高数 07.04 多元复合函数的求导法则

§第七章第四节多元复合函数的求导法则\color{blue}{\S 第七章第四节多元复合函数的求导法则}

一元复合函数y=f(u),u=φ(x)求导法则dydx=dydu⋅dudx微分法则dy=f′(u)du=f′(u)φ′(x)dx一元复合函数 y = f(u), u = \varphi(x) \\ 求导法则\quad \dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx} \\ 微分法则\quad dy = f^{\prime}(u)du = f^{\prime}(u)\varphi^{\prime}(x)dx

熟练掌握多元复合函数求导的链式法则

定理.若函数u=φ(t),v=ψ(t)在点t可导,z=f(u,v)在点(u,v)处偏导连续,则复合函数z=f(φ(t),ψ(t))在点t可导,且有链式法则dzdt=∂z∂u⋅dudt+∂z∂v⋅dvdt定理.若函数u = \varphi(t),v = \psi(t)在点t可导, z=f(u, v)\\ 在点(u, v)处偏导连续,则复合函数z = f(\varphi(t), \psi(t))\\ 在点t可导,且有链式法则\\ \dfrac{dz}{dt} = \dfrac{\partial z}{\partial u} \cdot \dfrac{du}{dt} + \dfrac{\partial z}{\partial v} \cdot \dfrac{dv}{dt}
证:设t取增量Δt,则相应中间变量有增量Δu,Δv,Δz=∂z∂uΔu+∂z∂vΔv+o(ρ)(ρ=(Δu)2+(Δv)2−−−−−−−−−−−−√)两边同时除以ΔtΔzΔt=∂z∂u⋅dudt+∂z∂v⋅dvdt+o(ρ)Δt(ρ=(Δu)2+(Δv)2−−−−−−−−−−−−√)令Δt→0,则有Δu→0,Δu→0ΔuΔt→dudt,ΔvΔt→dvdto(ρ)Δt=o(ρ)ρ(ΔuΔt)2+(ΔvΔt)2−−−−−−−−−−−−−√→0(Δt<0时，根式前加”−“号)(ΔuΔt)2+(ΔvΔt)2−−−−−−−−−−−−−√是常数，o(ρ)ρ→0∴dzdt=∂z∂u⋅dudt+∂z∂v⋅dvdt(全导数公式)证:\\ 设t取增量\Delta{t},则相应中间变量有增量\Delta{u}, \Delta{v}, \\ \Delta{z} = \dfrac{\partial z}{\partial u} \Delta{u} + \dfrac{\partial z}{\partial v} \Delta{v} + o(\rho) \quad (\rho = \sqrt{(\Delta{u})^2 + (\Delta{v})^2}) \\ 两边同时除以\Delta{t}\\ \dfrac{\Delta{z}}{\Delta{t}} = \dfrac{\partial z}{\partial u} \cdot \dfrac{du}{dt} + \dfrac{\partial z}{\partial v} \cdot \dfrac{dv}{dt} + \dfrac{o(\rho)}{\Delta{t}} \quad (\rho = \sqrt{(\Delta{u})^2 + (\Delta{v})^2}) \\ 令\Delta{t} \rightarrow 0, 则有\Delta{u} \rightarrow 0, \Delta{u} \rightarrow 0 \\ \dfrac{\Delta{u}}{\Delta{t}} \rightarrow \dfrac{du}{dt}, \dfrac{\Delta{v}}{\Delta{t}} \rightarrow \dfrac{dv}{dt} \\ \dfrac{o(\rho)}{\Delta{t}} = \dfrac{o(\rho)}{\rho} \sqrt{(\dfrac{\Delta{u}}{\Delta{t}})^2 + (\dfrac{\Delta{v}}{\Delta{t}})^2} \rightarrow 0 (\Delta{t}

