定义#

若 $z=f(x,y)$ 的全增量 $f(x+\Delta x, y+\Delta y)-f(x,y)=A(x,y)\Delta x+B(x,y)\Delta y+o(\rho)$，其中 $\rho =\sqrt{(\Delta x)^2+(\Delta y)^2}$，则称 $z=f(x,y)$ 在 $(x,y)$ 处可微，记

定理#

1. [可微的必要条件]可微 $\Rightarrow$ 偏导存在，且 $\mathrm dz=\dfrac{\partial z}{\partial x}\mathrm dx+\dfrac{\partial z}{\partial y}\mathrm dy$（可扩展至 $\mathrm du=\dfrac{\partial u}{\partial x}\mathrm dx+\dfrac{\partial u}{\partial y}\mathrm dy+\dfrac{\partial u}{\partial z}\mathrm dz$）

   证明：

   $$\begin{align*} f(x+\Delta x,y+\Delta y)-f(x,y)=A(x,y)\Delta x+B(x,y)\Delta y+o(\rho) \end{align*}$$

   令 $\Delta y=0$

   $$\begin{align*} f(x+\Delta x,y)-f(x,y)=A(x,y)\Delta x+o(|\Delta x|) \end{align*}$$

   两边同除以 $\Delta x$，再对 $\Delta x$ 求极限（$\Delta x\to 0$）

   $f_x(x,y)=A(x,y)+0$

   令 $\Delta x=0$，同理可得

   $$\begin{align*} f_y(x,y)=B(x,y) \end{align*}$$

   即得证

2. [可微的充分条件]偏导连续 $\Rightarrow$ 可微

   证明：

   $$\begin{align*} \Delta z&=f(x+\Delta x,y+\Delta y)-f(x,y+\Delta y)+f(x,y+\Delta y)-f(x,y)\\ &=f_x(x+\theta_1\Delta x,y+\Delta y)\Delta x+f_y(x+\Delta x,y+\theta_2\Delta y)\Delta y\qquad[0\lt \theta_1\lt 1, 0\lt \theta_2\lt 1,拉格朗日中值定理]\\ &=[f_x(x,y)+\epsilon_1]\Delta x+[f_y(x,y)+\epsilon_2]\Delta y\qquad[\lim_{(\Delta x,\Delta y)\to (0,0)}\epsilon_1=0,\lim_{(\Delta x,\Delta y)\to (0,0)}\epsilon_2=0] \end{align*}$$

   $\displaystyle \lim_{(\Delta x,\Delta y)\to (0,0)}\dfrac{\Delta z-(f_x\Delta x+f_y\Delta y)}{\sqrt{(\Delta x)^2+(\Delta y)^2}}=\lim_{(\Delta x,\Delta y)\to (0,0)}\dfrac{\epsilon_1\Delta x+\epsilon_2\Delta y}{\sqrt{(\Delta x)^2+(\Delta y)^2}}=0$

   $\Delta z=f_x\Delta x+f_y\Delta y+\underbrace{\epsilon_1\Delta x+\epsilon_2\Delta y}_{o(\rho)}$，证毕

例题#

1. $\displaystyle z=\int_x^ye^{-t^2}\mathrm dt$，求 $\mathrm dz|_{(0,1)}$

   解：$z_{x(0,1)}=-e^{-x^2}|_{(0,1)}=-1$

   $z_{y(0,1)}=e^{-y^2}|_{(0,1)}=e^{-1}$

   $\mathrm dz=-\mathrm dx+\dfrac{1}{e}\mathrm dy$

2. $f(x,y)=\begin{cases}\dfrac{xy}{x^2+y^2}&\quad (x,y)\ne(0,0)\\0&\quad(x,y)=(0,0)\end{cases}$ 在 $(0,0)$ 偏导存在，证明：在 $(0,0)$ 处不可微

