第10章 多元函数的微分

Note

例

判断 $1<x^2+y^2<2$$1 < x^2 + y^2 < 2$ 的内点, 外点和边界点.

Note

例

集合 $1<x^2+y^2<2$$1 < x^2 + y^2 < 2$ 为开集.

Note

例子

集合 $\{(x,y)\mid 1<x^2+y^2<2\}$$\{(x,y) \mid 1 < x^2 + y^2 < 2\}$ 是开区域,而集合 $\{(x,y)\mid 1\le x^2+y^2\le 2\}$$\{(x,y) \mid 1 \leq x^2 + y^2 \leq 2\}$ 是闭区域.

Note

例1: 设 $f(x,y)=(x^2+y^2){\displaystyle \sin \frac{1}{x^2+y^2}}$$f(x,y) = (x^2 + y^2) \displaystyle\sin \frac{1}{x^2 + y^2}$,求证：${\displaystyle \underset{(x,y)\to (0,0)}{lim}f(x,y)=0.}$$\displaystyle\lim_{(x,y) \to (0,0)} f(x,y) = 0.$

证: 这里函数 $f(x,y)$$f(x,y)$ 的定义域为 $D={\mathbb{R}}^2\mathrm{\setminus }\{(0,0)\}$$D = \mathbb{R}^2 \backslash \{(0,0)\}$,点 $O(0,0)$$O(0,0)$ 为 $D$$D$ 的聚点, 因为$|f(x,y)-0|={\displaystyle |(x^2+y^2)\sin {\displaystyle \frac{1}{x^2+y^2}-0}|\le x^2+y^2,}$$|f(x,y) - 0| = \displaystyle\left| (x^2 + y^2) \sin \displaystyle\frac{1}{x^2 + y^2} - 0 \right| \leq x^2 + y^2,$ 可见,$\mathrm{\forall }\epsilon >0$$\forall \varepsilon > 0$,取 $\delta =\sqrt{\epsilon }$$\delta = \sqrt{\varepsilon}$,则当$0<\sqrt{(x-0{)}^2+(y-0{)}^2}<\delta ,$$0 < \sqrt{(x-0)^2 + (y-0)^2} < \delta,$ 即 $P(x,y)\in D\cap U(0,\delta )$$P(x,y) \in D \cap U(0,\delta)$ 时,总有 $|f(x,y)-0|<\epsilon$$|f(x,y) - 0| < \varepsilon$ 成立,所以 ${\displaystyle \underset{(x,y)\to (0,0)}{lim}f(x,y)=0.}$$\displaystyle\lim_{(x,y) \to (0,0)} f(x,y) = 0.$

例2: 考察函数在 $(0,0)$$(0, 0)$ 处的极限, $\begin{array}{c}f(x,y)=\{\begin{array}{ll}{\displaystyle \frac{xy}{x^2+y^2},}& x^2+y^2\ne 0,\\ 0,& x^2+y^2=0.\end{array}\end{array}$$f(x,y) = \begin{cases} \displaystyle\frac{xy}{x^2 + y^2}, & x^2 + y^2 \neq 0, \\0, & x^2 + y^2 = 0.\end{cases}$.

解: 首先, 我们看当点 $P(x,y)$$P(x,y)$ 沿 $x$$x$ 轴趋于点 $(0,0)$$(0,0)$ 时有,

又当点 $P(x,y)$$P(x,y)$ 沿 $y$$y$ 轴趋于点 $(0,0)$$(0,0)$ 时,

虽然以上述两种特殊方式（沿 $x$$x$ 轴或沿 $y$$y$ 轴）趋于原点时函数的极限存在且相等,但是不能以此断言该函数的极限存在. 例如, 当点 $P(x,y)$$P(x,y)$ 沿着直线 $y=kx$$y = kx$ 趋于点 $(0,0)$$(0,0)$ 时,有

$k$$k$ 不同极限也不同, 由此可以断言函数在$(0,0)$$(0, 0)$ 处的极限不存在.

例3: 求 ${\displaystyle \underset{(x,y)\to (0,2)}{lim}{\displaystyle \frac{\sin \left(xy\right)}{x}}}$$\displaystyle\lim_{(x,y) \to (0,2)} \displaystyle\frac{\sin(xy)}{x}$.

解: 函数 ${\displaystyle \frac{\sin \left(xy\right)}{x}}$$\displaystyle\frac{\sin(xy)}{x}$ 的定义域为 $D=\{(x,y)|x\ne 0,y\in \mathbb{R}\}$$D = \{(x,y) | x \neq 0, y \in \mathbb{R}\}$,$P_0(0,2)$$P_0(0,2)$ 为 $D$$D$ 的聚点,

Note

例1:求分段函 $\begin{array}{c}z=f(x,y)={\displaystyle \{\begin{array}{ll}{\displaystyle \frac{xy}{x^2+y^2},}& x^2+y^2\ne 0\\ 0,& x^2+y^2=0\end{array}}\end{array}$$z = f(x, y) = \displaystyle\begin{cases} \displaystyle\frac{xy}{x^2 + y^2}, & x^2 + y^2 \neq 0 \\ 0, & x^2 + y^2 = 0 \end{cases}$ 在点 (0,0) 处的偏导数.

