例题#

1. 求以下各题的偏导

   1. $z=\arctan \dfrac{y}{x}$

      解：$\dfrac{\partial z}{\partial x}=\dfrac{1}{1+(\dfrac{y}{x})^2}\cdot y\cdot (-\dfrac{1}{x^2})=-\dfrac{y}{x^2+y^2}$

      $\dfrac{\partial z}{\partial y}=\dfrac{1}{1+(\dfrac{y}{x})^2}\cdot \dfrac{1}{x}=\dfrac{x}{x^2+y^2}$

   2. $z=x^y$（$x\gt 0$）

      解：$\dfrac{\partial z}{\partial x}=yx^{y-1}$，$\dfrac{\partial z}{\partial y}=x^y\ln x$

   3. $f(x,y)=\dfrac{xy}{x^2+y^2}$

      解：$f_x=\dfrac{y(x^2+y^2)-xy\cdot 2x}{(x^2+y^2)^2}=\dfrac{y^3-x^2y}{(x^2+y^2)^2}$

      $f_y=\dfrac{x(x^2+y^2)-xy\cdot 2y}{(x^2+y^2)^2}=\dfrac{x^3-y^2x}{(x^2+y^2)^2}$

      若 $f(x,y)=f(y,x)$，$f_x=\varphi(x,y)$，则 $f_y=\varphi(y,x)$

2. $f(x,y)=x+(y-1)\arctan \dfrac{y}{x}$，求 $f_x(5,1)$，$f_y(5,1)$

   解：$f_x(5,1)=\dfrac{\mathrm d}{\mathrm dx}f(x,1)|_{x=5}=\dfrac{\mathrm d}{\mathrm dx}x|_{x=5}=1$

   $f_y(5,1)=0+\arctan\dfrac{y}{x}+(y-1)\dfrac{x}{x^2+y^2}=\arctan\dfrac{1}{5}$

3. $f(x,y)=\begin{cases}\dfrac{xy}{x^2+y^2},(x,y)\ne(0,0)\\0,(x,y)=(0,0)\end{cases}$ 在 $(0,0)$ 间断，讨论 $f_x(0,0)$、$f_y(0,0)$ 是否存在

   解：

   $$\begin{align*} f_x(0,0)&=\lim_{\Delta x\to 0}\dfrac{f(0+\Delta x,0)-f(0,0)}{\Delta x}\\ &=\lim_{\Delta x\to 0}\dfrac{\frac{\Delta x\cdot 0}{(\Delta x)^2+0^2}-0}{\Delta x}\\ &=0 \end{align*}$$

   同理 $f_y(0,0)=0$，故两者均存在

4. $f(x,y)=\begin{cases}\dfrac{\sin(xy)-y}{xy^2},xy\ne 0\\0,xy=0\end{cases}$，求 $f_y(1,0)$

   解：

   $$\begin{align*} f_y(1,0)&=\lim_{\Delta y\to 0}\dfrac{f(1,0+\Delta y)-f(1,0)}{\Delta y}\\ &=\lim_{\Delta y\to 0}\dfrac{\sin \Delta y-\Delta y}{(\Delta y)^3}\\ &=\lim_{\Delta y\to 0}\dfrac{\Delta y-\frac{(\Delta y)^3}{3!}+o[(\Delta y)^3]-\Delta y}{(\Delta y)^3}\\ &=-\dfrac{1}{6} \end{align*}$$

5. $\displaystyle u=\int_{xz}^{yz}e^{t^2}\mathrm dt$，求 $\dfrac{\partial u}{\partial x}$、$\dfrac{\partial u}{\partial y}$、$\dfrac{\partial u}{\partial z}$

   解： $\dfrac{\partial u}{\partial x}=-e^{(xz)^2}\cdot z$

   $\dfrac{\partial u}{\partial y}=e^{(yz)^2}\cdot z$

   $\dfrac{\partial u}{\partial z}=e^{(yz)^2}\cdot y-e^{(xz)^2}\cdot x$

6. $PV=RT$（R为常数），证明：$\dfrac{\partial P}{\partial V}\cdot \dfrac{\partial V}{\partial T}\cdot \dfrac{\partial T}{\partial P}=-1$

   解： $P=R\dfrac{T}{V}\Rightarrow \dfrac{\partial P}{\partial V}=RT\cdot (-\dfrac{1}{V^2})$