推广:中间变量是多元函数的情形.例如,z=f(u,v),u=φ(x,y),v=ψ(x,y)∂z∂x=∂z∂u⋅∂u∂x+∂z∂v⋅∂v∂x=f′1φ′1+f′2ψ′1∂z∂y=∂z∂u⋅∂u∂y+∂z∂v⋅∂v∂y=f′1φ′2+f′2ψ′2推广:\\ 中间变量是多元函数的情形.例如,\\ z = f(u, v), u = \varphi(x, y), v = \psi(x, y)\\ \dfrac{\partial z}{\partial x} = \dfrac{\partial z}{\partial u} \cdot \dfrac{\partial u}{\partial x} + \dfrac{\partial z}{\partial v} \cdot \dfrac{\partial v}{\partial x} = f_1^{\prime}\varphi_1^{\prime} + f_2^{\prime}\psi_1^{\prime} \\ \dfrac{\partial z}{\partial y} = \dfrac{\partial z}{\partial u} \cdot \dfrac{\partial u}{\partial y} + \dfrac{\partial z}{\partial v} \cdot \dfrac{\partial v}{\partial y} = f_1^{\prime}\varphi_2^{\prime} + f_2^{\prime}\psi_2^{\prime}

又如,z=f(x,y),v=ψ(x,y)当他们都具有可微条件时,有∂z∂x=∂f∂x+∂f∂v⋅∂v∂x=f′1+f′2ψ′1∂z∂y=∂f∂v⋅∂v∂y=f′2ψ′2又如,z = f(x, y), v = \psi(x, y) \\ 当他们都具有可微条件时,有\\ \dfrac{\partial z}{\partial x} = \dfrac{\partial f}{\partial x} + \dfrac{\partial f}{\partial v} \cdot \dfrac{\partial v}{\partial x} = f_1^{\prime} + f_2^{\prime}\psi_1^{\prime} \\ \dfrac{\partial z}{\partial y} = \dfrac{\partial f}{\partial v} \cdot \dfrac{\partial v}{\partial y} = f_2^{\prime}\psi_2^{\prime}
注意：这里∂z∂x与∂f∂x不同,∂z∂x表示固定y对x求导,∂f∂x表示固定v对x求导注意：这里\dfrac{\partial z}{\partial x} 与\dfrac{\partial f}{\partial x}不同, \\ \dfrac{\partial z}{\partial x}表示固定y对x求导, \dfrac{\partial f}{\partial x}表示固定v对x求导

口诀：分段用乘，分叉用加，单路全导，叉路偏导口诀：分段用乘，分叉用加，单路全导，叉路偏导

例1.设z=eusinv,u=xy,v=x+y,求∂z∂x,∂z∂y例1.设z = e^u \sin{v}, u = xy, v = x + y,求\dfrac{\partial z}{\partial x}, \dfrac{\partial z}{\partial y}
解:∂z∂x=∂z∂u⋅∂u∂x+∂z∂v⋅∂v∂x=eusinv⋅y+eucosv⋅1=eu(ysinv+cosv)=exy[ysin(x+y)+cos(x+y)]∂z∂y=∂z∂u⋅∂u∂y+∂z∂v⋅∂v∂y=eusinv⋅x+eucosv⋅1=eu(xsinv+cosv)=exy[xsin(x+y)+cos(x+y)]解:\\ \dfrac{\partial z}{\partial x} = \dfrac{\partial z}{\partial u} \cdot \dfrac{\partial u}{\partial x} + \dfrac{\partial z}{\partial v} \cdot \dfrac{\partial v}{\partial x} \\ = e^u\sin{v} \cdot y + e^u \cos{v} \cdot 1 \\ = e^u(y\sin{v} + \cos{v}) \\ = e^{xy}[y\sin{(x + y)} + \cos{(x + y)}] \\ \dfrac{\partial z}{\partial y} = \dfrac{\partial z}{\partial u} \cdot \dfrac{\partial u}{\partial y} + \dfrac{\partial z}{\partial v} \cdot \dfrac{\partial v}{\partial y} \\ = e^u\sin{v} \cdot x + e^u \cos{v} \cdot 1 \\ = e^u(x\sin{v} + \cos{v}) \\ = e^{xy}[x\sin{(x + y)} + \cos{(x + y)}]