解:

• 计算 $f_x(0,0)$$f_x(0, 0)$

$f_x(0,0)={\displaystyle \underset{\mathrm{\Delta }x\to 0}{lim}{\displaystyle \frac{f(0+\mathrm{\Delta }x,0)-f(0,0)}{\mathrm{\Delta }x}={\displaystyle \underset{\mathrm{\Delta }x\to 0}{lim}{\displaystyle \frac{0-0}{\mathrm{\Delta }x}=0}}}}$$f_x(0, 0) = \displaystyle\lim_{\Delta x \to 0} \displaystyle\frac{f(0+\Delta x, 0) - f(0, 0)}{\Delta x} = \displaystyle\lim_{\Delta x \to 0} \displaystyle\frac{0 - 0}{\Delta x} = 0$

• 计算 $f_y(0,0)$$f_y(0, 0)$

$f_y(0,0)={\displaystyle \underset{\mathrm{\Delta }y\to 0}{lim}{\displaystyle \frac{f(0,0+\mathrm{\Delta }y)-f(0,0)}{\mathrm{\Delta }y}={\displaystyle \underset{\mathrm{\Delta }y\to 0}{lim}{\displaystyle \frac{0-0}{\mathrm{\Delta }y}=0}}}}$$f_y(0, 0) = \displaystyle\lim_{\Delta y \to 0} \displaystyle\frac{f(0, 0+\Delta y) - f(0, 0)}{\Delta y} = \displaystyle\lim_{\Delta y \to 0} \displaystyle\frac{0 - 0}{\Delta y} = 0$

Note

例：设函数 $z=x^3{y}^2-3x{y}^3-xy+1$$z = x^3 y^2 - 3xy^3 - xy + 1$,求下列高阶偏导数： ${\displaystyle \frac{{\partial }^{2}z}{\partial x^2},{\displaystyle \frac{{\partial }^{2}z}{\partial y\partial x},{\displaystyle \frac{{\partial }^{2}z}{\partial x\partial y},{\displaystyle \frac{{\partial }^{2}z}{\partial y^2}\text{及}{\displaystyle \frac{{\partial }^{3}z}{\partial x^3}.}}}}}$$\displaystyle\frac{\partial^2 z}{\partial x^2},\ \displaystyle\frac{\partial^2 z}{\partial y \partial x},\ \displaystyle\frac{\partial^2 z}{\partial x \partial y},\ \displaystyle\frac{\partial^2 z}{\partial y^2} \ \text{及}\ \displaystyle\frac{\partial^3 z}{\partial x^3} .$

解:
先求一阶偏导数: ${\displaystyle \frac{\partial z}{\partial x}=3x^2{y}^2-3y^3-y}$$\displaystyle\frac{\partial z}{\partial x} = 3x^2 y^2 - 3y^3 - y$ ${\displaystyle \frac{\partial z}{\partial y}=2x^{3}y-9x{y}^2-x}$$\displaystyle\frac{\partial z}{\partial y} = 2x^3 y - 9xy^2 - x$ 然后求二阶偏导数： ${\displaystyle \frac{{\partial }^{2}z}{\partial x^2}={\displaystyle \frac{\partial }{\partial x}(3x^2{y}^2-3y^3-y)=6x{y}^2}}$$\displaystyle\frac{\partial^2 z}{\partial x^2} = \displaystyle\frac{\partial}{\partial x}(3x^2 y^2 - 3y^3 - y) = 6xy^2$ ${\displaystyle \frac{{\partial }^{2}z}{\partial y\partial x}={\displaystyle \frac{\partial }{\partial y}(3x^2{y}^2-3y^3-y)=6x^{2}y-9y^2-1}}$$\displaystyle\frac{\partial^2 z}{\partial y \partial x} = \displaystyle\frac{\partial}{\partial y}(3x^2 y^2 - 3y^3 - y) = 6x^2 y - 9y^2 - 1$ ${\displaystyle \frac{{\partial }^{2}z}{\partial x\partial y}={\displaystyle \frac{\partial }{\partial x}(2x^{3}y-9x{y}^2-x)=6x^{2}y-9y^2-1}}$$\displaystyle\frac{\partial^2 z}{\partial x \partial y} = \displaystyle\frac{\partial}{\partial x}(2x^3 y - 9xy^2 - x) = 6x^2 y - 9y^2 - 1$ ${\displaystyle \frac{{\partial }^{2}z}{\partial y^2}={\displaystyle \frac{\partial }{\partial y}(2x^{3}y-9x{y}^2-x)=2x^3-18xy}}$$\displaystyle\frac{\partial^2 z}{\partial y^2} = \displaystyle\frac{\partial}{\partial y}(2x^3 y - 9xy^2 - x) = 2x^3 - 18xy$ 最后求三阶偏导数 ${\displaystyle \frac{{\partial }^{3}z}{\partial x^3}={\displaystyle \frac{\partial }{\partial x}(6x{y}^2)=6y^2}}$$\displaystyle\frac{\partial^3 z}{\partial x^3} = \displaystyle\frac{\partial}{\partial x}(6xy^2) = 6y^2$

