^{t+\sin t} \end{align*}$$

2. $y=x^x$（$x\gt 0$），求 $\dfrac{\mathrm dy}{\mathrm dx}$

   解:

   • 法一：

     $y=e^{x\ln x}$，$\dfrac{\mathrm dy}{\mathrm dx}=e^{x\ln x}(1+\ln x)=x^x(\ln x+1)$

   • 法二：

     $y=u^v$，$u=x$，$v=x$

     $$\begin{align*} \dfrac{\mathrm dy}{\mathrm dx}&=\dfrac{\partial y}{\partial u}\cdot \dfrac{\mathrm du}{\mathrm dx}+\dfrac{\partial y}{\partial v}\cdot \dfrac{\mathrm dv}{\mathrm dx}\\ &=vu^{v-1}\cdot 1+u^v\ln u\cdot 1\\ &=x^x(\ln x+1) \end{align*}$$

3. $z=f(u,v)=\begin{cases}\dfrac{u^2v}{u^2+v^2}&(u,v)\ne(0,0)\\0&(u,v)=(0,0)\end{cases}$，$u=t$，$v=t$，求 $\dfrac{\mathrm dz}{\mathrm dx}|_{t=0}$

   解：【本题不可使用链式法则，因为 $f(u,v)$ 在 $(0,0)$ 偏导不连续】

   $$\begin{align*} z=\begin{cases} \dfrac{t^3}{t^2+t^2}=\dfrac{t}{2} & t\ne 0\\ 0 & t=0 \end{cases} \end{align*}$$

   $\therefore z=\dfrac{t}{2},t\in R$

   $\dfrac{\mathrm dz}{\mathrm dx}|_{t=0}=\dfrac{1}{2}$

4. $z=e^u\cos v$，$u=2x-y$，$v=xy$，求 $\dfrac{\partial z}{\partial x}$、$\dfrac{\partial z}{\partial y}$

解：

• 法1：$z=e^{2x-y}\cos(xy)$，$\dfrac{\partial z}{\partial x}=2e^{2x-y}\cos(xy)+e^{2x-y}[-\sin(xy)]\cdot y$，$\dfrac{\partial z}{\partial y}=-e^{2x-y}\cos(xy)+e^{2x-y}[-\sin(xy)]\cdot x$

• 法2：

  $$\begin{align*} \dfrac{\partial z}{\partial x}&=\dfrac{\partial f}{\partial u}\cdot \dfrac{\partial u}{\partial x}+\dfrac{\partial f}{\partial x}\cdot \dfrac{\partial v}{\partial x}\\ &=e^u\cos v\cdot 2+(-\sin v)e^uy\\ &=2e^{2x-y}\cos(xy)+[-\sin(xy)]e^{2x-y}y \end{align*}$$ $$\begin{align*} \dfrac{\partial z}{\partial y}&=\dfrac{\partial f}{\partial u}\cdot \dfrac{\partial u}{\partial y}+\dfrac{\partial f}{\partial v}\cdot \dfrac{\partial v}{\partial y}\\ &=e^u\cos v\cdot(-1)+e^u(-\sin v)x\\ &=-e^{2x-y}\cos(xy)-e^{2x-y}\sin(xy)\cdot x \end{align*}$$

5. $z=f(x,u)=x^2+u$，$u=\cos (xy)$，求 $\dfrac{\partial z}{\partial x}$、$\dfrac{\partial f}{\partial x}$

   解：$\dfrac{\partial z}{\partial x}=\dfrac{\partial f}{\partial u}\cdot \dfrac{\partial u}{\partial x}+\dfrac{\partial f}{\partial x}\cdot 1=1\cdot(-\sin xy)\cdot y+2x$，$\dfrac{\partial f}{\partial x}=2x$

6. $z=f(u,v)$ 在 $(u,v)$ 偏导连续，求 $\dfrac{\partial z}{\partial x}$、$\dfrac{\partial z}{\partial y}$，其中 $u=3x+2y$，$v=x^2+y^2$

   解：$\dfrac{\partial z}{\partial x}=\dfrac{\partial f}{\partial u}\cdot 3+\dfrac{\partial f}{\partial v}\cdot 2x=3f_1'+2xf_2'$，$\dfrac{\partial z}{\partial y}=\dfrac{\partial f}{\partial u}\cdot 2+\dfrac{\partial f}{\partial v}\cdot 2y=2f_1'+2yf_2'$

