文章目录
- 1 复习一元函数复合函数求导
- 2 一元函数与多元函数复合的情形
- 3 多元函数与多元函数复合的情形
- 4 其他情形
- 5 抽象复合函数求导
- 6 全微分不变性
- 结语
1 复习一元函数复合函数求导
y = f ( u ) , u = ϕ ( x ) ⇒ f [ ϕ ( x ) ] d y d x = d y d u ⋅ d u d x = f ′ ( u ) ⋅ ϕ ′ ( x ) = f ′ [ ϕ ( x ) ] ⋅ ϕ ′ ( x ) 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) y=f(u),u=ϕ(x)⇒f[ϕ(x)]dxdy=dudy⋅dxdu=f′(u)⋅ϕ′(x)=f′[ϕ(x)]⋅ϕ′(x)
注：
- 复合关系图： d y d u ⋅ d u d x = d y d x x → u → y \frac{dy}{du}\cdot\frac{du}{dx}=\frac{dy}{dx}\hskip1em x\rightarrow u\rightarrow y dudy⋅dxdu=dxdyx→u→y
2 一元函数与多元函数复合的情形
定理1 如果函数 u = ϕ ( t ) 及 v = ψ ( t ) u=\phi(t)及v=\psi(t) u=ϕ(t)及v=ψ(t)都在点t处可导，函数 z = f ( u , v ) z=f(u,v) z=f(u,v)在对应点 ( u , v ) (u,v) (u,v)具有连续偏导数，那么复合函数 z = f [ ϕ ( t ) , ψ ( t ) ] 在点 t z=f[\phi(t),\psi(t)]在点t z=f[ϕ(t),ψ(t)]在点t可导，且有
d z d t = ∂ z ∂ u d u d t + ∂ z ∂ v d v d t \frac{dz}{dt}=\frac{\partial z}{\partial u}\frac{du}{dt}+\frac{\partial z}{\partial v}\frac{dv}{dt} dtdz=∂u∂zdtdu+∂v∂zdtdv
设 t 获得增量 Δ t , 此时 u = ϕ ( t ) , v = ψ ( t ) 获得增量 Δ u , Δ v 由此 z = f ( u , v ) 相应获得增量 Δ z 函数 z = f ( u , v ) 在点 ( u , v ) 具有连续偏导， Δ z = ∂ z ∂ u Δ u + ∂ v ∂ v Δ v + ϵ 1 Δ u + ϵ 2 Δ v 当 Δ u → 0 , Δ v → 0 时， ϵ 1 → 0 , ϵ 2 → 0 上式两边各除以 Δ t ，得 Δ z Δ t = ∂ z ∂ u Δ u Δ t + ∂ z ∂ v Δ v Δ t + ϵ 1 Δ u Δ t + ϵ 2 Δ v Δ t ∵ 当 Δ → 0 , Δ u → 0 , Δ v → 0 , Δ u Δ t → d u d t , Δ v Δ t = d v d t ∴ lim Δ t → 0 Δ z Δ t = ∂ z ∂ u d u d t + ∂ z ∂ v d v d t 设t获得增量\Delta t,此时u=\phi(t),v=\psi(t)获得增量\Delta u,\Delta v\\ 由此z=f(u,v)相应获得增量\Delta z\\ 函数z=f(u,v)在点(u,v)具有连续偏导，\\ \Delta z=\frac{\partial z}{\partial u}\Delta u+\frac{\partial v}{\partial v}\Delta v +\epsilon_1\Delta u+\epsilon_2\Delta v\\ 当\Delta u\to 0,\Delta v\to 0时，\epsilon_1\to0,\epsilon_2\to0\\ 上式两边各除以\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 当\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} 设t获得增量Δt,此时u=ϕ(t),v=ψ(t)获得增量Δu,Δv由此z=f(u,v)相应获得增量Δz函数z=f(u,v)在点(u,v)具有连续偏导，Δz=∂u∂zΔu+∂v∂vΔv+ϵ1Δu+ϵ2Δv当Δu→0,Δv→0时，ϵ1→0,ϵ2→0上式两边各除以Δt，得ΔtΔz=∂u∂zΔtΔu+∂v∂zΔtΔv+ϵ1ΔtΔu+ϵ2ΔtΔv∵当Δ→0,Δu→0,Δv→0,ΔtΔu→dtdu,ΔtΔv=dtdv∴Δt→0limΔtΔz=∂u∂zdtdu+∂v∂zdtdv