Note

例1:验证函数 $z=\ln \sqrt{x^2+y^2}$$z = \ln \sqrt{x^2 + y^2}$ 满足拉普拉斯方程 ${\displaystyle \frac{{\partial }^{2}z}{\partial x^2}+{\displaystyle \frac{{\partial }^{2}z}{\partial y^2}=0.}}$$\displaystyle\frac{\partial^2 z}{\partial x^2} + \displaystyle\frac{\partial^2 z}{\partial y^2} = 0 .$

证明:
首先将函数化简为： $z={\displaystyle \frac{1}{2}\ln (x^2+y^2)}$$z = \displaystyle\frac{1}{2} \ln(x^2 + y^2)$
${\displaystyle \frac{\partial z}{\partial x}={\displaystyle \frac{x}{x^2+y^2}}}$$\displaystyle\frac{\partial z}{\partial x} = \displaystyle\frac{x}{x^2 + y^2}$ ${\displaystyle \frac{\partial z}{\partial y}={\displaystyle \frac{y}{x^2+y^2}}}$$\displaystyle\frac{\partial z}{\partial y} = \displaystyle\frac{y}{x^2 + y^2}$ ${\displaystyle \frac{{\partial }^{2}z}{\partial x^2}={\displaystyle \frac{y^2-x^2}{(x^2+y^2{)}^2}}}$$\displaystyle\frac{\partial^2 z}{\partial x^2} = \displaystyle\frac{y^2 - x^2}{(x^2 + y^2)^2}$ ${\displaystyle \frac{{\partial }^{2}z}{\partial y^2}={\displaystyle \frac{x^2-y^2}{(x^2+y^2{)}^2}}}$$\displaystyle\frac{\partial^2 z}{\partial y^2} = \displaystyle\frac{x^2 - y^2}{(x^2 + y^2)^2}$ 验证方程 将二阶偏导数相加： ${\displaystyle \frac{{\partial }^{2}z}{\partial x^2}+{\displaystyle \frac{{\partial }^{2}z}{\partial y^2}={\displaystyle \frac{y^2-x^2+x^2-y^2}{(x^2+y^2{)}^2}=0}}}$$\displaystyle\frac{\partial^2 z}{\partial x^2} + \displaystyle\frac{\partial^2 z}{\partial y^2} = \displaystyle\frac{y^2 - x^2 + x^2 - y^2}{(x^2 + y^2)^2} = 0$

例2:证明函数 $u={\displaystyle \frac{1}{r}}$$u = \dfrac{1}{r}$ 满足拉普拉斯方程${\displaystyle \frac{{\partial }^{2}u}{\partial x^2}+{\displaystyle \frac{{\partial }^{2}u}{\partial y^2}+{\displaystyle \frac{{\partial }^{2}u}{\partial z^2}=0}}}$$\displaystyle\frac{\partial^2 u}{\partial x^2} + \displaystyle\frac{\partial^2 u}{\partial y^2} + \displaystyle\frac{\partial^2 u}{\partial z^2} = 0$,其中 $r=\sqrt{x^2+y^2+z^2}$$r = \sqrt{x^2 + y^2 + z^2}$.