例2.设z=uv+sint,u=et,v=cost,求全导数dzdt例2.设z = uv + \sin{t}, u = e^t, v = cos{t},求全导数\dfrac{dz}{dt}
解:dzdt=∂z∂u⋅dudt+∂z∂v⋅dvdt+∂z∂t=v⋅et+u⋅(−sint)+cost=et(cost−sint)+cost解:\\ \dfrac{dz}{dt} = \dfrac{\partial z}{\partial u} \cdot \dfrac{du}{dt} + \dfrac{\partial z}{\partial v} \cdot \dfrac{dv}{dt} + \dfrac{\partial z}{\partial t} \\ = v \cdot e^t + u \cdot (-\sin{t}) + \cos{t}\\ = e^t(\cos{t} - \sin{t}) + \cos{t}
注意：多元抽象复合函数求偏微分方程变形与验证解的问题中经常遇到,下列两个例题有助于掌握这方面问题和常用的导数符号注意：多元抽象复合函数求偏微分方程变形与验证解的问题中经常遇到,下列两个例题有助于掌握这方面问题和常用的导数符号

内容小结
复合函数求导的链式法则
弄清结构，选对公式

练习
1.已知f(x,y)|y=x2=1,f′1(x,y)|y=x2=2x,求f′2(x,y)|y=x2.1.已知f(x, y)| _ {y = x^2} = 1, f_1^{\prime}(x, y) | _ {y = x^2} = 2x,求f_2^{\prime}(x, y) | _ {y = x^2}.
解:由f(x,x2)=1两边对x求导,得f′1(x,x2)+f′2(x,x2)⋅2x=02x[1+f′2(x,x2)]=0f′2(x,x2)=−1即f′2(x,y)|y=x2=−1解:\\ 由f(x, x^2) = 1两边对x求导,得\\ f_1^{\prime}(x, x^2) + f_2^{\prime}(x, x^2)\cdot 2x = 0\\ 2x[1 + f_2^{\prime}(x, x^2)] = 0 \\ f_2^{\prime}(x, x^2) = -1 \\ 即f_2^{\prime}(x, y) | _ {y = x^2} = -1

2.设z=sin(xy2),求∂z∂x,∂z∂y2.设z = \sin(xy^2),求\dfrac{\partial z}{\partial x}, \dfrac{\partial z}{\partial y}
解:令u=xy2,则z=sinu∂z∂x=dzdu⋅∂u∂x=cosu⋅y2=y2cos(xy2)∂z∂y=dzdu⋅∂u∂y=cosu⋅2xy=2xycos(xy2)解:\\ 令u = xy^2, 则z = \sin{u} \\ \dfrac{\partial z}{\partial x} = \dfrac{dz}{du} \cdot \dfrac{\partial u}{\partial x} = \cos{u} \cdot y^2 = y^2 \cos(xy^2) \\ \dfrac{\partial z}{\partial y} = \dfrac{dz}{du} \cdot \dfrac{\partial u}{\partial y} = \cos{u} \cdot 2xy = 2xy \cos{(xy^2)}

3.设z=f(x2y,y2),求∂z∂x,∂z∂y3.设z = f(x^2y, y^2),求\dfrac{\partial z}{\partial x}, \dfrac{\partial z}{\partial y}
解:令u=x2y,v=y2∂z∂x=∂z∂u⋅∂u∂x+∂z∂v⋅∂v∂x=f′1(u,v)⋅2xy+f′2(u,v)⋅0=2xy⋅f′1(x2y,y2)∂z∂y=∂z∂u⋅∂u∂y+∂z∂v⋅∂v∂y=f′2(u,v)⋅x2+f′2(u,v)⋅2y=f′2(x2y,y2)⋅x2+f′2(x2y,y2)⋅2y解:\\ 令u = x^2y, v = y^2 \\ \dfrac{\partial z}{\partial x} = \dfrac{\partial z}{\partial u} \cdot \dfrac{\partial u}{\partial x} + \dfrac{\partial z}{\partial v} \cdot \dfrac{\partial v}{\partial x} \\ = f_1^{\prime}(u, v) \cdot 2xy + f_2^{\prime}(u, v) \cdot 0 \\ = 2xy \cdot f_1^{\prime}(x^2y, y^2) \\ \dfrac{\partial z}{\partial y} = \dfrac{\partial z}{\partial u} \cdot \dfrac{\partial u}{\partial y} + \dfrac{\partial z}{\partial v} \cdot \dfrac{\partial v}{\partial y} \\ = f_2^{\prime}(u, v) \cdot x^2 + f_2^{\prime}(u, v) \cdot 2y \\ = f_2^{\prime}(x^2y, y^2) \cdot x^2+ f_2^{\prime}(x^2y, y^2) \cdot 2y

