偏导数与全微分

偏导数

函数$f(x,y)$在点$x_0,y_0$关于$x$的偏导数：

\begin{align}f_{x}\left(x_{0}, y_{0}\right) =\left.\frac{\partial f}{\partial x}\right|_{(x_0,y_0)}=\left.\frac{\mathrm{d}}{\mathrm{d}x}f(x,y_0)\right|_{x=x_0}=\lim _{\Delta x \rightarrow 0} \frac{f\left(x_{0}+\Delta x, y_{0}\right)-f\left(x_{0}, y_{0}\right)}{\Delta x}\end{align}

关于$y$的偏导数为：

\begin{align}f_{y}\left(x_{0}, y_{0}\right)=\left.\frac{\partial f}{\partial y}\right|_{(x_0,y_0)}=\left.\frac{\mathrm{d}}{\mathrm{d}y}f(x_0,y)\right|_{y=y_0}=\lim _{\Delta y \rightarrow 0} \frac{f\left(x_{0}, y_{0}+\Delta y\right)-f\left(x_{0}, y_{0}\right)}{\Delta y}\end{align}

偏导数的几何意义

偏导数为函数在每个位置处沿着自变量坐标轴方向上的导数（切线斜率）。如果对$x$求导就是$y$值不变，$x$变化率的切线斜率，例子如下图

以二元函数为例，令$z=f(x,y)$，绘制在3维坐标系如下图所示

在分别固定$y$和$x$的取值后得到下图中的黑色曲线——“退化”为一元函数，二维坐标系中的曲线——则偏导数$\frac{\partial z}{\partial x},\frac{\partial z}{\partial y}$分别为曲线的导数（切线斜率）。

由上可知，一个变量对应一个坐标轴，偏导数为函数在每个位置处沿着自变量坐标轴方向上的导数（切线斜率）

视频的话可以看下面

偏导数的求法

对于$z=f(x,y)$求偏导$\frac{\partial z}{\partial x},\frac{\partial z}{\partial y}$

$\frac{\partial z}{\partial x}$：将$y$看做常量，而对$x$求导

$\frac{\partial z}{\partial y}$：将$x$看做常量，而对$y$求导

N元函数的偏导数

偏导数的概念还可推广到二元以上的函数。例如三元函数$u=f(x,y,z)$在点$(x,y,z)$处对$x$的偏导数定义为

\begin{align}f_{x}(x,y,z)=\lim_{\Delta x \rightarrow 0}\frac{f(x+\Delta x,y,z)-f(x,y,z)}{\Delta x}\end{align}

其中$(x,y,z)$是函数$u=f(x,y,z)$的定义域的内点。它们的求法也仍旧是一元函数的微分法问题。

求导方法

$z=f(x,y)$，偏导数有2个：$\frac{\partial z}{\partial x},\frac{\partial z}{\partial y}$

$u=f(x,y,z)$，偏导数有3个：$\frac{\partial u}{\partial x},\frac{\partial u}{\partial y},\frac{\partial u}{\partial z}$

偏导数例题

例1：求$z=x^{2}+3xy+y^{2}$在点$(1,2)$处的偏导数。

解：把$y$看做常量，得

\begin{align}\frac{\partial z}{\partial x}=2x+3y\end{align}

把$x$看做常量，得

\begin{align}\frac{\partial z}{\partial y}=3x+2y\end{align}

将$(1,2)$代人上面的结果，就得

\begin{align}\left.\frac{\partial z}{\partial x}\right|_{\substack{x=1\\y=2}}=2\cdot 1+3\cdot2=8\\\left.\frac{\partial z}{\partial y}\right|_{\substack{x=1\\y=2}}=3\cdot 1+2\cdot 2=7\end{align}

例2：求$z=x^{2}\sin2y$的偏导数。

解：

\begin{align}&\frac{\partial z}{\partial x}=2x\sin2y\\&\frac{\partial z}{\partial y}=2x^{2}\cos2y\end{align}

例3：求$r=\sqrt{x^{2}+y^{2}+z^{2}}$的偏导数.