证明: ${\displaystyle \frac{\partial u}{\partial x}=-{\displaystyle \frac{1}{r^2}\cdot {\displaystyle \frac{\partial r}{\partial x}=-{\displaystyle \frac{x}{r^3}}}}}$$\displaystyle\frac{\partial u}{\partial x} = -\displaystyle\frac{1}{r^2} \cdot \displaystyle\frac{\partial r}{\partial x} = -\displaystyle\frac{x}{r^3}$ ${\displaystyle \frac{{\partial }^{2}u}{\partial x^2}=-{\displaystyle \frac{1}{r^3}+{\displaystyle \frac{3x^2}{r^5}}}}$$\displaystyle\frac{\partial^2 u}{\partial x^2} = -\displaystyle\frac{1}{r^3} + \displaystyle\frac{3x^2}{r^5}$ 由对称性可得： ${\displaystyle \frac{{\partial }^{2}u}{\partial y^2}=-{\displaystyle \frac{1}{r^3}+{\displaystyle \frac{3y^2}{r^5},{\displaystyle \frac{{\partial }^{2}u}{\partial z^2}=-{\displaystyle \frac{1}{r^3}+{\displaystyle \frac{3z^2}{r^5}}}}}}}$$\displaystyle\frac{\partial^2 u}{\partial y^2} = -\displaystyle\frac{1}{r^3} + \displaystyle\frac{3y^2}{r^5}, \quad \displaystyle\frac{\partial^2 u}{\partial z^2} = -\displaystyle\frac{1}{r^3} + \displaystyle\frac{3z^2}{r^5}$ 验证方程
$\begin{array}{rl}{\displaystyle \frac{{\partial }^{2}u}{\partial x^2}+{\displaystyle \frac{{\partial }^{2}u}{\partial y^2}+{\displaystyle \frac{{\partial }^{2}u}{\partial z^2}}}}& =-{\displaystyle \frac{3}{r^3}+{\displaystyle \frac{3(x^2+y^2+z^2)}{r^5}}}\\ & =-{\displaystyle \frac{3}{r^3}+{\displaystyle \frac{3r^2}{r^5}}}\\ & =0\end{array}$$\begin{aligned} \displaystyle\frac{\partial^2 u}{\partial x^2} + \displaystyle\frac{\partial^2 u}{\partial y^2} + \displaystyle\frac{\partial^2 u}{\partial z^2} &= -\displaystyle\frac{3}{r^3} + \displaystyle\frac{3(x^2 + y^2 + z^2)}{r^5} \\&= -\displaystyle\frac{3}{r^3} + \displaystyle\frac{3r^2}{r^5} \\ &= 0 \end{aligned}$

Note

例1:$z(x,y)=x+y$$z(x,y) = x + y$.

解: 变量变化：$x\to x+\mathrm{\Delta }x,y\to y+\mathrm{\Delta }y$$x \to x + \Delta x, \quad y \to y + \Delta y$ 函数增量计算： $z(x+\mathrm{\Delta }x,y+\mathrm{\Delta }y)=(x+\mathrm{\Delta }x)+(y+\mathrm{\Delta }y)$$z(x + \Delta x, y + \Delta y) = (x + \Delta x) + (y + \Delta y)$ 增量分解： $\mathrm{\Delta }z=z(x+\mathrm{\Delta }x,y+\mathrm{\Delta }y)-z(x,y)=\mathrm{\Delta }x+\mathrm{\Delta }y$$\Delta z = z(x + \Delta x, y + \Delta y) - z(x, y) = \Delta x + \Delta y$ 偏导数表示： ${\displaystyle \frac{\partial z}{\partial x}=1,{\displaystyle \frac{\partial z}{\partial y}=1}}$$\displaystyle\frac{\partial z}{\partial x} = 1, \quad \displaystyle\frac{\partial z}{\partial y} = 1$ 全微分公式： $dz=dx+dy$$dz = dx + dy$

例2:函数 $z(x,y)=x^2+2y^3$$z(x,y) = x^2 + 2y^3$.

解: 考虑自变量的微小变化： $\{\begin{array}{l}x\to x+dx\\ y\to y+dy\end{array}$$\begin{cases} x \rightarrow x + dx \\ y \rightarrow y + dy \end{cases}$ 函数增量计算： $\begin{array}{rl}z(x+dx,y+dy)& =(x+dx{)}^2+2(y+dy{)}^3\\ & =x^2+2xdx+d{x}^2+2(y^3+3y^{2}dy+3yd{y}^2+d{y}^3)\\ & =x^2+2y^3+2xdx+6y^{2}dy+\underset{\text{高阶无穷小项}}{\underbrace{d{x}^2+6yd{y}^2+2d{y}^3}}\end{array}$$\begin{aligned} z(x+dx,y+dy) &= (x+dx)^2 + 2(y+dy)^3 \\ &= x^2 + 2xdx + dx^2 + 2(y^3 + 3y^2dy + 3ydy^2 + dy^3) \\ &= x^2 + 2y^3 + 2xdx + 6y^2dy + \underbrace{dx^2 + 6ydy^2 + 2dy^3}_{\text{高阶无穷小项}} \end{aligned}$ 线性主部提取： 保留一阶增量项 $\mathrm{\Delta }z\approx 2xdx+6y^{2}dy$$\Delta z \approx 2xdx + 6y^2dy$ 偏导数计算： ${\displaystyle \frac{\partial z}{\partial x}=2x\text{和}{\displaystyle \frac{\partial z}{\partial y}=6y^2}}$$\displaystyle\frac{\partial z}{\partial x} = 2x \quad \text{和} \quad \displaystyle\frac{\partial z}{\partial y} = 6y^2$ 全微分公式：$dz={\displaystyle \frac{\partial z}{\partial x}dx+{\displaystyle \frac{\partial z}{\partial y}dy=2xdx+6y^{2}dy}}$$dz = \displaystyle\frac{\partial z}{\partial x}dx + \displaystyle\frac{\partial z}{\partial y}dy = 2xdx + 6y^2dy$

例3:$S={\displaystyle \frac{1}{2}(L-2x+L-2x+2x\cos \theta )x\sin \theta }$$S = \displaystyle\frac{1}{2} \left( L - 2x + L - 2x + 2x \cos \theta \right) x \sin \theta$.