注：
- d z d t 称为全导数 \frac{dz}{dt}称为全导数 dtdz称为全导数
- 复合关系图
例1 z = u 2 e v , u = sin t , v = e t ，求 d z d t z=u^2e^v,u=\sin t,v=e^t，求\frac{dz}{dt} z=u2ev,u=sint,v=et，求dtdz
解： d z d t = ∂ z ∂ u d u d t + ∂ z ∂ v d v d t = 2 sin t ⋅ e e t cos t + ( sin t ) 2 ⋅ e e t ⋅ e t = sin t e e t ( 2 cos t + sin t e t ) 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}) 解：dtdz=∂u∂zdtdu+∂v∂zdtdv=2sint⋅eetcost+(sint)2⋅eet⋅et=sinteet(2cost+sintet)
3 多元函数与多元函数复合的情形
定理2 如果函数 u = ϕ ( x , y ) 及 v = ψ ( x , y ) 都在点 ( x , y ) u=\phi(x,y)及v=\psi(x,y)都在点(x,y) u=ϕ(x,y)及v=ψ(x,y)都在点(x,y)具有对x及对y的偏导数，函数 z = f ( u , v ) 在对应点 ( u , v ) z=f(u,v)在对应点(u,v) z=f(u,v)在对应点(u,v)具有连续偏导数，那么复合函数 z = f [ ϕ ( x , y ) , ψ ( x , y ) ] 在点 ( x , y ) z=f[\phi(x,y),\psi(x,y)]在点(x,y) z=f[ϕ(x,y),ψ(x,y)]在点(x,y)的连个偏导数都存在，且有
∂ z ∂ x = ∂ z ∂ u ∂ u ∂ x + ∂ z ∂ v ∂ v ∂ x ∂ z ∂ y = ∂ z ∂ u ∂ u ∂ y + ∂ z ∂ v ∂ v ∂ y \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} \\ ∂x∂z=∂u∂z∂x∂u+∂v∂z∂x∂v∂y∂z=∂u∂z∂y∂u+∂v∂z∂y∂v
注：
- 复合关系图：
例3 z = e u sin v , u = x y , v = x + y , 求 ∂ z ∂ x , ∂ z ∂ y z=e^u\sin v,u=xy,v=x+y,求\frac{\partial z}{\partial x},\frac{\partial z}{\partial y} z=eusinv,u=xy,v=x+y,求∂x∂z,∂y∂z
解： ∂ z ∂ x = ∂ z ∂ u ∂ u ∂ x + ∂ z ∂ v ∂ v ∂ x = e x y sin ( x + y ) y + e x y cos ( x + y ) ( 1 + 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)\\ 解：∂x∂z=∂u∂z∂x∂u+∂v∂z∂x∂v=exysin(x+y)y+exycos(x+y)(1+y)
4 其他情形
定理3 如果函数 u = ϕ ( x , y ) 在点 ( x , y ) u=\phi(x,y)在点(x,y) u=ϕ(x,y)在点(x,y)具有对x及对y的偏导数，函数 v = ψ ( y ) 在点 y v=\psi(y)在点y v=ψ(y)在点y具有对点y可导，函数 z = f ( u , v ) 在对应点 ( u , v ) z=f(u,v)在对应点(u,v) z=f(u,v)在对应点(u,v)具有连续偏导数，那么复合函数 z = f [ ϕ ( x , y ) , ψ ( y ) ] 在点 ( x , y ) z=f[\phi(x,y),\psi(y)]在点(x,y) z=f[ϕ(x,y),ψ(y)]在点(x,y)的两个偏导数存在，且有
∂ z ∂ x = ∂ z ∂ u ∂ u ∂ x ∂ z ∂ y = ∂ z ∂ u ∂ u ∂ y + ∂ z ∂ v d v d y \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} \\ ∂x∂z=∂u∂z∂x∂u∂y∂z=∂u∂z∂y∂u+∂v∂zdydv