   $V=\dfrac{RT}{P}\Rightarrow \dfrac{\partial V}{\partial T}=\dfrac{R}{P}$

   $T=\dfrac{PV}{R}\Rightarrow \dfrac{\partial T}{\partial P}=\dfrac{V}{R}$

   以上三者相乘得 $-\dfrac{RT}{PV}=-1$，得证

一个二元函数连续但偏导不存在的例子：$z=\sqrt{x^2+y^2}$

• $\displaystyle \lim_{(x,y)\to (0,0)}\sqrt{x^2+y^2}=0=f(0,0)$，连续
• $\displaystyle z_x(0,0)=\lim_{\Delta x\to 0}\dfrac{\sqrt{(0+(\Delta x)^2)+0^2}-0}{\Delta x}=\lim_{\Delta x\to 0}\dfrac{|\Delta x|}{\Delta x}$，极限不存在
• $z_y(0,0)$ 同理

例题#

1. 求 $u=e^{x+2y}$ 的所有二阶偏导

   解：

   $\dfrac{\partial u}{\partial x}=e^{x+2y}\cdot 1=e^{x+2y}$，$\dfrac{\partial u}{\partial y}=2\cdot e^{x+2y}$

   $\dfrac{\partial ^2u}{\partial x^2}=e^{x+2y}$，$\dfrac{\partial ^2u}{\partial y^2}=4e^{x+2y}$

   $\dfrac{\partial ^2u}{\partial x\partial y}=2e^{x+2y}$，$\dfrac{\partial ^2u}{\partial y\partial x}=2e^{x+2y}$

2. $f(x,y)=\begin{cases}xy\cdot \dfrac{x^2-y^2}{x^2+y^2},x^2+y^2\ne 0\\0,x^2+y^2=0\end{cases}$，求 $f_{xy}(0,0)$、$f_{yx}(0,0)$

   解：

   $$\begin{align*} f_x=\begin{cases} \dfrac{y(x^4+4x^2y^2-y^4)}{(x^2+y^2)^2}&\quad x^2+y^2\ne 0\\ 0&\quad x^2+y^2=0 \end{cases} \end{align*}$$

   $\displaystyle f_{xy}(0,0)=\lim_{\Delta y\to 0}\dfrac{f_x(0,0+\Delta y)-f_x(0,0)}{\Delta y}=-1$

   同理

   $$\begin{align*} f_y=\begin{cases} y\cdot \dfrac{x^2-y^2}{x^2+y^2}+xy\dfrac{-4xy^4}{(x^2+y^2)^2} &\quad x^2+y^2\ne 0\\ 0&\quad x^2+y^2=0 \end{cases} \end{align*}$$

   $\displaystyle f_{yx}(0,0)=\lim_{\Delta x\to 0}\dfrac{f_y(0+\Delta x,0)-f_y(0,0)}{\Delta x}=1$

3. $u=x^2y^3e^{y+z}-yz\ln(1+x^2+y^2+z^2)$，求 $u_{xx}(0,1,-1)$、$u_{xy}(1,-1,0)$

   解： $u_{xx}(0,1,-1)=\dfrac{\mathrm d^2}{\mathrm dx^2}u(x,1,-1)|_{x=0}=\dfrac{\mathrm d^2}{\mathrm dx^2}[x^2+\ln(3+x^2)]|_{x=0}=\dfrac{8}{3}$

   $u_{xy}(1,-1,0)=\dfrac{\partial ^2}{\partial x\partial y}u(x,y,0)|_{x=1,y=-1}=\dfrac{\partial ^2}{\partial x\partial y}(x^2y^3e^y)|_{x=1,y=-1}=\dfrac{4}{e}$

4. $u=\dfrac{1}{\sqrt{x^2+y^2+z^2}}$，证明 $u_{xx}+u_{yy}+u_{zz}=0$

   证：$u_x=(-\dfrac{1}{2})(x^2+y^2+z^2)^{-\frac{3}{2}}\cdot 2x=-x(x^2+y^2+z^2)^{-\frac{3}{2}}$

   $u_{xx}=-(x^2+y^2+z^2)^{-\frac{3}{2}}-x(-\dfrac{3}{2})(x^2+y^2+z^2)^{-\frac{5}{2}}\cdot 2x=\dfrac{3x^2-(x^2+y^2+z^2)}{(x^2+y^2+z^2)^{\frac{5}{2}}}$

   $u_{yy}=\dfrac{3y^2-(x^2+y^2+z^2)}{(x^2+y^2+z^2)^{\frac{5}{2}}}$

   $u_{zz}=\dfrac{3z^2-(x^2+y^2+z^2)}{(x^2+y^2+z^2)^{\frac{5}{2}}}$

   $u_{xx}+u_{yy}+u_{zz}=0$，得证