解: 化简：$S(x,\theta )=Lx\sin \theta -2x^2\sin \theta +x^2\sin \theta \cos \theta$$S(x, \theta) = L x \sin \theta - 2x^2 \sin \theta + x^2 \sin \theta \cos \theta$ 变量代换过程：$x\to x+\mathrm{\Delta }x,\theta \to \theta +\mathrm{\Delta }\theta$$x \to x + \Delta x, \quad \theta \to \theta + \Delta \theta$ 函数增量展开： $\begin{array}{rl}S(x+\mathrm{\Delta }x,\theta +\mathrm{\Delta }\theta )& =L(x+\mathrm{\Delta }x)\sin (\theta +\mathrm{\Delta }\theta )\\ & -2(x+\mathrm{\Delta }x{)}^2\sin (\theta +\mathrm{\Delta }\theta )\\ & +(x+\mathrm{\Delta }x{)}^2\sin (\theta +\mathrm{\Delta }\theta )\cos (\theta +\mathrm{\Delta }\theta )\end{array}$$\begin{aligned} S(x+\Delta x, \theta+\Delta \theta) &= L (x+\Delta x) \sin(\theta+\Delta \theta) \\ &\quad -2(x+\Delta x)^2 \sin(\theta+\Delta \theta) \\ &\quad + (x+\Delta x)^2 \sin(\theta+\Delta \theta)\cos(\theta+\Delta \theta) \end{aligned}$ 线性近似处理（保留一阶项）： $\begin{array}{rl}\mathrm{\Delta }S& \approx (L\sin \theta -4x\sin \theta +2x\sin \theta \cos \theta )\mathrm{\Delta }x\\ & +(Lx\cos \theta -2x^2\cos \theta -x^2{\sin }^2\theta +x^2{\cos }^2\theta )\mathrm{\Delta }\theta \end{array}$$\begin{aligned} \Delta S &\approx (L \sin \theta - 4x \sin \theta + 2x \sin \theta \cos \theta)\Delta x \\ &\quad + (L x \cos \theta - 2x^2 \cos \theta - x^2 \sin^2 \theta + x^2 \cos^2 \theta)\Delta \theta \end{aligned}$ 偏导数提取： $\{\begin{array}{l}{\displaystyle \frac{\partial S}{\partial x}=L\sin \theta -4x\sin \theta +x\sin \theta \cos \theta }\\ {\displaystyle \frac{\partial S}{\partial \theta }=Lx\cos \theta -2x^2\cos \theta +x^2({\cos }^2\theta -{\sin }^2\theta )}\end{array}$$\begin{cases} \displaystyle\frac{\partial S}{\partial x} = L \sin \theta - 4x \sin \theta + x \sin \theta \cos \theta \\ \displaystyle\frac{\partial S}{\partial \theta} = L x \cos \theta - 2x^2 \cos \theta + x^2 (\cos^2 \theta - \sin^2 \theta) \end{cases}$ 全微分公式：$dS={\displaystyle \frac{\partial S}{\partial x}dx+{\displaystyle \frac{\partial S}{\partial \theta }d\theta }}$$dS = \displaystyle\frac{\partial S}{\partial x}dx + \displaystyle\frac{\partial S}{\partial \theta}d\theta$ 最终结果：$dS=\sin \theta (L-4x+2x\cos \theta )dx+[x\cos \theta (L-2x)+x^2\cos 2\theta ]d\theta$$dS = \sin\theta (L - 4x + 2x \cos\theta)dx + [x \cos\theta (L - 2x) + x^2 \cos 2\theta]d\theta$

Note

同济教材 P78 例3: 设 $z=uv+\sin t$$z = uv + \sin t$，而 $u=e^t,v=\cos t$$u = e^t, v = \cos t$，求全导数 ${\displaystyle \frac{dz}{dt}}$$\displaystyle\frac{dz}{dt}$.