注：
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复合关系图：
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推广：复合函数的某些中间变量又是复合函数的自变量，比如 z = f ( u , x , y ) ， u = ϕ ( x , y ) z=f(u,x,y)，u=\phi(x,y) z=f(u,x,y)，u=ϕ(x,y)
令 v = x , w = y ，则 ∂ v ∂ x = 1 ， ∂ v ∂ y = 0 ∂ w ∂ x = 0 , ∂ w ∂ y = 1 则 ∂ z ∂ x = ∂ f ∂ u ∂ u ∂ x + ∂ f ∂ x ∂ z ∂ y = ∂ f ∂ u ∂ u ∂ y + ∂ f ∂ y 令v=x,w=y，则\\ \frac{\partial v}{\partial x}=1，\frac{\partial v}{\partial y}=0\\ \frac{\partial w}{\partial x}=0,\frac{\partial w}{\partial y}=1 \\ 则 \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} 令v=x,w=y，则∂x∂v=1，∂y∂v=0∂x∂w=0,∂y∂w=1则∂x∂z=∂u∂f∂x∂u+∂x∂f∂y∂z=∂u∂f∂y∂u+∂y∂f
例5 u = f ( x , y , z ) = e x 2 + y 2 + z 2 , z = x 2 sin y u=f(x,y,z)=e^{x^2+y^2+z^2},z=x^2\sin y u=f(x,y,z)=ex2+y2+z2,z=x2siny
∂ u ∂ x = ∂ f ∂ x + ∂ f ∂ z ∂ z ∂ x = e x 2 + y 2 + ( x 2 sin y ) 2 2 x + e x 2 + y 2 + ( x 2 sin y ) 2 2 x 2 sin y 2 x sin y = 2 x e x 2 + y 2 + x 4 sin 2 y ( 1 + 2 x 2 sin 2 y ) ∂ z ∂ y = ∂ f ∂ y + ∂ f ∂ z ∂ z ∂ y = 2 ( y + x 4 sin y cos y ) e x 2 + y 2 + x 4 sin 2 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} ∂x∂u=∂x∂f+∂z∂f∂x∂z=ex2+y2+(x2siny)22x+ex2+y2+(x2siny)22x2siny2xsiny=2xex2+y2+x4sin2y(1+2x2sin2y)∂y∂z=∂y∂f+∂z∂f∂y∂z=2(y+x4sinycosy)ex2+y2+x4sin2y
5 抽象复合函数求导
为了写法和计算的f方便，引入记号
f ( u , v ) : { f ′ u → f 1 ′ , f v ′ → f 2 ′ f ′ u ′ u → f 1 ′ ′ 1 , f u ′ ′ v → f 1 ′ ′ 2 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} f(u,v):{f′u→f1′,fv′→f2′f′u′u→f1′′1,fu′′v→f1′′2
例7 w = f ( x + y + z , x y z ) , f 具有二阶连续偏导数，求 ∂ w ∂ x , ∂ 2 w ∂ x ∂ z w=f(x+y+z,xyz),f具有二阶连续偏导数，求\frac{\partial w}{\partial x},\frac{\partial^2 w}{\partial x\partial z} w=f(x+y+z,xyz),f具有二阶连续偏导数，求∂x∂w,∂x∂z∂2w
