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8.5 隐函数的求导公式
2025-12-03
高等数学
数学

§\S8.5 隐函数的求导公式#

一、一个方程的情形#

  1. [定理1]设 F(x,y)=0F(x,y)=0，若在 P0(x0,y0)P_0(x_0,y_0) 某邻域内偏导连续，且 F(x0,y0)=0F(x_0,y_0)=0，Fy(x0,y0)≠0F_y(x_0,y_0)\ne 0，则 F(x,y)=0F(x,y)=0 在 P0P_0 某邻域内可唯一确定 y=f(x)y=f(x)、y0=f(x0)y_0=f(x_0) 连续，导数连续。 dydx=−F1′F2′=−FxFy\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{F_1'}{F_2'}=-\dfrac{F_x}{F_y}
  2. [定理2]设 F(x,y,z)=0F(x,y,z)=0 在 P(x0,y0,z0)P(x_0,y_0,z_0) 某一邻域内偏导连续，且 F(x0,y0,z0)=0F(x_0,y_0,z_0)=0，Fz(x0,y0,z0)≠0F_z(x_0,y_0,z_0)\ne 0，则 F(x,y,z)=0F(x,y,z)=0 可确定唯一 z=f(x,y)z=f(x,y)（其连续，一阶偏导连续）且 ∂z∂x=−FxFz∂z∂y=−FyFz\dfrac{\partial z}{\partial x}=-\dfrac{F_x}{F_z}\\ \dfrac{\partial z}{\partial y}=-\dfrac{F_y}{F_z}
  3. [2的扩展]F(x,y,z,t)=0F(x,y,z,t)=0，t=f(x,y,z)t=f(x,y,z)，偏导连续且 Ft≠0F_t\ne 0，有 ∂t∂x=−FxFt∂t∂y=−FyFt∂t∂z=−FzFt\dfrac{\partial t}{\partial x}=-\dfrac{F_x}{F_t}\\ \dfrac{\partial t}{\partial y}=-\dfrac{F_y}{F_t}\\ \dfrac{\partial t}{\partial z}=-\dfrac{F_z}{F_t}

例题#

  1. y=x2−12sin⁡yy=x^2-\dfrac{1}{2}\sin y，求 dydx\dfrac{\mathrm dy}{\mathrm dx} 和 d2ydx2\dfrac{\mathrm d^2y}{\mathrm dx^2}

    解：

    • 直接法

      dydx=2x−12cos⁡ydydxdydx=4x2+cos⁡y\begin{align*} \dfrac{\mathrm dy}{\mathrm dx}&=2x-\dfrac{1}{2}\cos y\dfrac{\mathrm dy}{\mathrm dx}\\ \dfrac{\mathrm dy}{\mathrm dx}&=\dfrac{4x}{2+\cos y} \end{align*} d2ydx2=ddx(4x2+cos⁡y)=4(2+cos⁡y)−4x(−sin⁡y)dydx(2+cos⁡y)2=4(2+cos⁡y)2+16x2sin⁡y(2+cos⁡y)3\begin{align*} \dfrac{\mathrm d^2y}{\mathrm dx^2}&=\dfrac{\mathrm d}{\mathrm dx}(\dfrac{4x}{2+\cos y})\\ &=\dfrac{4(2+\cos y)-4x(-\sin y)\dfrac{\mathrm dy}{\mathrm dx}}{(2+\cos y)^2}\\ &=\dfrac{4(2+\cos y)^2+16x^2\sin y}{(2+\cos y)^3} \end{align*}

    • 公式法 F(x,y)=y−x2+12sin⁡y=0F(x,y)=y-x^2+\dfrac{1}{2}\sin y=0

      dydx=−FxFy=−−2x1+12cos⁡y=4x2+cos⁡y\begin{align*} \dfrac{\mathrm dy}{\mathrm dx}&=-\dfrac{F_x}{F_y}\\ &=-\dfrac{-2x}{1+\frac{1}{2}\cos y}\\ &=\dfrac{4x}{2+\cos y} \end{align*} d2ydx2=ddx(4x2+cos⁡y)=4(2+cos⁡y)2+16x2sin⁡y(2+cos⁡y)3\begin{align*} \dfrac{\mathrm d^2y}{\mathrm dx^2}&=\dfrac{\mathrm d}{\mathrm dx}(\dfrac{4x}{2+\cos y})\\ &=\dfrac{4(2+\cos y)^2+16x^2\sin y}{(2+\cos y)^3} \end{align*}