解：因为函数是复合函数，先把根号里面的内容看做一个整体，进行求导

$(\sqrt{u})^{\prime}=\frac{1}{2\sqrt{u}}$，然后所以需要把$y$和$z$都看做常量，再次对$u$求导$u=x^2+y^2+z^2$，得到下面公式

\begin{align}\frac{\partial r}{\partial x}=\frac{2x}{2\sqrt{x^{2}+y^{2}+z^{2}}}=\frac{x}{\sqrt{x^{2}+y^{2}+z^{2}}}=\frac{x}{r}\end{align}

同理其他偏导数也可以一样带入，得出$\frac{\partial r}{\partial y}=\frac{y}{r},\frac{\partial r}{\partial z}=\frac{z}{r}$

高阶偏导数

设函数$z=f(x,y)$在区域$D$内具有偏导数

$\frac{\partial z}{\partial x}=f_{x}(x,y),\quad\frac{\partial z}{\partial y}=f_{y}(x,y)$

于是在$D$内$f_{x}(x,y),f_{y}(x,y)$都是$x,y$的函数。如果这两个函数的偏导数也存在，那么称它们是函数$z=f(x,y)$的二阶偏导数。按照对变量求导次序的不同有下列四个二阶偏导数：

\begin{align}&\frac{\partial}{\partial x}\left(\frac{\partial z}{\partial x}\right)=\frac{\partial(\frac{\partial z}{\partial x})}{\partial x}=\frac{\partial^{2} z}{\partial x^{2}}=f_{x x}(x, y)\\& \frac{\partial}{\partial y}\left(\frac{\partial z}{\partial x}\right)=\frac{\partial(\frac{\partial z}{\partial x})}{\partial y}=\frac{\partial^{2} z}{\partial x \partial y}=f_{x y}(x, y)\\& \frac{\partial}{\partial x}\left(\frac{\partial z}{\partial y}\right)=\frac{\partial(\frac{\partial z}{\partial y})}{\partial x}=\frac{\partial^{2} z}{\partial y \partial x}=f_{y x}(x, y)\\ &\frac{\partial}{\partial y}\left(\frac{\partial z}{\partial y}\right)=\frac{\partial(\frac{\partial z}{\partial y})}{\partial y}=\frac{\partial^{2} z}{\partial y^{2}}=f_{y y}(x, y)\end{align}

上面第二个和第三个公式为混合偏导

例题

设$z=x^{3}y^{2}-3xy^{3}-xy+1$，求$\frac{\partial^{2} z}{\partial x^{2}},\frac{\partial^{2} z}{\partial y\partial x},\frac{\partial^{2} z}{\partial x\partial y},\frac{\partial^{2} z}{\partial y^{2}},\frac{\partial^{3} z}{\partial x^{3}}$

\begin{align} &\frac{\partial z}{\partial x}=3 x^{2} y^{2}-3 y^{3}-y, \frac{\partial z}{\partial y}=2 x^{3} y-9 x y^{2}-x \\&\frac{\partial^{2} z}{\partial x^{2}}=6 x y^{2}, \frac{\partial^{2} z}{\partial y \partial x}=6 x^{2} y-9 y^{2}-1 \\& \frac{\partial^{2} z}{\partial x \partial y}=6 x^{2} y-9 y^{2}-1, \frac{\partial^{2} z}{\partial y^{2}}=2 x^{3}-18 x y \\& \frac{\partial^{3} z}{\partial x^{3}}=6 y^{2} \end{align}

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验证函数$z=\ln \sqrt{x^{2}+y^{2}}$，满足方程$\frac{\partial^{2} z}{\partial x^{2}}+\frac{\partial^{2} z}{\partial y^{2}}=0$

