3-19 微积分笔记

复合函数求导法则

【定理1】设$z=f(u,v)$在$(u,v)$处可微，在$(x,y)$处有偏导数，则复合函数$z=f(u(x,y),v(x,y))$在$(u,v)$对应的$(x,y)$处偏导存在，且

$$\frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial x},\qquad \frac{\partial z}{\partial y}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial y}$$

证：固定$y$（即$\Delta y=0$），考虑

$$\Delta_x z =\frac{\partial z}{\partial u}\Delta _x u + \frac{\partial z}{\partial v}\Delta _xv + o(\rho) \quad where~\rho = \sqrt{(\Delta _xu)^2+(\Delta _xu)^2}$$

故有

$$\lim_{\Delta x\to 0}\frac{\Delta _xz}{\Delta x}=\lim (\frac{\partial z}{\partial u}\frac{\Delta _x u}{\Delta x} + \frac{\partial z}{\partial v}\frac{\Delta _xv}{\Delta x} + \frac{o(\rho)}{\Delta x})$$

来看最后那项。$\rho=0$时，明显$\lim \frac{o(\rho)}{\Delta x}=0$；
若$\rho \neq 0$，则

$$\lim_{\Delta x \to 0}\frac{o(\rho)}{\Delta x}=\lim_{\Delta x \to 0}\frac{o(\rho)}{\rho}\frac{\sqrt{(\Delta _xu)^2+(\Delta _xv)^2}}{\Delta x} = 0$$
故原命题成立。满足链式法则。

$z=f(u(x),v(x))$

$$\frac{dz}{dx}=f'_u\frac{du}{dx}+f_v'\frac{dv}{dx}$$

例1、设$z=e^u\cos v$，$u=x+y,v=xy$，求$z'_x,z'_y$.
解：$z_x'=z'_uu'_x+z'_vv'_x=e^{u}\cos v - y\cdot e^u\sin v$
$z_y'=z'_uu'_y + z'_vv'_y = e^u\cos v -xe^u\sin v$

例2、设$u=f(x,y,z)$可微，$y=y(x),z=z(x)$可导，求$\frac{du}{dx}$
解：$u=f(x,y(x),z(x))$

$$\frac{du}{dx}=\frac{\partial f}{\partial x} \cdot 1 + \frac{\partial f}{\partial y}\cdot y' +\frac{\partial f}{\partial z}\cdot z'$$

例3、设$z=f(x,y,u)$可微，$u=u(x,y)$存在偏导，求$\frac{\partial z}{\partial x},\frac{\partial z}{\partial y}$
解：$u=f(x,y,u(x,y))$

$$z'_x=\frac{\partial f}{\partial x}\cdot 1+\frac{\partial y}{\partial x}\cdot 0 +\frac{\partial f}{\partial u}\cdot \frac{\partial u}{\partial x}=f'_x+f'_u\cdot u'_x$$
$$z'_y = \frac{\partial f}{\partial y}+\frac{\partial f}{\partial u}\cdot \frac{\partial u}{\partial y}$$

例4、设$z=f(e^x\sin y,x^2+y^2,xy)$，其中$f$可微，求$z'_x,z'_y$.

解：

$$\frac{\partial z}{\partial x}=f'_1\cdot e^x\sin y+f_2' \cdot 2x +f_3'\cdot y$$
其中$f_1$是指$f$对第一个变量$e^x\sin y$求偏导的结果。

$$\frac{\partial z}{\partial y}=f_1'e^x\cos y + 2y f'_2+xf_3'$$

例5、设$u=f(x,y,z),y=g(x,t),t=h(x,z)$，其中$f,g,h$均可微，求$u'_x$

$$\frac{\partial u}{\partial x}=f'_x\cdot 1 +f'_y\frac{\partial y}{\partial x}+f'_z\cdot 0\\=f'_x +f'_y(\frac{\partial g}{\partial x}\cdot 1 + \frac{\partial g}{\partial t}\frac{\partial h}{\partial x}) \\ =f'_x+f'_yg'_x + f'_yg'_t(\frac{\partial h}{\partial x}\cdot 1 + \frac{\partial h}{\partial z}\cdot 0)\\ = f'_x + f'_yg'_x + f'_yg'_th'_x$$