  2. z=x3y−ezz=x^3y-e^z，求 ∂2z∂x∂y\dfrac{\partial ^2z}{\partial x \partial y}

    解：F(x,y,z)=x3y−ez−zF(x,y,z)=x^3y-e^z-z

    ∂z∂x=−FxFz=−3x2y−ez−1=3x2yez+1\dfrac{\partial z}{\partial x}=-\dfrac{F_x}{F_z}=-\dfrac{3x^2y}{-e^z-1}=\dfrac{3x^2y}{e^z+1}

    ∂∂y(∂z∂x)=∂∂y(3x2yez+1)=3x2(ez+1)−3x2y(ez+1)⋅∂z∂y(ez+1)2=3x2(1+ez−x3y)(ez+1)2\begin{align*} \dfrac{\partial }{\partial y}(\dfrac{\partial z}{\partial x})&=\dfrac{\partial }{\partial y}(\dfrac{3x^2y}{e^z+1})\\ &=\dfrac{3x^2(e^z+1)-3x^2y(e^z+1)\cdot \frac{\partial z}{\partial y}}{(e^z+1)^2}\\ &=\dfrac{3x^2(1+e^z-x^3y)}{(e^z+1)^2} \end{align*}

  3. ϕ(x−az,y−bz)=0=F(x,y,z)\phi(x-az,y-bz)=0=F(x,y,z)，z=f(x,y)z=f(x,y)，求证 a∂z∂x+b∂z∂y=1a\dfrac{\partial z}{\partial x}+b\dfrac{\partial z}{\partial y}=1

    证明：

    • 公式法 ∂z∂x=−FxFz=−ϕ1′⋅1+ϕ2′⋅0ϕ1′(−a)+ϕ2′(−b)=ϕ1′aϕ1′+bϕ2′\dfrac{\partial z}{\partial x}=-\dfrac{F_x}{F_z}=-\dfrac{\phi_1'\cdot 1+\phi_2'\cdot 0}{\phi_1'(-a)+\phi_2'(-b)}=\dfrac{\phi_1'}{a\phi_1'+b\phi_2'}

      ∂z∂y=−FyFz=−ϕ1′⋅0+ϕ2′⋅1ϕ1′(−a)+ϕ2′(−b)=ϕ2′aϕ1′+bϕ2′\dfrac{\partial z}{\partial y}=-\dfrac{F_y}{F_z}=-\dfrac{\phi_1'\cdot 0+\phi_2'\cdot 1}{\phi_1'(-a)+\phi_2'(-b)}=\dfrac{\phi_2'}{a\phi_1'+b\phi_2'}

      a⋅ϕ1′aϕ1′+bϕ2′+b⋅ϕ2′aϕ1′+bϕ2′=1a\cdot \dfrac{\phi_1'}{a\phi_1'+b\phi_2'}+b\cdot \dfrac{\phi_2'}{a\phi_1'+b\phi_2'}=1，得证

    • 微分法 F(x,y,z)=ϕ(x−az,y−bz)F(x,y,z)=\phi(x-az,y-bz)，dϕ(x−az,y−bz=0)\mathrm d \phi(x-az,y-bz=0)