因为$z=\ln \sqrt{x^{2}+y^{2}}=\frac{1}{2} \ln \left(x^{2}+y^{2}\right)$，所以

\begin{align} &\frac{\partial z}{\partial x}=\frac{x}{x^{2}+y^{2}}, \quad \frac{\partial z}{\partial y}=\frac{y}{x^{2}+y^{2}} \\ &\frac{\partial^{2} z}{\partial x^{2}}=\frac{\left(x^{2}+y^{2}\right)-x \cdot 2 x}{\left(x^{2}+y^{2}\right)^{2}}=\frac{y^{2}-x^{2}}{\left(x^{2}+y^{2}\right)^{2}} \\& \frac{\partial^{2} z}{\partial y^{2}}=\frac{\left(x^{2}+y^{2}\right)-y \cdot 2 y}{\left(x^{2}+y^{2}\right)^{2}}=\frac{x^{2}-y^{2}}{\left(x^{2}+y^{2}\right)^{2}} \end{align}

因此

\begin{align}\frac{\partial^{2} z}{\partial x^{2}}+\frac{\partial^{2} z}{\partial y^{2}}=\frac{y^{2}-x^{2}}{\left(x^{2}+y^{2}\right)^{2}}+\frac{x^{2}-y^{2}}{\left(x^{2}+y^{2}\right)^{2}}=0\end{align}

全微分

以上关于二元函数全微分的定义及可微分的必要条件和充分条件，可以完全类似地推广到三元和三元以上的多元函数。
习惯上，我们将自变量的增量$\Delta x$与$\Delta y$分别记作$\mathrm{d} x$与$\mathrm{d} y$，并分别称为自变量$x$与$y$的微分。这样, 函数$z=f(x, y)$的全微分就可写为

\begin{align}\mathrm{d} z=\frac{\partial z}{\partial x} \mathrm{~d} x+\frac{\partial z}{\partial y} \mathrm{~d} y\end{align}

通常把二元函数的全微分等于它的两个偏微分之和这件事称为二元函数的微分符合叠加原理。
叠加原理也适用于二元以上的函数.。例如，如果三元函数$u=f(x, y, z)$可微分，那么它的全微分就等于它的三个偏微分之和，即

\begin{align}\mathrm{d} u=\frac{\partial u}{\partial x} \mathrm{~d} x+\frac{\partial u}{\partial y} \mathrm{~d} y+\frac{\partial u}{\partial z} \mathrm{~d} z\end{align}

例题

计算函数$z=x^{2} y+y^{2}$的全微分

因为

\begin{align}\frac{\partial z}{\partial x}=2 x y, \frac{\partial z}{\partial y}=x^{2}+2 y\end{align}

所以

\begin{align}\mathrm{d} z=2 x y \mathrm{~d} x+\left(x^{2}+2 y\right) \mathrm{d} y\end{align}

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计算函数$z=\mathrm{e}^{x y}$在点$(2,1)$处的全微分

因为

\begin{align}\frac{\partial z}{\partial x}=y \mathrm{e}^{x y}, \frac{\partial z}{\partial y}=x \mathrm{e}^{x y} ;\left.\frac{\partial z}{\partial x}\right|_{\substack{x=2 \\ y=1}}=\mathrm{e}^{2},\left.\frac{\partial z}{\partial y}\right|_{\substack{x=2 \\ y=1}}=2 \mathrm{e}^{2}\end{align}

所以

\begin{align}\left.\mathrm{d} z\right|_{\substack{x=2 \\ y=1}}=\mathrm{e}^{2} \mathrm{~d} x+2 \mathrm{e}^{2} \mathrm{~d} y\end{align}

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计算函数$u=x+\sin \frac{y}{2}+\mathrm{e}^{y z}$的全微分

因为

\begin{align}\frac{\partial u}{\partial x}=1, \frac{\partial u}{\partial y}=\frac{1}{2} \cos \frac{y}{2}+z \mathrm{e}^{y z}, \frac{\partial u}{\partial z}=y \mathrm{e}^{y z}\end{align}

所以

\begin{align}\mathrm{d} u=\mathrm{d} x+\left(\frac{1}{2} \cos \frac{y}{2}+z \mathrm{e}^{y z}\right) \mathrm{d} y+y \mathrm{e}^{y z} \mathrm{~d} z\end{align}