多元隐函数的偏导数

隐函数的偏导数求法,与一元函数隐函数的导数求法类似,就是将方程两边对自变量求偏导数,在求导的过程中,将因变量看成自变量的函数,运算复合函数的求导法则,得到一个含有偏导数的方程,最后利用这个方程解出偏导数。如果是两个方程的隐函数,那么得到的是方程组,我们从方程组里解出偏导数即可。

笔记下载:多元隐函数的偏导数

1,方程 \(F(x,y,z)=0\) 给出 \(z\) 是 \(x,y\) 的函数,则

\[\frac{\partial z}{\partial x}=-\frac{F_x}{F_z},\quad \frac{\partial z}{\partial y}=-\frac{F_y}{F_z}\]

这是因为,对方程两边同时对 \(x\) 求导和同时对 \(y\) 求导,就有

\[F_x+F_z\frac{\partial z}{\partial x}=0,\quad F_y+F_z\frac{\partial z}{\partial y}=0\]

解出 \(\displaystyle \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}\) 就得到了前面的公式。

2,由方程组

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

给出 \(u,v\) 是 \(x,y\) 的函数,确定 \(\displaystyle\frac{\partial u}{\partial x},\frac{\partial u}{\partial y},\frac{\partial v}{\partial x},\frac{\partial v}{\partial y}\)。

方程组两边对 \(x\) 求导,得到方程组

\begin{cases}F_x+F_u\frac{\partial u}{\partial x}+F_v\frac{\partial v}{\partial x}=0\\ G_x+G_u\frac{\partial u}{\partial x}+G_v\frac{\partial v}{\partial x}=0\end{cases}

方程组两边对 \(y\) 求导,得到方程组

\begin{cases}F_y+F_u\frac{\partial u}{\partial y}+F_v\frac{\partial v}{\partial y}=0\\ G_y+G_u\frac{\partial u}{\partial y}+G_v\frac{\partial v}{\partial y}=0\end{cases}

在实际问题中,可以用这两个方程组求出几个偏导数。我们这里给出几个偏导数的公式。

\[\frac{\partial u}{\partial x}=\frac{\begin{vmatrix}-F_x&F_v\\ -G_x&G_v\end{vmatrix}}{\begin{vmatrix}F_u&F_v\\ G_u&G_v\end{vmatrix}},\qquad \frac{\partial v}{\partial x}=\frac{\begin{vmatrix}F_u&-F_x\\ G_u&-G_x\end{vmatrix}}{\begin{vmatrix}F_u&F_v\\ G_u&G_v\end{vmatrix}}\]

\[\frac{\partial u}{\partial y}=\frac{\begin{vmatrix}-F_y&F_v\\ -G_y&G_v\end{vmatrix}}{\begin{vmatrix}F_u&F_v\\ G_u&G_v\end{vmatrix}},\qquad \frac{\partial v}{\partial y}=\frac{\begin{vmatrix}F_u&-F_y\\ G_u&-G_y\end{vmatrix}}{\begin{vmatrix}F_u&F_v\\ G_u&G_v\end{vmatrix}}\]

例1,设 \(xy+yz-xz=0\),求 \(\displaystyle \frac{\partial z}{\partial x},\frac{\partial z}{\partial y}\)。

解:令 \(F(x,y,z)=xy+yz-xz\),

\[F_x=y-z, F_y=x+z, F_z=y-x\]

所以

\[\frac{\partial z}{\partial x}=-\frac{F_x}{F_z}=-\frac{y-z}{y-x},\quad \frac{\partial z}{\partial y}=-\frac{F_y}{F_z}=-\frac{x+z}{y-x}\]

例2,设 \(xy^2z^3+x^3y^2z=x+y+z\),求 \(\displaystyle \frac{\partial z}{\partial x},\frac{\partial z}{\partial y}\)。

解:令 \(F(x,y,z)=xy^2z^3+x^3y^2z-x-y-z=0\),

\begin{array}{l}F_x=y^2z^3+3x^2x^2z-1\\ F_y=2xyz^3+2x^2yz-1\\ F_z=3xy^2z^2+x^3y^-1\end{array}

所以

\begin{array}{l}\frac{\partial z}{\partial x}=-\frac{F_x}{F_z}=-\frac{y^2z^3+3x^2x^2z-1}{3xy^2z^2+x^3y^-1}\\ \frac{\partial z}{\partial y}=-\frac{F_y}{F_z}=-\frac{2xyz^3+2x^2yz-1}{3xy^2z^2+x^3y^-1}\end{array}

例3,设 \begin{cases}xu-yv=0\\ yu+xv=1\end{cases},求 \(\displaystyle\frac{\partial u}{\partial x},\frac{\partial u}{\partial y},\frac{\partial v}{\partial x},\frac{\partial v}{\partial y}\)。

解:\[F_x=u,F_y=-y, F_u=x, F_v=-y\]

\[G_x=v,G_y=u,G_u=y,G_v=x\]

所以

\begin{align*}\frac{\partial u}{\partial x}&=\frac{\begin{vmatrix}-F_x&F_v\\ -G_x&G_v\end{vmatrix}}{\begin{vmatrix}F_u&F_v\\ G_u&G_v\end{vmatrix}}=\frac{\begin{vmatrix}-u&-y\\ -v&x\end{vmatrix}}{\begin{vmatrix}x&-y\\ y&x\end{vmatrix}}=\frac{-xu-yv}{x^2+y^2}\end{align*}

\[\frac{\partial v}{\partial x}=\frac{\begin{vmatrix}F_u&-F_x\\ G_u&-G_x\end{vmatrix}}{\begin{vmatrix}F_u&F_v\\ G_u&G_v\end{vmatrix}}=\frac{\begin{vmatrix}x&-u\\ y&-v\end{vmatrix}}{\begin{vmatrix}x&-y\\y&x\end{vmatrix}}=\frac{-xv+yu}{x^2+y^2}\]

\[\frac{\partial u}{\partial y}=\frac{\begin{vmatrix}-F_y&F_v\\ -G_y&G_v\end{vmatrix}}{\begin{vmatrix}F_u&F_v\\ G_u&G_v\end{vmatrix}}=\frac{\begin{vmatrix}v&-y\\ -u&x\end{vmatrix}}{\begin{vmatrix}x&-y\\y&x\end{vmatrix}}=\frac{xv-yu}{x^2+y^2}\]

\[\frac{\partial v}{\partial y}=\frac{\begin{vmatrix}F_u&-F_y\\ G_u&-G_y\end{vmatrix}}{\begin{vmatrix}F_u&F_v\\ G_u&G_v\end{vmatrix}}=\frac{\begin{vmatrix}x&v\\ y&-u\end{vmatrix}}{\begin{vmatrix}x&-y\\y&x\end{vmatrix}}=\frac{-xu-yv}{x^2+y^2}\]