   6-1 $z=f(u,v)$ 在 $(u,v)$ 二阶偏导也连续，求 $\dfrac{\partial ^2z}{\partial x^2}$、$\dfrac{\partial ^2z}{\partial y^2}$、$\dfrac{\partial ^2z}{\partial x \partial y}$，其他同上

   解：

   $$\begin{align*} \dfrac{\partial ^2z}{\partial x^2}&=\dfrac{\partial }{\partial x}(\dfrac{\partial z}{\partial x})\\ &=\dfrac{\partial }{\partial x}(3f_1'+2xf_2')\\ &=3\dfrac{\partial }{\partial x}f_1'+2(f_2'+xf_{21}'')\\ &=3f_{11}''+2xf_{21}'' \end{align*}$$ $$\begin{align*} \dfrac{\partial ^2z}{\partial y^2}&=\dfrac{\partial }{\partial y}(\dfrac{\partial z}{\partial y})\\ &=\dfrac{\partial }{\partial y}(2f_1'+2yf_2')\\ &=2f_{12}''+2(f_2'+yf_{22}'')\\ &=2f_{12}''+2yf_{22}'' \end{align*}$$ $$\begin{align*} \dfrac{\partial ^2z}{\partial x \partial y}&=\dfrac{\partial }{\partial y}(\dfrac{\partial z}{\partial x})\\ &=\dfrac{\partial }{\partial y}(3f_1'+2xf_2')\\ &=3f_{12}''+2xf_{22}'' \end{align*}$$

7. $z=f(e^x\sin y)$，求 $\dfrac{\partial ^2z}{\partial x^2}$、$\dfrac{\partial ^2z}{\partial x \partial y}$

   解： $\dfrac{\partial z}{\partial x}=\dfrac{\mathrm df}{\mathrm du}\cdot \dfrac{\partial u}{\partial x}=f'e^x\sin y$

   $\dfrac{\partial z}{\partial y}=\dfrac{\mathrm df}{\mathrm du}\cdot \dfrac{\partial u}{\partial y}=f'e^x\cos y$

   $$\begin{align*} \dfrac{\partial ^2z}{\partial x^2}&=\dfrac{\partial }{\partial x}(f'e^x\sin y)\\ &=\sin y[(\dfrac{\partial }{\partial x}f)'e^x+f'e^x]\\ &=\sin y(f''e^x\sin y\cdot e^2+f'e^x)\\ &=\sin y(f''e^{2x}\sin y+f'e^x) \end{align*}$$ $$\begin{align*} \dfrac{\partial ^2z}{\partial x \partial y}&=\dfrac{\partial }{\partial y}(f'e^x\sin y)\\ &=e^x[(\dfrac{\partial }{\partial y}f)'\sin y+f'\cos y]\\ &=e^x(f''\cos y\sin y+f'\cos y) \end{align*}$$

8. $u=f(x,y)$ 二阶偏导连续，证明经过 $x=\rho \cos\theta$、$y=\rho \sin \theta$ 变换后的方程满足如下关系：

   • $(\dfrac{\partial u}{\partial x})^2+(\dfrac{\partial u}{\partial y})^2=(\dfrac{\partial u}{\partial \rho})^2+\dfrac{1}{\rho^2}(\dfrac{\partial u}{\partial \theta})^2$
   • $\dfrac{\partial ^2u}{\partial x^2}+\dfrac{\partial ^2y}{\partial y^2}=\dfrac{\partial ^2u}{\partial \rho^2}+\dfrac{1}{\rho}\cdot \dfrac{\partial u}{\partial \rho}+\dfrac{1}{\rho^2}(\dfrac{\partial ^2u}{\partial \theta^2})$

   解： $\dfrac{\partial u}{\partial \rho}=\dfrac{\partial }{\partial \rho}f(\rho \cos\theta, \rho \sin\theta)=f_1'\cos\theta+f_2'\sin\theta$

   $\dfrac{\partial u}{\partial \theta}=\dfrac{\partial }{\partial \theta}f(\rho \cos\theta, \rho \sin\theta)=-\rho f_1'\sin \theta+\rho f_2'\cos \theta$