    令 x−az=ux-az=u，y−bz=vy-bz=v

    ∂ϕ∂udu+∂ϕ∂vdv=0ϕ1′d(x−az)+ϕ2′d(y−bz)=0ϕ1′(dx−adz)+ϕ2′(dy−bdz)=0ϕ1′dx+ϕ2′dy−(aϕ1′+bϕ2′)dz=0dz=ϕ1′aϕ1′+bϕ2′dx+ϕ2′1aϕ1′+bϕ2′dy\begin{align*} \dfrac{\partial \phi}{\partial u}\mathrm du+\dfrac{\partial \phi}{\partial v}\mathrm dv&=0\\ \phi_1'\mathrm d(x-az)+\phi_2'\mathrm d(y-bz)&=0\\ \phi_1'(\mathrm dx-a \mathrm dz)+\phi_2'(\mathrm dy-b \mathrm dz)&=0\\ \phi_1'\mathrm dx+\phi_2'\mathrm dy-(a\phi_1'+b\phi_2')\mathrm dz&=0\\ \mathrm dz&=\dfrac{\phi_1'}{a\phi_1'+b\phi_2'}\mathrm dx+\phi_2'\dfrac{1}{a\phi_1'+b\phi_2'}\mathrm dy \end{align*}

    同理可证

二、方程组的情形#

[定理3] {F(x,y,u,v)=0G(x,y,u,v)=0\begin{cases}F(x,y,u,v)=0\\G(x,y,u,v)=0\end{cases}，若在 P0P_0 某邻域内偏导连续，F(x0,y0,u0,v0)=0F(x_0,y_0,u_0,v_0)=0 且 G(x0,y0,u0,v0)=0G(x_0,y_0,u_0,v_0)=0，J=∂(F,G)∂(u,v)=∣FuFvGuGv∣J=\dfrac{\partial (F,G)}{\partial (u,v)}=\begin{vmatrix}F_u & F_v\\ G_u&G_v\end{vmatrix} 在 P0P_0 处不等于0，则可确定一个隐函数组 {u=u(x,y)v=v(x,y)\begin{cases}u=u(x,y)\\v=v(x,y)\end{cases}，其唯一、连续、偏导也连续，且

  • ∂u∂x=−1J∣FxFvGxGv∣=−1J∂(F,G)∂(x,v)\dfrac{\partial u}{\partial x}=-\dfrac{1}{J}\begin{vmatrix}F_x&F_v\\G_x&G_v\end{vmatrix}=-\dfrac{1}{J}\dfrac{\partial (F,G)}{\partial (x,v)}
  • ∂u∂y=−1J∣FyFvGyGv∣=−1J∂(F,G)∂(y,v)\dfrac{\partial u}{\partial y}=-\dfrac{1}{J}\begin{vmatrix}F_y&F_v\\G_y&G_v\end{vmatrix}=-\dfrac{1}{J}\dfrac{\partial (F,G)}{\partial (y,v)}
  • ∂v∂x=−1J∣FuFxGuGx∣=−1J∂(F,G)∂(u,x)\dfrac{\partial v}{\partial x}=-\dfrac{1}{J}\begin{vmatrix}F_u&F_x\\G_u&G_x\end{vmatrix}=-\dfrac{1}{J}\dfrac{\partial (F,G)}{\partial (u,x)}
  • ∂v∂y=−1J∣FuFyGuGy∣=−1J∂(F,G)∂(u,y)\dfrac{\partial v}{\partial y}=-\dfrac{1}{J}\begin{vmatrix}F_u&F_y\\G_u&G_y\end{vmatrix}=-\dfrac{1}{J}\dfrac{\partial (F,G)}{\partial (u,y)}
证明：

  1. [直接法]

    {F(x,y,u,v)=0G(x,y,u,v)=0\begin{align*} \begin{cases} F(x,y,u,v)=0\\ G(x,y,u,v)=0 \end{cases} \end{align*}