解: 注意这里 $z$$z$ 不仅通过 $u,v$$u, v$ 依赖于 $t$$t$，还直接依赖于 $t$$t$。根据链式法则： $\begin{array}{rl}\frac{dz}{dt}& =\frac{\partial z}{\partial u}\frac{du}{dt}+\frac{\partial z}{\partial v}\frac{dv}{dt}+\frac{\partial z}{\partial t}\\ & =v\cdot e^t+u\cdot (-\sin t)+\cos t\\ & =e^t\cos t-e^t\sin t+\cos t\\ & =e^t(\cos t-\sin t)+\cos t.\end{array}$$\begin{aligned} \frac{dz}{dt} &= \frac{\partial z}{\partial u}\frac{du}{dt} + \frac{\partial z}{\partial v}\frac{dv}{dt} + \frac{\partial z}{\partial t} \\ &= v \cdot e^t + u \cdot (-\sin t) + \cos t \\ &= e^t \cos t - e^t \sin t + \cos t \\ &= e^t(\cos t - \sin t) + \cos t. \end{aligned}$

同济教材 P78 例2: 设 $u=e^{x^2+y^2+z^2}$$u = e^{x^2+y^2+z^2}$，而 $z=x^2\sin y$$z = x^2 \sin y$，求 ${\displaystyle \frac{\partial u}{\partial x}}$$\displaystyle\frac{\partial u}{\partial x}$ 与 ${\displaystyle \frac{\partial u}{\partial y}}$$\displaystyle\frac{\partial u}{\partial y}$.

解: 这里的 $z$$z$ 是中间变量，将其视为 $z(x,y)$$z(x,y)$。应用链式法则： $\begin{array}{rl}\frac{\partial u}{\partial x}& =\frac{\partial f}{\partial x}+\frac{\partial f}{\partial z}\frac{\partial z}{\partial x}\\ & =2x{e}^{x^2+y^2+z^2}+2z{e}^{x^2+y^2+z^2}\cdot (2x\sin y)\\ & =2x(1+2z\sin y)e^{x^2+y^2+z^2}.\end{array}$$\begin{aligned} \frac{\partial u}{\partial x} &= \frac{\partial f}{\partial x} + \frac{\partial f}{\partial z}\frac{\partial z}{\partial x} \\ &= 2x e^{x^2+y^2+z^2} + 2z e^{x^2+y^2+z^2} \cdot (2x \sin y) \\ &= 2x(1 + 2z \sin y)e^{x^2+y^2+z^2}. \end{aligned}$

同理求对 $y$$y$ 的偏导： $\begin{array}{rl}\frac{\partial u}{\partial y}& =\frac{\partial f}{\partial y}+\frac{\partial f}{\partial z}\frac{\partial z}{\partial y}\\ & =2y{e}^{x^2+y^2+z^2}+2z{e}^{x^2+y^2+z^2}\cdot (x^2\cos y)\\ & =2(y+x^{2}z\cos y)e^{x^2+y^2+z^2}.\end{array}$$\begin{aligned} \frac{\partial u}{\partial y} &= \frac{\partial f}{\partial y} + \frac{\partial f}{\partial z}\frac{\partial z}{\partial y} \\ &= 2y e^{x^2+y^2+z^2} + 2z e^{x^2+y^2+z^2} \cdot (x^2 \cos y) \\ &= 2(y + x^2 z \cos y)e^{x^2+y^2+z^2}. \end{aligned}$

Note

同济教材 P81 例6: 设 $z=e^u\sin v$$z = e^u \sin v$，而 $u=xy,v=x+y$$u = xy, v = x+y$，利用全微分形式不变性求偏导数.

解: 先写出外层函数的全微分： $dz=d(e^u\sin v)=e^u\sin vdu+e^u\cos vdv$$dz = d(e^u \sin v) = e^u \sin v \, du + e^u \cos v \, dv$

再计算中间变量的微分： $du=d(xy)=ydx+xdy$$du = d(xy) = y \, dx + x \, dy$ $dv=d(x+y)=dx+dy$$dv = d(x+y) = dx + dy$

将 $du,dv$$du, dv$ 直接代入 $dz$$dz$ 并合并同类项（像初中代数一样合并 $dx$$dx$ 和 $dy$$dy$）： $\begin{array}{rl}dz& =e^u\sin v(ydx+xdy)+e^u\cos v(dx+dy)\\ & =[y{e}^u\sin v+e^u\cos v]dx+[x{e}^u\sin v+e^u\cos v]dy\end{array}$$\begin{aligned} dz &= e^u \sin v (y \, dx + x \, dy) + e^u \cos v (dx + dy) \\ &= [y e^u \sin v + e^u \cos v] dx + [x e^u \sin v + e^u \cos v] dy \end{aligned}$

比较等式两端 $dx$$dx$ 和 $dy$$dy$ 的系数，直接就能读出偏导数： $\frac{\partial z}{\partial x}=e^{xy}[y\sin (x+y)+\cos (x+y)]$$\frac{\partial z}{\partial x} = e^{xy}[y \sin(x+y) + \cos(x+y)]$ $\frac{\partial z}{\partial y}=e^{xy}[x\sin (x+y)+\cos (x+y)]$$\frac{\partial z}{\partial y} = e^{xy}[x \sin(x+y) + \cos(x+y)]$ 这个方法比死记链式法则公式要直观、安全得多！

Note

同济教材 P84 例1 (修改版): 验证方程 $x^2+y^2-1=0$$x^2 + y^2 - 1 = 0$ 确定了一个隐函数，并求其一阶与二阶导数 ${\displaystyle \frac{dy}{dx}}$$\displaystyle\frac{dy}{dx}$ 和 ${\displaystyle \frac{d^{2}y}{d{x}^2}}$$\displaystyle\frac{d^2y}{dx^2}$.