   解：

   $$\begin{align*} \lim_{(\Delta x,\Delta y)\to (0,0)}\dfrac{\frac{\Delta x\cdot \Delta y}{(\Delta x)^2+(\Delta y)^2}-(0\cdot \Delta x+0\cdot \Delta y)}{\sqrt{(\Delta x)^2+(\Delta y)^2}}&=\lim_{(\Delta x,\Delta y)\to (0,0)}\dfrac{\Delta x\Delta y}{[(\Delta x)^2+(\Delta y)^2]^\frac{3}{2}}\\ &=\lim_{(\Delta x,\Delta y)\to (0,0),\Delta y=\Delta x}\dfrac{(\Delta x)^\frac{3}{2}}{2^\frac{3}{2}(\Delta x)^3}，不存在 \end{align*}$$

   $\therefore$ 在 $(0,0)$ 处不可微

3. $(axy^3-y^2\cos x)\mathrm dx+(1+by\sin x+3x^2y^2)\mathrm dy$ 是某 $f(x,y)$ 的全微分，求 $a,b$

   解： $\mathrm df=\dfrac{\partial f}{\partial x}\mathrm dx+\dfrac{\partial f}{\partial y}\mathrm dy$

   $\dfrac{\partial f}{\partial x}=axy^3-y^2\cos x$

   $\dfrac{\partial f}{\partial y}=1+by\sin x+3x^2y^2$

   $\dfrac{\partial ^2f}{\partial x\partial y}=\dfrac{\partial ^2f}{\partial y\partial x}$

   $3axy^2-2y\cos x=by\cos x+6xy^2\Rightarrow\begin{cases}3a=6\\-2=b\end{cases}\Rightarrow \begin{cases}a=2\\b=-2\end{cases}$

4. $f(x,y)=\begin{cases}(x^2+y^2)\sin\dfrac{1}{x^2+y^2}&\quad (x,y)\ne (0,0)\\0&\quad (x,y)=(0,0)\end{cases}$，讨论 $f(x,y)$ 在 $(0,0)$：连续性、偏导是否存在、偏导是否连续、可微性

   解：$\displaystyle \lim_{(x,y)\to(0,0)}(x^2+y^2)\sin\dfrac{1}{x^2+y^2}=0=f(0,0)$，连续

   $\displaystyle f_x(0,0)=\lim_{\Delta x\to 0}\dfrac{f(0+\Delta x, 0)-f(0,0)}{\Delta x}=0$，$f_y(0,0)=0$，偏导存在

   当 $(x,y)\ne(0,0)$ 时，$\displaystyle f_x=2x\cdot \sin\dfrac{1}{x^2+y^2}+(x^2+y^2)\cos\dfrac{1}{x^2+y^2}[\dfrac{-2x}{(x^2+y^2)^2}]$

   $\displaystyle \lim_{(x,y)\to(0,0)}f_x(x,y)$ 不存在，$\displaystyle \lim_{(x,y)\to(0,0)}f_y(x,y)$ 也不存在，偏导不连续

   $\displaystyle \lim_{(x,y)\to(0,0)}\dfrac{[(\Delta x)^2+(\Delta y)^2]\sin\frac{1}{(\Delta x)^2+(\Delta y)^2}-(0\cdot \Delta x+0\cdot \Delta y)}{\sqrt{(\Delta x)^2+(\Delta y)^2}}=0$ 可微

判断 $z=f(x,y)$ 是否可微

式1指：$\displaystyle \lim_{\Delta x\to 0,\Delta y\to 0}\dfrac{(f(x_0+\Delta x,y_0+\Delta y)-f(x_0,y_0))-(f'_x(x_0,y_0)\Delta x+(f'_y(x_0,y_0)\Delta y)}{\sqrt{(\Delta x)^2+(\Delta y)^2}}$

例题#

求 $(1.02)^{2.05}$ 的近似值

解： 法一：$f(x,y)=x^y$，$\Delta x=0.02$，$\Delta y=0.05$

法二： $(1+0.02)^{2.05}\approx 1+\dfrac{2.05}{1!}\times 0.02=1.041$