    把 u,vu,v 看作 x,yx,y 的函数，两边关于 x,yx,y 求偏导得

    {Fx⋅1+Fy⋅0+Fu⋅∂u∂x+Fv⋅∂v∂x=0Gx⋅1+Gy⋅0+Gu⋅∂u∂x+Gv⋅∂v∂x=0\begin{align*} \begin{cases} F_x\cdot 1+F_y\cdot 0+F_u\cdot \dfrac{\partial u}{\partial x}+F_v\cdot \dfrac{\partial v}{\partial x}=0\\ G_x\cdot 1+G_y\cdot 0+G_u\cdot \dfrac{\partial u}{\partial x}+G_v\cdot \dfrac{\partial v}{\partial x}=0 \end{cases} \end{align*} {Fu⋅∂u∂x+Fv⋅∂u∂x=−FxGu⋅∂u∂x+Gv⋅∂u∂x=−Gx\begin{align*} \begin{cases} F_u\cdot \dfrac{\partial u}{\partial x}+F_v\cdot \dfrac{\partial u}{\partial x}=-F_x\\ G_u\cdot \dfrac{\partial u}{\partial x}+G_v\cdot \dfrac{\partial u}{\partial x}=-G_x \end{cases} \end{align*}

    根据 Cramer 法则

    ∂u∂x=D1D=∣−FxFv−GxGv∣∣FuFvGuGv∣=−1J∣FxFvGxGv∣∂v∂x=D2D=∣Fu−FxGu−Gx∣∣FuFvGuGv∣=−1J∣FuFxGuGx∣\begin{align*} \dfrac{\partial u}{\partial x}=\dfrac{D_1}{D}=\dfrac{\begin{vmatrix}-F_x&F_v\\-G_x&G_v\end{vmatrix}}{\begin{vmatrix}F_u&F_v\\G_u&G_v\end{vmatrix}}=-\dfrac{1}{J}\begin{vmatrix}F_x&F_v\\G_x&G_v\end{vmatrix}\\ \dfrac{\partial v}{\partial x}=\dfrac{D_2}{D}=\dfrac{\begin{vmatrix}F_u&-F_x\\G_u&-G_x\end{vmatrix}}{\begin{vmatrix}F_u&F_v\\G_u&G_v\end{vmatrix}}=-\dfrac{1}{J}\begin{vmatrix}F_u&F_x\\G_u&G_x\end{vmatrix} \end{align*}

    同理可得

    {Fu⋅∂u∂y+Fv⋅∂u∂y=−FyGu⋅∂u∂y+Gv⋅∂u∂y=−Gy\begin{align*} \begin{cases} F_u\cdot \dfrac{\partial u}{\partial y}+F_v\cdot \dfrac{\partial u}{\partial y}=-F_y\\ G_u\cdot \dfrac{\partial u}{\partial y}+G_v\cdot \dfrac{\partial u}{\partial y}=-G_y \end{cases} \end{align*} ∂u∂y=D1D=∣−FyFv−GyGv∣∣FuFvGuGv∣=−1J∣FyFvGyGv∣∂v∂y=D2D=∣Fu−FyGu−Gy∣∣FuFvGuGv∣=−1J∣FuFyGuGy∣\begin{align*} \dfrac{\partial u}{\partial y}=\dfrac{D_1}{D}=\dfrac{\begin{vmatrix}-F_y&F_v\\-G_y&G_v\end{vmatrix}}{\begin{vmatrix}F_u&F_v\\G_u&G_v\end{vmatrix}}=-\dfrac{1}{J}\begin{vmatrix}F_y&F_v\\G_y&G_v\end{vmatrix}\\ \dfrac{\partial v}{\partial y}=\dfrac{D_2}{D}=\dfrac{\begin{vmatrix}F_u&-F_y\\G_u&-G_y\end{vmatrix}}{\begin{vmatrix}F_u&F_v\\G_u&G_v\end{vmatrix}}=-\dfrac{1}{J}\begin{vmatrix}F_u&F_y\\G_u&G_y\end{vmatrix} \end{align*}

  2. [微分法]

    {F(x,y,u,v)=0G(x,y,u,v)=0\begin{align*} \begin{cases} F(x,y,u,v)=0\\ G(x,y,u,v)=0 \end{cases} \end{align*}