   右边 $=(f_1'\cos\theta+f_2'\sin\theta)^2+\dfrac{1}{\rho^2}(-\rho f_1'\sin\theta+\rho f_2'\cos\theta)=f_1'^2+f_2'^2=$ 左边，1式得证

   $$\begin{align*} \dfrac{\partial ^2u}{\partial \rho^2}&=\dfrac{\partial }{\partial \rho}(\dfrac{\partial u}{\partial \rho})\\ &=\cos\theta \dfrac{\partial }{\partial \rho}f_1'+\sin\theta\dfrac{\partial }{\partial \rho}f_2'\\ &=\cos\theta(f_{11}''\cos\theta+f_{12}''\sin\theta)+\sin\theta(f_{21}''\cos\theta+f_{22}''\sin\theta)\\ &=\cos^2\theta f_{11}''+\sin^2\theta f_{12}''+\sin 2\theta f_{12}'' \end{align*}$$ $$\begin{align*} \dfrac{\partial ^2u}{\partial \theta^2}&=\dfrac{\partial }{\partial \theta}(\dfrac{\partial u}{\partial \theta})\\ &=\dfrac{\partial }{\partial \theta}(-\rho f_1'\sin\theta+\rho f_2'\cos\theta)\\ &=-\rho [\dfrac{\partial }{\partial \theta}(f_1'\sin\theta)-\dfrac{\partial }{\partial \theta}(f_2'\cos \theta)]\\ &=-\rho\{\dfrac{\partial }{\partial \theta}f_1'\sin\theta+f_1'\cos\theta-[\dfrac{\partial }{\partial \theta}f_2'\cos\theta+f_2'(-\sin\theta)]\}\\ &=-\rho[\sin\theta(-\rho f_{11}''\sin\theta+\rho f_{12}''\cos\theta)-\cos\theta(-\rho f_{21}''\sin\theta+\rho f_{22}''\cos\theta)]-\rho\cdot \dfrac{\partial u}{\partial \rho}\\ &=\rho^2 \sin^2\theta f_{11}''+\rho^2 \cos^2\theta f_{22}''-\rho^2 \sin 2\theta f_{12}''-\rho\cdot \dfrac{\partial u}{\partial \rho} \end{align*}$$

   右边 $=\cos^2\theta f_{11}''+\sin^2\theta f_{22}''+\sin 2\theta f_{12}''+\dfrac{1}{\rho}(f_1'\cos\theta+f_2'\sin\theta)+\dfrac{1}{\rho^2}(\rho^2\sin^2\theta f_{11}''+\rho^2\cos^2\theta f_{22}''-\rho^2\sin 2\theta f_{12}''-\rho\cdot \dfrac{\partial u}{\partial \rho})=f_{11}''+f_{12}''=$ 左边，2式得证

例题#

1. $z=e^{\sin(xy)}$，求 $\dfrac{\partial z}{\partial x}$、$\dfrac{\partial z}{\partial y}$

   解：

   $$\begin{align*} \mathrm dz&=\mathrm d[e^{\sin(xy)}]\\ &=e^{\sin (xy)}\mathrm d\sin(xy)\\ &=e^{\sin (xy)}\cos (xy)\mathrm d(xy)\\ &=e^{\sin (xy)}\cos (xy)(y \mathrm dx+ x \mathrm dy) \end{align*}$$

   $\dfrac{\partial z}{\partial x}=y\cdot e^{\sin (xy)}\cos (xy)$

   $\dfrac{\partial z}{\partial y}=x\cdot e^{\sin (xy)}\cos (xy)$

2. $u=f(x,y,z)$，$y=\phi(x,t)$，$t=\psi(x,z)$，求 $\dfrac{\partial u}{\partial x}$、$\dfrac{\partial u}{\partial z}$

   解：

   $$\begin{align*} \mathrm du&=\mathrm df(x,y,z)\\ &=f_1'\mathrm dx+f_2'\mathrm dy+f_3'\mathrm dz\\ &=f_1'\mathrm dx+f_2'\mathrm d\phi(x,t)+f_3'\mathrm dz\\ &=f_1'\mathrm dx+f_2'\mathrm d(\phi_1'\mathrm dx+\phi_2'\mathrm dt)+f_3'\mathrm dz\\ &=f_1'\mathrm dx+f_2'[\phi_1'\mathrm dx+\phi_2'(\psi_1' \mathrm dx+\psi_2' \mathrm dz)]+f_3'\mathrm dz\\ &=(f_1'+f_2'\phi_1'+f_2'\phi_2'\psi_1')\mathrm dx+(f_3'+f_2'\phi_2'\psi_2')\mathrm dz \end{align*}$$

   $\dfrac{\partial u}{\partial x}=f_1'+f_2'\phi_1'+f_2'\phi_2'\psi_1'$

   $\dfrac{\partial u}{\partial z}=f_3'+f_2'\phi_2'\psi_2'$