0904多元復合函數求導-多元函數微分法及其應用

文章目錄

- 1 復習一元函數復合函數求導
- 2 一元函數與多元函數復合的情形
- 3 多元函數與多元函數復合的情形
- 4 其他情形
- 5 抽象復合函數求導
- 6 全微分不變性
- 結語

1 復習一元函數復合函數求導

$y=f(u),u=\phi(x) \Rightarrow f[\phi(x)]\\ \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=f{'}(u)\cdot \phi{'}(x)=f{'}[\phi(x)]\cdot\phi{'}(x)$

注：

復合關系圖： $\frac{dy}{du}\cdot\frac{du}{dx}=\frac{dy}{dx}\hskip1em x\rightarrow u\rightarrow y$

2 一元函數與多元函數復合的情形

定理1 如果函數 $u=\phi(t)及v=\psi(t)$ 都在點t處可導，函數 $z = f (u, v)$ 在對應點 $(u, v)$ 具有連續偏導數，那么復合函數 $z=f[\phi(t),\psi(t)]在點t$ 可導，且有

$\frac{dz}{dt}=\frac{\partial z}{\partial u}\frac{du}{dt}+\frac{\partial z}{\partial v}\frac{dv}{dt}$

$設t獲得增量\Delta t,此時u=\phi(t),v=\psi(t)獲得增量\Delta u,\Delta v\\ 由此z=f(u,v)相應獲得增量\Delta z\\ 函數z=f(u,v)在點(u,v)具有連續偏導，\\ \Delta z=\frac{\partial z}{\partial u}\Delta u+\frac{\partial v}{\partial v}\Delta v +\epsilon_1\Delta u+\epsilon_2\Delta v\\ 當\Delta u\to 0,\Delta v\to 0時，\epsilon_1\to0,\epsilon_2\to0\\ 上式兩邊各除以\Delta t，得\\ \frac{\Delta z}{\Delta t}=\frac{\partial z}{\partial u}\frac{\Delta u}{\Delta t}+\frac{\partial z}{\partial v}\frac{\Delta v}{\Delta t}+\epsilon_1\frac{\Delta u}{\Delta t}+\epsilon_2\frac{\Delta v}{\Delta t}\\ \because 當\Delta \to0,\Delta u\to0,\Delta v\to0,\frac{\Delta u}{\Delta t}\to\frac{du}{dt},\frac{\Delta v}{\Delta t}=\frac{dv}{dt}\\ \therefore \lim\limits_{\Delta t\to0}{\frac{\Delta z}{\Delta t}}=\frac{\partial z}{\partial u}\frac{du}{dt}+\frac{\partial z}{\partial v}\frac{dv}{dt}$

注：

$\frac{dz}{dt}稱為全導數$
復合關系圖

例1 $z=u^2e^v,u=\sin t,v=e^t，求\frac{dz}{dt}$
$解：\\ \frac{dz}{dt}=\frac{\partial z}{\partial u}\frac{du}{dt}+\frac{\partial z}{\partial v}\frac{dv}{dt}\\ =2\sin t\cdot e^{e^t}\cos t+(\sin t)^2\cdot e^{e^t}\cdot e^t\\ =\sin t e^{e^t}(2\cos t+\sin t e^{t})$

3 多元函數與多元函數復合的情形

定理2 如果函數 $u=\phi(x,y)及v=\psi(x,y)都在點(x,y)$ 具有對x及對y的偏導數，函數 $z = f (u, v) 在對應點 (u, v)$ 具有連續偏導數，那么復合函數 $z=f[\phi(x,y),\psi(x,y)]在點(x,y)$ 的連個偏導數都存在，且有
$\frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial x} \\ \frac{\partial z}{\partial y}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial y} \\$

注：

復合關系圖：

例3 $z=e^u\sin v,u=xy,v=x+y,求\frac{\partial z}{\partial x},\frac{\partial z}{\partial y}$
$解：\\ \frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial x}\\ =e^{xy}\sin(x+y)y+e^{xy}\cos(x+y)(1+y)\\$

4 其他情形

定理3 如果函數 $u=\phi(x,y)在點(x,y)$ 具有對x及對y的偏導數，函數 $v=\psi(y)在點y$ 具有對點y可導，函數 $z = f (u, v) 在對應點 (u, v)$ 具有連續偏導數，那么復合函數 $z=f[\phi(x,y),\psi(y)]在點(x,y)$ 的兩個偏導數存在，且有