    两边微分得

    {Fxdx+Fydy+Fudu+Fvdv=0Gxdx+Gydy+Gudu+Gvdv=0\begin{align*} \begin{cases} F_x \mathrm dx+F_y \mathrm dy+F_u \mathrm du+F_v \mathrm dv=0\\ G_x \mathrm dx+G_y \mathrm dy+G_u \mathrm du+G_v \mathrm dv=0\\ \end{cases} \end{align*} {Fudu+Fvdv=−Fxdx−FydyGudu+Gvdv=−Gxdx−Gydy\begin{align*} \begin{cases} F_u \mathrm du+F_v \mathrm dv=-F_x \mathrm dx-F_y \mathrm dy\\ G_u \mathrm du+G_v \mathrm dv=-G_x \mathrm dx-G_y \mathrm dy \end{cases} \end{align*}

    根据 Cramer 法则

    du=∣−Fxdx−FydyFv−Gxdx−GydyGv∣∣FuFvGuGv∣=−1J∣Fxdx+FydyFvGxdx+GydyGv∣=−1J(∣FxFvGxGv∣dx+∣FyFvGyGv∣dy)\begin{align*} \mathrm du=\dfrac{\begin{vmatrix}-F_x \mathrm dx-F_y \mathrm dy&F_v\\-G_x \mathrm dx-G_y \mathrm dy & G_v\end{vmatrix}}{\begin{vmatrix}F_u&F_v\\G_u&G_v\end{vmatrix}}&=-\dfrac{1}{J}\left | \begin{matrix} F_x \mathrm dx+F_y \mathrm dy & F_v \\ G_x \mathrm dx+G_y \mathrm dy & G_v \\ \end{matrix} \right |\\ &=-\dfrac{1}{J}(\begin{vmatrix}F_x&F_v\\G_x&G_v\end{vmatrix}\mathrm dx+\begin{vmatrix}F_y&F_v\\G_y&G_v\end{vmatrix}\mathrm dy) \end{align*} dv=∣Fu−Fxdx−FydyGu−Gxdx−Gydy∣∣FuFvGuGv∣=−1J∣FuFxdx+FydyGuGxdx+Gydy∣=−1J(∣FuFxGuGx∣dx+∣FuFyGuGy∣dy)\begin{align*} \mathrm dv=\dfrac{\begin{vmatrix}F_u&-F_x \mathrm dx-F_y \mathrm dy\\G_u &-G_x \mathrm dx-G_y \mathrm dy\end{vmatrix}}{\begin{vmatrix}F_u&F_v\\G_u&G_v\end{vmatrix}}&=-\dfrac{1}{J}\left | \begin{matrix} F_u & F_x \mathrm dx+F_y \mathrm dy \\ G_u & G_x \mathrm dx+G_y \mathrm dy \\ \end{matrix} \right |\\ &=-\dfrac{1}{J}(\begin{vmatrix}F_u&F_x\\G_u&G_x\end{vmatrix}\mathrm dx+\begin{vmatrix}F_u&F_y\\G_u&G_y\end{vmatrix}\mathrm dy) \end{align*}

    根据一阶微分的形式不变性，可得到与法一相同的结果

扩展情形#

  • 5元方程组~2个中间变量，3个自变量

    {F(x,y,z,u,v)=0G(x,y,z,u,v)=0→{u=u(x,y,z)v=v(x,y,z)\begin{align*} \begin{cases} F(x,y,z,u,v)=0\\ G(x,y,z,u,v)=0 \end{cases} \rightarrow \begin{cases} u=u(x,y,z)\\ v=v(x,y,z) \end{cases} \end{align*}

    其中 J=∣FuFvGuGv∣J=\begin{vmatrix}F_u&F_v\\G_u&G_v\end{vmatrix}（隐函数中因变量求偏导）

    • ∂u∂z=−1J∣FzFvGzGv∣\dfrac{\partial u}{\partial z}=-\dfrac{1}{J}\begin{vmatrix}F_z&F_v\\G_z&G_v\end{vmatrix}

    • ∂v∂z=−1J∣FuFzGuGz∣\dfrac{\partial v}{\partial z}=-\dfrac{1}{J}\begin{vmatrix}F_u&F_z\\G_u&G_z\end{vmatrix}