一阶微分的形式不变性

若$z=f(u,v)$可微，$u=u(x,y),v=v(x,y)$在$(x,y)$处可微，那么

$$dz = z'_x dx + z'_y dy$$
而我们考虑
$$dz = (z'_uu'_x + z'_vv'_x)dx+(z'_uu'_y+z'_vv'_y)dy\\= z'_u(u'_xdx+u'_ydy)+z'_v(v'_xdx+v'_ydy) \\=z_u'du + z'_vdv$$

这就是一阶微分的形式不变性。

另一个例子：$z=f(u,v,w),u=u(x,y),v=v(x,y),w=w(x,y),x=x(s,t),y=y(s,t)$

$$dz=f'_udu+f'_vdv+f'_wdw$$

例7、设$z=\frac1xf(xy)+yg(x+y)$，其中$f,g$具有二阶连续导数，求$\frac{\partial ^2 z}{\partial x\partial y}$

解：$\frac{\partial z}{\partial x}=f'(xy)\frac{y}{x}-\frac{1}{x^2}f(xy)+yg'(x+y)$

$$\frac{\partial }{\partial y}z'_x = \frac{1}{x}f'(xy)+yf''(xy)-\frac1x f'(xy)+g'(x+y)+y\cdot g''(x+y)$$

例8、设$z=f(x+y,z-y,xy)$，其中$f$具有二阶连续偏导数。求$dz$和$z''_{xy}$.
解：

$$dz=z'_xdx+z'_ydy=(f_1'+f_2'+yf_3')dx+(f_1'-f_2'+xf_3')dy$$
$$\frac{\partial z}{\partial x}=f'_1+f'_2+yf'_3 \\ \frac{\partial ^2z}{\partial x\partial y} = (f_{11}''-f''_{12}+xf_{13}'')+(f''_{21}-f_{22}''+xf_{23}'') + f'_3 +y(f''_{31}-f_{32}''+xf_{33}'')$$

例9、设$z=f(x,u,v),u=g(x,y),v=h(x,y,u)$，其中$f,g,h$均可微。求$z'_x,z'_y$
解：$z=f(x,g(x,y),h(x,y,g(x,y)))$

$$\frac{\partial z}{\partial x}=f'_x + f_u'u'_x +f'_vv'_x \\ =f'_x +f'_ug'_x+f'_v(h'_x+h'_ug'_x)$$

$$\frac{\partial z}{\partial y}=f'_ug'_y+f'_vv'_y=f'_ug'_y+f'_v(h'_y+h'_ug'_y)$$

例10、证明：若$u=u(x,y)$具有二阶连续导数，$x=x(r,\theta)=r\cos \theta,y=y(r,\theta)=r\sin \theta$，则

$$\frac{\partial ^2 u}{\partial x^2}+\frac{\partial ^2u}{\partial y^2}=\frac{\partial ^2u}{\partial r^2}+\frac{1}{r^2}\frac{\partial ^2u}{\partial \theta^2}+\frac1r \frac{\partial u}{\partial r}$$

分析：平面直角坐标系和极坐标系上的点明显是一一对应的，也就是说存在反函数，根据$x,y$得到$r,\theta$：$r=r(x,y),\theta =\theta(x,y)$

证：

$$\frac{\partial u}{\partial r}=\frac{\partial u}{\partial x}\cos \theta+\frac{\partial u}{\partial y}\sin \theta \\ \frac{\partial ^2u}{\partial r^2}=(\frac{\partial ^2 u}{\partial x^2}\cos \theta + \frac{\partial ^2u}{\partial x\partial y}\sin \theta)\cos \theta + (\frac{\partial ^2 u}{\partial y\partial x}\cos \theta + \frac{\partial ^2u}{\partial y^2}\sin \theta)\sin\theta \\ \frac{\partial u}{\partial \theta}=\frac{\partial u}{\partial x}(-r\sin \theta)+\frac{\partial u}{\partial y}(r\cos \theta)$$
代入之后会发现和左边相等。