解: 设 $F(x,y)=x^2+y^2-1$$F(x,y) = x^2 + y^2 - 1$。 容易算得 $F_x=2x,F_y=2y$$F_x = 2x, \quad F_y = 2y$。 根据一元隐函数求导公式，得一阶导数： $\frac{dy}{dx}=-\frac{F_x}{F_y}=-\frac{2x}{2y}=-\frac{x}{y}$$\frac{dy}{dx} = -\frac{F_x}{F_y} = -\frac{2x}{2y} = -\frac{x}{y}$

进一步求二阶导数（对一阶导数关于 $x$$x$ 再求一次导，注意 $y$$y$ 是 $x$$x$ 的函数）： $\begin{array}{rl}\frac{d^{2}y}{d{x}^2}& =\frac{d}{dx}(-\frac{x}{y})=-\frac{1\cdot y-x\cdot \frac{dy}{dx}}{y^2}\\ & =-\frac{y-x(-\frac{x}{y})}{y^2}=-\frac{y^2+x^2}{y^3}\end{array}$$\begin{aligned} \frac{d^2y}{dx^2} &= \frac{d}{dx}\left(-\frac{x}{y}\right) = -\frac{1 \cdot y - x \cdot \frac{dy}{dx}}{y^2} \\ &= -\frac{y - x(-\frac{x}{y})}{y^2} = -\frac{y^2 + x^2}{y^3} \end{aligned}$ 因为原方程 $x^2+y^2=1$$x^2+y^2=1$，所以代入后得到最简结果：$\frac{d^{2}y}{d{x}^2}=-\frac{1}{y^3}$$\frac{d^2y}{dx^2} = -\frac{1}{y^3}$。

同济教材 P85 例2: 设 $x^2+y^2+z^2-4z=0$$x^2 + y^2 + z^2 - 4z = 0$，求 ${\displaystyle \frac{\partial z}{\partial x}}$$\displaystyle\frac{\partial z}{\partial x}$ 及 ${\displaystyle \frac{{\partial }^{2}z}{\partial x^2}}$$\displaystyle\frac{\partial^2 z}{\partial x^2}$.

解: 设 $F(x,y,z)=x^2+y^2+z^2-4z$$F(x,y,z) = x^2 + y^2 + z^2 - 4z$。 分别求偏导数：$F_x=2x$$F_x = 2x$，$F_z=2z-4$$F_z = 2z - 4$。 当 $F_z\ne 0$$F_z \neq 0$（即 $z\ne 2$$z \neq 2$）时，利用公式得： $\frac{\partial z}{\partial x}=-\frac{F_x}{F_z}=-\frac{2x}{2z-4}=\frac{x}{2-z}$$\frac{\partial z}{\partial x} = -\frac{F_x}{F_z} = -\frac{2x}{2z - 4} = \frac{x}{2 - z}$

再对 $x$$x$ 求一次偏导数得到二阶导（注意把 $y$$y$ 看作常量，而 $z$$z$ 是 $x$$x$ 的函数）： $\begin{array}{rl}\frac{{\partial }^{2}z}{\partial x^2}& =\frac{\partial }{\partial x}\left(\frac{x}{2-z}\right)\\ & =\frac{1\cdot (2-z)-x\cdot (-\frac{\partial z}{\partial x})}{(2-z{)}^2}\\ & =\frac{(2-z)+x(\frac{x}{2-z})}{(2-z{)}^2}=\frac{(2-z{)}^2+x^2}{(2-z{)}^3}\end{array}$$\begin{aligned} \frac{\partial^2 z}{\partial x^2} &= \frac{\partial}{\partial x}\left(\frac{x}{2-z}\right) \\ &= \frac{1 \cdot (2-z) - x \cdot (- \frac{\partial z}{\partial x})}{(2-z)^2} \\ &= \frac{(2-z) + x(\frac{x}{2-z})}{(2-z)^2} = \frac{(2-z)^2 + x^2}{(2-z)^3} \end{aligned}$

Note

例3:求 $\mathrm{grad}{\displaystyle \frac{1}{x^2+y^2}}$$\mathrm{grad}\ \dfrac{1}{x^2 + y^2}$.