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

注：

復合關系圖：
推廣：復合函數的某些中間變量又是復合函數的自變量，比如 $z=f(u,x,y)，u=\phi(x,y)$
$令v=x,w=y，則\\ \frac{\partial v}{\partial x}=1，\frac{\partial v}{\partial y}=0\\ \frac{\partial w}{\partial x}=0,\frac{\partial w}{\partial y}=1 \\ 則 \frac{\partial z}{\partial x}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial x}\\ \frac{\partial z}{\partial y}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial f}{\partial y}$

例5 $u=f(x,y,z)=e^{x^2+y^2+z^2},z=x^2\sin y$
$\frac{\partial u}{\partial x}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial z}\frac{\partial z}{\partial x}\\ =e^{x^2+y^2+(x^2\sin y)^2}2x+e^{x^2+y^2+(x^2\sin y)^2}2x^2\sin y2x\sin y\\ =2xe^{x^2+y^2+x^4\sin^2y}(1+2x^2\sin^2y)\\ \frac{\partial z}{\partial y}=\frac{\partial f}{\partial y}+\frac{\partial f}{\partial z}\frac{\partial z}{\partial y}\\ =2(y+x^4\sin y\cos y)e^{x^2+y^2+x^4\sin^2 y}$

5 抽象復合函數求導

為了寫法和計算的f方便，引入記號
$f(u,v):\\ \begin{cases} f{'}_u\rightarrow f^{'}_1,f^{'}_v\rightarrow f^{'}_2\\ f{'}_u{'}_u\rightarrow f^{'}_1{'}_1,f^{'}_u{'}_v\rightarrow f^{'}_1{'}_2\\ \end{cases}$
例7 $w=f(x+y+z,xyz),f具有二階連續偏導數，求\frac{\partial w}{\partial x},\frac{\partial^2 w}{\partial x\partial z}$
$解：\\ 令u=x+y+z,v=xyz\\ \frac{\partial w}{\partial x}=\frac{\partial w}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial w}{\partial v}\frac{\partial v}{\partial x}\\ =f^{'}_1+yzf{'}_2\\ \frac{\partial^2 w}{\partial x\partial z}=\frac{\partial f^{'}_1}{\partial z}+yf^{'}_2+yz\frac{\partial f^{'}_2}{\partial z}\\ \frac{\partial f_u}{\partial z}=f_{uu}+xyf_{uv}\\ \frac{\partial f_v}{\partial z}=f_{vu}+xyf_{vv}\\ \frac{\partial^2 w}{\partial x\partial z}=f_{uu}+xyf_{uv}+yf_v+yz(f_{vu}+xyf_{vv})\\ =f_{uu}+y(x+z)f_{uv}+xy^2zf_{vv}+yf_v$

6 全微分不變性

設函數 $z = f (u, v) 具有連續偏導數$ ，則有全微分

$dz=\frac{\partial z}{\partial u}du+\frac{\partial z}{\partial v}dv$

如果 $u=\phi(x,y),v=\psi(x,y)$ ,且兩個函數具有連續偏導數，那么復合函數 $z=f[\phi(x,y),\psi(x,y)]$ 的全微分

$dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy$

$dz=(\frac{\partial z}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial x})dx+(\frac{\partial z}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial y})dy\\ =\frac{\partial z}{\partial u}(\frac{\partial u}{\partial x}dx+\frac{\partial u}{\partial y}dy)+\frac{\partial z}{\partial v}(\frac{\partial u}{\partial x}dx+\frac{\partial v}{\partial y}dy) =\frac{\partial z}{\partial u}du+\frac{\partial z}{\partial v}dv$

由此可見無論u和v是自變量還中間變量，函數 $z = f (u, v)$ 的全微分形式都是一樣的。這個性質叫做全微分形式不變性。

例8 $z=f(x^2-y^2,e^{xy}),求dz$
$解：\\ 令u=x^2-y^2,v=e^{xy} dz=df(u,v)=f_udu+f_vdv\\ =f_ud(x^2-y^2)+f_v(e^{xy})\\ =f_u(2xdx-2ydy)+f_v(ye^{xy}dx+xe^{xy}dy)\\ =(2xf_u+ye^{xy}f_v)dx+(xe^{xy}v_v-2yf_u)dy$

結語

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??文檔筆記地址：https://gitee.com/gaogzhen/math

參考：

[1]同濟大學數學系.高等數學第七版下冊[M].北京:高等教育出版社,2014.7.p78-84.

[2]同濟七版《高等數學》全程教學視頻[CP/OL].2020-04-16.p67.

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