    • ……

  • 4元方程组~3个中间变量，1个自变量

    {F(x,u,v,w)=0G(x,u,v,w)=0H(x,u,v,w)=0→{u=u(x)v=v(x)w=w(x)\begin{align*} \begin{cases} F(x,u,v,w)=0\\ G(x,u,v,w)=0\\ H(x,u,v,w)=0 \end{cases} \rightarrow \begin{cases} u=u(x)\\ v=v(x)\\ w=w(x) \end{cases} \end{align*}

    其中 J=∣FuFvFwGuGvGwHuHvHw∣J=\begin{vmatrix}F_u&F_v&F_w\\G_u&G_v&G_w\\H_u&H_v&H_w\end{vmatrix}

    • dudx=−1J∣FxFvFwGxGvGwHxHvHw∣\dfrac{\mathrm du}{\mathrm dx}=-\dfrac{1}{J}\begin{vmatrix}F_x&F_v&F_w\\G_x&G_v&G_w\\H_x&H_v&H_w\end{vmatrix}

    • dvdx=−1J∣FuFxFwGuGxGwHuHxHw∣\dfrac{\mathrm dv}{\mathrm dx}=-\dfrac{1}{J}\begin{vmatrix}F_u&F_x&F_w\\G_u&G_x&G_w\\H_u&H_x&H_w\end{vmatrix}

    • dwdx=−1J∣FuFvFxGuGvGxHuHvHx∣\dfrac{\mathrm dw}{\mathrm dx}=-\dfrac{1}{J}\begin{vmatrix}F_u&F_v&F_x\\G_u&G_v&G_x\\H_u&H_v&H_x\end{vmatrix}

例题#

  1. {xu2−2yv2=0y2u+3x2v=3\begin{cases}xu^2-2yv^2=0\\y^2u+3x^2v=3\end{cases}，求 ∂u∂x\dfrac{\partial u}{\partial x}、∂u∂y\dfrac{\partial u}{\partial y}、∂v∂x\dfrac{\partial v}{\partial x}、∂v∂y\dfrac{\partial v}{\partial y}

    解：

    • F(x,y,u,v)=xu2−2yv2F(x,y,u,v)=xu^2-2yv^2，Fx=u2F_x=u^2，Fy=−2v2F_y=-2v^2，Fu=2xuF_u=2xu，Fv=−4vyF_v=-4vy
    • G(x,y,u,v)=y2u+3x2v−3G(x,y,u,v)=y^2u+3x^2v-3，Gx=6xG_x=6x，Gy=2yuG_y=2yu，Gu=2yG_u=2y，Gv=3x2G_v=3x^2
    • ∂u∂x=−3x2u2+24xyv22(3x3u+2y3v)\dfrac{\partial u}{\partial x}=-\dfrac{3x^2u^2+24xyv^2}{2(3x^3u+2y^3v)}
    • ∂u∂y=3x3v2−4y2uv3x3u+2y3v\dfrac{\partial u}{\partial y}=\dfrac{3x^3v^2-4y^2uv}{3x^3u+2y^3v}
    • ∂v∂x=−12x2uv−y2u22(3x3u+2y3v)\dfrac{\partial v}{\partial x}=-\dfrac{12x^2uv-y^2u^2}{2(3x^3u+2y^3v)}
    • ∂v∂y=−2xyu2+y2v23x3u+2y3v\dfrac{\partial v}{\partial y}=-\dfrac{2xyu^2+y^2v^2}{3x^3u+2y^3v}

  2. y=f(x,t)y=f(x,t)，t是 F(x,y,t)=0F(x,y,t)=0 确定的关于x、y的函数，求 dydx\dfrac{\mathrm dy}{\mathrm dx}、dtdx\dfrac{\mathrm dt}{\mathrm dx}

    解：令 {G(x,y,t)=y−f(x,t)F(x,y,t)=0\begin{cases}G(x,y,t)=y-f(x,t)\\F(x,y,t)=0\end{cases}，其中，Gx=−f1′G_x=-f_1'，Gy=1G_y=1，Gt=−f2′G_t=-f_2'，Fx=F1′F_x=F_1'，Fy=F2′F_y=F_2'，Ft=F3′F_t=F_3'