解： 令 u = x + y + z , v = x y z ∂ w ∂ x = ∂ w ∂ u ∂ u ∂ x + ∂ w ∂ v ∂ v ∂ x = f 1 ′ + y z f ′ 2 ∂ 2 w ∂ x ∂ z = ∂ f 1 ′ ∂ z + y f 2 ′ + y z ∂ f 2 ′ ∂ z ∂ f u ∂ z = f u u + x y f u v ∂ f v ∂ z = f v u + x y f v v ∂ 2 w ∂ x ∂ z = f u u + x y f u v + y f v + y z ( f v u + x y f v v ) = f u u + y ( x + z ) f u v + x y 2 z f v v + y f v 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 解：令u=x+y+z,v=xyz∂x∂w=∂u∂w∂x∂u+∂v∂w∂x∂v=f1′+yzf′2∂x∂z∂2w=∂z∂f1′+yf2′+yz∂z∂f2′∂z∂fu=fuu+xyfuv∂z∂fv=fvu+xyfvv∂x∂z∂2w=fuu+xyfuv+yfv+yz(fvu+xyfvv)=fuu+y(x+z)fuv+xy2zfvv+yfv
6 全微分不变性
设函数 z = f ( u , v ) 具有连续偏导数 z=f(u,v)具有连续偏导数 z=f(u,v)具有连续偏导数，则有全微分
d z = ∂ z ∂ u d u + ∂ z ∂ v d v dz=\frac{\partial z}{\partial u}du+\frac{\partial z}{\partial v}dv dz=∂u∂zdu+∂v∂zdv
如果 u = ϕ ( x , y ) , v = ψ ( x , y ) u=\phi(x,y),v=\psi(x,y) u=ϕ(x,y),v=ψ(x,y),且两个函数具有连续偏导数 ，那么复合函数 z = f [ ϕ ( x , y ) , ψ ( x , y ) ] z=f[\phi(x,y),\psi(x,y)] z=f[ϕ(x,y),ψ(x,y)]的全微分
d z = ∂ z ∂ x d x + ∂ z ∂ y d y dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy dz=∂x∂zdx+∂y∂zdy
d z = ( ∂ z ∂ u ∂ u ∂ x + ∂ z ∂ v ∂ v ∂ x ) d x + ( ∂ z ∂ u ∂ u ∂ y + ∂ z ∂ v ∂ v ∂ y ) d y = ∂ z ∂ u ( ∂ u ∂ x d x + ∂ u ∂ y d y ) + ∂ z ∂ v ( ∂ u ∂ x d x + ∂ v ∂ y d y ) = ∂ z ∂ u d u + ∂ z ∂ v d v 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 dz=(∂u∂z∂x∂u+∂v∂z∂x∂v)dx+(∂u∂z∂y∂u+∂v∂z∂y∂v)dy=∂u∂z(∂x∂udx+∂y∂udy)+∂v∂z(∂x∂udx+∂y∂vdy)=∂u∂zdu+∂v∂zdv
由此可见无论u和v是自变量还中间变量，函数 z = f ( u , v ) z=f(u,v) z=f(u,v)的全微分形式都是一样的。这个性质叫做全微分形式不变性。
例8 z = f ( x 2 − y 2 , e x y ) , 求 d z z=f(x^2-y^2,e^{xy}),求dz z=f(x2−y2,exy),求dz
解： 令 u = x 2 − y 2 , v = e x y d z = d f ( u , v ) = f u d u + f v d v = f u d ( x 2 − y 2 ) + f v ( e x y ) = f u ( 2 x d x − 2 y d y ) + f v ( y e x y d x + x e x y d y ) = ( 2 x f u + y e x y f v ) d x + ( x e x y v v − 2 y f u ) d y 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 解：令u=x2−y2,v=exydz=df(u,v)=fudu+fvdv=fud(x2−y2)+fv(exy)=fu(2xdx−2ydy)+fv(yexydx+xexydy)=(2xfu+yexyfv)dx+(xexyvv−2yfu)dy
结语
❓QQ:806797785
⭐️文档笔记地址：https://gitee.com/gaogzhen/math
参考：
[1]同济大学数学系.高等数学 第七版 下册[M].北京:高等教育出版社,2014.7.p78-84.
[2]同济七版《高等数学》全程教学视频[CP/OL].2020-04-16.p67.
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