解:这里 $f(x,y)={\displaystyle \frac{1}{x^2+y^2}}$$f(x, y) = \dfrac{1}{x^2 + y^2}$,因为 ${\displaystyle \frac{\partial f}{\partial x}=-{\displaystyle \frac{2x}{(x^2+y^2{)}^2},{\displaystyle \frac{\partial f}{\partial y}=-{\displaystyle \frac{2y}{(x^2+y^2{)}^2},}}}}$$\displaystyle\frac{\partial f}{\partial x} = -\displaystyle\frac{2x}{(x^2 + y^2)^2}, \quad \displaystyle\frac{\partial f}{\partial y} = -\displaystyle\frac{2y}{(x^2 + y^2)^2},$ 所以
$\mathrm{grad}{\displaystyle \frac{1}{x^2+y^2}}=-{\displaystyle \frac{2x}{(x^2+y^2{)}^2}\mathbf{i}-{\displaystyle \frac{2y}{(x^2+y^2{)}^2}\mathbf{j}.}}$$\mathrm{grad}\ \dfrac{1}{x^2 + y^2} = -\displaystyle\frac{2x}{(x^2 + y^2)^2} \mathbf{i} - \displaystyle\frac{2y}{(x^2 + y^2)^2} \mathbf{j}.$

例5:设 $f(x,y,z)=x^3-x{y}^2-z^2,P_0(1,1,0)$$f(x, y, z) = x^3 - x y^2 - z^2,\ P_0(1,1,0)$,问 $f(x,y,z)$$f(x,y,z)$ 在 $P_0$$P_0$ 处沿什么方向变化最快, 在这个方向的变化率是多少?

解: $\mathrm{\nabla }f={\displaystyle \frac{\partial f}{\partial x}}\mathbf{i}+{\displaystyle \frac{\partial f}{\partial y}}\mathbf{j}+{\displaystyle \frac{\partial f}{\partial z}}\mathbf{k}=(3x^2-y^2)\mathbf{i}-2xy\mathbf{j}-2z\mathbf{k},\mathrm{\nabla }f(1,1,0)=2\mathbf{i}-2\mathbf{j}-\mathbf{k}$$\nabla f = \dfrac{\partial f}{\partial x}\mathbf{i} + \dfrac{\partial f}{\partial y}\mathbf{j} + \dfrac{\partial f}{\partial z}\mathbf{k} = (3x^2 - y^2)\mathbf{i} - 2xy\mathbf{j} - 2z\mathbf{k},\ \nabla f(1,1,0) = 2\mathbf{i} - 2\mathbf{j} - \mathbf{k}$.
$f(x,y,z)$$f(x,y,z)$ 在 $P_0$$P_0$ 处沿 $\mathrm{\nabla }f(1,1,0)$$\nabla f(1,1,0)$ 的方向增加最快,沿 $-\mathrm{\nabla }f(1,1,0)$$-\nabla f(1,1,0)$ 的方向减少最快,在这两个方向的变化率分别是
$|\mathrm{\nabla }f(1,1,0)|=\sqrt{2^2+(-2{)}^2+1^2}=3,-|\mathrm{\nabla }f(1,1,0)|=-3.$$|\nabla f(1,1,0)| = \sqrt{2^2 + (-2)^2 + 1^2} = 3, \quad -|\nabla f(1,1,0)| = -3.$

例6:求曲面 $x^2+y^2+z=9$$x^2 + y^2 + z = 9$ 在点 $P_0(1,2,4)$$P_0(1,2,4)$ 的切平面和法线方程.

解: 设 $f(x,y,z)=x^2+y^2+z$$f(x, y, z) = x^2 + y^2 + z$,由梯度与等值面的关系可知,梯度
$\mathrm{\nabla }f{|}_{P_0}=(2x\mathbf{i}+2y\mathbf{j}+\mathbf{k}){|}_{(1,2,4)}=2\mathbf{i}+4\mathbf{j}+\mathbf{k}$$\nabla f \Big|_{P_0} = (2x\mathbf{i} + 2y\mathbf{j} + \mathbf{k})\Big|_{(1,2,4)} = 2\mathbf{i} + 4\mathbf{j} + \mathbf{k}$的方向是等值面 $f(x,y,z)=9$$f(x,y,z)=9$ 在点 $P_0$$P_0$ 的法线方向,因此切平面方程是 $2(x-1)+4(y-2)+(z-4)=0,$$2(x - 1) + 4(y - 2) + (z - 4) = 0,$ 即
$2x+4y+z=14,$$2x + 4y + z = 14,$ 曲面在 $P_0$$P_0$ 处的法线方程是
$x=1+2t,y=2+4t,z=4+t(t\text{为任意常数}).$$x = 1 + 2t,\ y = 2 + 4t,\ z = 4 + t \quad (t\ \text{为任意常数}).$