4.设z=f(yx),f(u)为可微函数，证明:x∂z∂x+y∂z∂y=04.设z = f(\dfrac{y}{x}), f(u)为可微函数，\\ 证明: x\dfrac{\partial z}{\partial x} + y\dfrac{\partial z}{\partial y} = 0
证:u=yxx∂z∂x+y∂z∂y=x∂z∂u⋅∂u∂x+y∂z∂u⋅∂u∂y=x∂z∂u⋅−yx2+y∂z∂u⋅1x=∂z∂u⋅y−yx=0证:\\ u = \dfrac{y}{x} \\ x\dfrac{\partial z}{\partial x} + y\dfrac{\partial z}{\partial y} \\ = x \dfrac{\partial z}{\partial u} \cdot \dfrac{\partial u}{\partial x} + y \dfrac{\partial z}{\partial u} \cdot \dfrac{\partial u}{\partial y} \\ = x \dfrac{\partial z}{\partial u} \cdot \dfrac{-y}{x^2} + y \dfrac{\partial z}{\partial u} \cdot \dfrac{1}{x} \\ = \dfrac{\partial z}{\partial u} \cdot \dfrac{y - y}{x} \\ = 0

5.设z=(x+2y)(x+2y),求∂z∂x,∂z∂y5.设z = (x + 2y)^{(x + 2y)}, 求\dfrac{\partial z}{\partial x}, \dfrac{\partial z}{\partial y}
解:令u=x+2y,v=x+2y,z=uv∂z∂x=∂z∂u⋅∂u∂x+∂z∂v⋅∂v∂x=vuv−1⋅1+uvlnu⋅1=(x+2y)(x+2y)(x+2y−1)+(x+2y)(x+2y)ln(x+2y)=(x+2y)(x+2y)(1+ln(x+2y))∂z∂y=∂z∂u⋅∂u∂y+∂z∂v⋅∂v∂y=vuv−1⋅2+uvlnu⋅2=2(x+2y)(x+2y)(x+2y−1)+2(x+2y)(x+2y)ln(x+2y)=2(x+2y)(x+2y)(1+ln(x+2y))解:令u = x + 2y, v = x + 2y, z = u^v\\ \dfrac{\partial z}{\partial x} = \dfrac{\partial z}{\partial u} \cdot \dfrac{\partial u}{\partial x} + \dfrac{\partial z}{\partial v} \cdot \dfrac{\partial v}{\partial x} \\ = vu^{v - 1} \cdot 1 + u^v \ln{u} \cdot 1 \\ = (x +2y)(x + 2y)^{(x + 2y - 1)} + (x + 2y)^{(x + 2y)} \ln{(x + 2y)} \\ = (x + 2y)^{(x + 2y)} (1 + ln{(x + 2y)}) \\ \dfrac{\partial z}{\partial y} = \dfrac{\partial z}{\partial u} \cdot \dfrac{\partial u}{\partial y} + \dfrac{\partial z}{\partial v} \cdot \dfrac{\partial v}{\partial y} \\ = vu^{v - 1} \cdot 2 + u^v \ln{u} \cdot 2 \\ = 2(x +2y)(x + 2y)^{(x + 2y - 1)} + 2(x + 2y)^{(x + 2y)} \ln{(x + 2y)} \\ = 2(x + 2y)^{(x + 2y)} (1 + ln{(x + 2y)})