    代入公式得

    dydx=−1J∣GxGtFxFt∣=−1F3′−F2′f2′⋅(F1′f2′−F3′f1′)\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{1}{J}\begin{vmatrix}G_x&G_t\\F_x&F_t\end{vmatrix}=-\dfrac{1}{F_3'-F_2'f_2'}\cdot (F_1'f_2'-F_3'f_1')

    dtdx=−1J∣GyGxFyFx∣=−1F3′−F2′f2′⋅(F1′+f1′F2′)\dfrac{\mathrm dt}{\mathrm dx}=-\dfrac{1}{J}\begin{vmatrix}G_y&G_x\\F_y&F_x\end{vmatrix}=-\dfrac{1}{F_3'-F_2'f_2'}\cdot (F_1'+f_1'F_2')

    其中 J=∣GyGtFyFt∣J=\begin{vmatrix}G_y&G_t\\F_y&F_t\end{vmatrix}

  3. u=f(x,y,xyz)u=f(x,y,xyz)、z=z(x,y)z=z(x,y) 由 ∫xyzg(xy+z−t)dt=exyz\int_{xy}^zg(xy+z-t)\mathrm dt=e^{xyz} 确定，ff 可微，gg 连续，求 x⋅∂u∂x−y⋅∂u∂yx\cdot \dfrac{\partial u}{\partial x}-y\cdot \dfrac{\partial u}{\partial y}

    解：

    ∂u∂x=f1′⋅1+f2′⋅0+f3′(yz+xy⋅∂z∂x)\dfrac{\partial u}{\partial x}=f_1'\cdot 1+f_2'\cdot 0+f_3'(yz+xy\cdot \dfrac{\partial z}{\partial x})

    ∂u∂y=f1′⋅0+f2′⋅1+f3′(xz+xy⋅∂z∂y)\dfrac{\partial u}{\partial y}=f_1'\cdot 0+f_2'\cdot 1+f_3'(xz+xy\cdot \dfrac{\partial z}{\partial y})

    令 s=xy+z−ts=xy+z-t

    ∫zxyg(s)(−ds)=∫xyzg(s)ds\begin{align*} \int_z^{xy}g(s)(-\mathrm ds)=\int_{xy}^zg(s)\mathrm ds \end{align*} ∂∂x[∫xyzg(s)ds]=g(z)⋅∂z∂x−g(xy)⋅y左边=exyz(yz+xy∂z∂x)右边\begin{align*} \dfrac{\partial }{\partial x}[\int_{xy}^zg(s)\mathrm ds]=\underset{左边}{g(z)\cdot \dfrac{\partial z}{\partial x}-g(xy)\cdot y} =\underset{右边}{e^{xyz}(yz+xy\dfrac{\partial z}{\partial x})} \end{align*}

    ∂z∂x=yzexyz+g(xy)⋅yg(z)−xyexyz\dfrac{\partial z}{\partial x}=\dfrac{yze^{xyz}+g(xy)\cdot y}{g(z)-xye^{xyz}}，对y求偏导得 g(z)⋅∂z∂y−g(xy)⋅x=exyz(xz+xy∂z∂y)g(z)\cdot \dfrac{\partial z}{\partial y}-g(xy)\cdot x=e^{xyz}(xz+xy\dfrac{\partial z}{\partial y})

    ∂z∂y=xzexyz+g(xy)⋅yg(z)−xzexyz\dfrac{\partial z}{\partial y}=\dfrac{xze^{xyz}+g(xy)\cdot y}{g(z)-xze^{xyz}}，原式 =xf1′−yf2′=xf_1'-yf_2'

8.5 隐函数的求导公式
https://gitee.com/jason_ren/advanced-math-note
作者
Jason Ren
发布于
2025-12-03
许可协议
CC BY-SA 4.0

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8.4 复合函数求导
8.6 多元函数微分学的几何应用
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