52 多元复合函数求导、隐函数求导

多元复合函数求导

链式法则（Chain Rule）

设 $z=f(u,v)$，其中 $u=u(x,y)$，$v=v(x,y)$，则复合函数 $z=f(u(x,y),v(x,y))$ 的偏导数为：

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

一般情况的链式法则

设 $z=f(u_1,u_2,\dots ,u_m)$，其中 $u_i=u_i(x_1,x_2,\dots ,x_n)$（$i=1,2,\dots ,m$），则：

$$\frac{\partial z}{\partial x_j}=\sum _{i=1}^m\frac{\partial f}{\partial u_i}\cdot \frac{\partial u_i}{\partial x_j},j=1,2,\dots ,n$$

特殊情况

情况1：一元函数的复合

设 $z=f(u,v)$，其中 $u=u(t)$，$v=v(t)$，则：

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

情况2：混合复合

设 $z=f(u,v)$，其中 $u=u(x,y)$，$v=v(x)$，则：

$$\frac{\partial z}{\partial x}=\frac{\partial f}{\partial u}\cdot \frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\cdot \frac{dv}{dx}$$ $$\frac{\partial z}{\partial y}=\frac{\partial f}{\partial u}\cdot \frac{\partial u}{\partial y}$$

全微分形式的链式法则

设 $z=f(u,v)$，其中 $u=u(x,y)$，$v=v(x,y)$，则：

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

其中：

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

因此：

$$dz=\left(\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x}\right)dx+\left(\frac{\partial f}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial y}\right)dy$$

隐函数求导

一个方程确定的隐函数

一元隐函数

设方程 $F(x,y)=0$ 确定隐函数 $y=y(x)$，则：

$$\frac{dy}{dx}=-\frac{F_x}{F_y}=-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}$$

条件：$F_y\ne 0$

二元隐函数

设方程 $F(x,y,z)=0$ 确定隐函数 $z=z(x,y)$，则：

$$\frac{\partial z}{\partial x}=-\frac{F_x}{F_z}=-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial z}}$$ $$\frac{\partial z}{\partial y}=-\frac{F_y}{F_z}=-\frac{\frac{\partial F}{\partial y}}{\frac{\partial F}{\partial z}}$$

条件：$F_z\ne 0$

方程组确定的隐函数

两个方程确定两个隐函数

设方程组：

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

确定隐函数 $u=u(x,y)$，$v=v(x,y)$。

设雅可比行列式：

$$J=\frac{\partial (F,G)}{\partial (u,v)}=\left|\begin{array}{cc}{F}_u& F_v\\ G_u& G_v\end{array}\right|=F_u{G}_v-F_v{G}_u$$

当 $J\ne 0$ 时，有：

$$\frac{\partial u}{\partial x}=-\frac{1}{J}\left|\begin{array}{cc}{F}_x& F_v\\ G_x& G_v\end{array}\right|=-\frac{F_x{G}_v-F_v{G}_x}{J}$$ $$\frac{\partial u}{\partial y}=-\frac{1}{J}\left|\begin{array}{cc}{F}_y& F_v\\ G_y& G_v\end{array}\right|=-\frac{F_y{G}_v-F_v{G}_y}{J}$$ $$\frac{\partial v}{\partial x}=-\frac{1}{J}\left|\begin{array}{cc}{F}_u& F_x\\ G_u& G_x\end{array}\right|=-\frac{F_u{G}_x-F_x{G}_u}{J}$$ $$\frac{\partial v}{\partial y}=-\frac{1}{J}\left|\begin{array}{cc}{F}_u& F_y\\ G_u& G_y\end{array}\right|=-\frac{F_u{G}_y-F_y{G}_u}{J}$$

参数方程确定的函数的导数

参数方程确定的一元函数

设参数方程：

$$\{\begin{array}{l}x=x(t)\\ y=y(t)\end{array}$$

确定函数 $y=y(x)$，则：

$$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{y^{\prime }(t)}{x^{\prime }(t)}$$

条件：$x^{\prime }(t)\ne 0$

二阶导数：

$$\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dt}\left(\frac{dy}{dx}\right)\cdot \frac{dt}{dx}=\frac{\frac{d}{dt}\left(\frac{y^{\prime }(t)}{x^{\prime }(t)}\right)}{x^{\prime }(t)}$$ $$=\frac{y^{\prime \prime }(t)x^{\prime }(t)-y^{\prime }(t)x^{\prime \prime }(t)}{[x^{\prime }(t){]}^3}$$

参数方程确定的二元函数

设参数方程：

$$\{\begin{array}{l}x=x(u,v)\\ y=y(u,v)\\ z=z(u,v)\end{array}$$

确定函数 $z=z(x,y)$，则：

$$\frac{\partial z}{\partial x}=\frac{\frac{\partial (z,y)}{\partial (u,v)}}{\frac{\partial (x,y)}{\partial (u,v)}}=\frac{\left|\begin{array}{cc}{z}_u& z_v\\ y_u& y_v\end{array}\right|}{\left|\begin{array}{cc}{x}_u& x_v\\ y_u& y_v\end{array}\right|}$$ $$\frac{\partial z}{\partial y}=\frac{\frac{\partial (x,z)}{\partial (u,v)}}{\frac{\partial (x,y)}{\partial (u,v)}}=\frac{\left|\begin{array}{cc}{x}_u& x_v\\ z_u& z_v\end{array}\right|}{\left|\begin{array}{cc}{x}_u& x_v\\ y_u& y_v\end{array}\right|}$$

条件：$\frac{\partial (x,y)}{\partial (u,v)}\ne 0$

典型例题

例1：复合函数求导

题目：设 $z=e^{u^2+v^2}$，其中 $u=x\cos y$，$v=x\sin y$，求 $\frac{\partial z}{\partial x}$ 和 $\frac{\partial z}{\partial y}$。

解：

$$\frac{\partial z}{\partial u}=2u{e}^{u^2+v^2},\frac{\partial z}{\partial v}=2v{e}^{u^2+v^2}$$ $$\frac{\partial u}{\partial x}=\cos y,\frac{\partial u}{\partial y}=-x\sin y$$ $$\frac{\partial v}{\partial x}=\sin y,\frac{\partial v}{\partial y}=x\cos y$$

因此：

$$\frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial x}$$ $$=2u{e}^{u^2+v^2}\cos y+2v{e}^{u^2+v^2}\sin y$$ $$=2e^{u^2+v^2}(u\cos y+v\sin y)$$ $$=2e^{x^2}\cdot x=2x{e}^{x^2}$$ $$\frac{\partial z}{\partial y}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial y}$$ $$=2u{e}^{u^2+v^2}(-x\sin y)+2v{e}^{u^2+v^2}(x\cos y)$$ $$=2x{e}^{u^2+v^2}(-u\sin y+v\cos y)$$ $$=2x{e}^{x^2}\cdot 0=0$$

例2：隐函数求导

题目：设 $x^2+y^2+z^2=3xyz$，求 $\frac{\partial z}{\partial x}$ 和 $\frac{\partial z}{\partial y}$。

解：

设 $F(x,y,z)=x^2+y^2+z^2-3xyz=0$

$$F_x=2x-3yz,F_y=2y-3xz,F_z=2z-3xy$$ $$\frac{\partial z}{\partial x}=-\frac{F_x}{F_z}=-\frac{2x-3yz}{2z-3xy}$$ $$\frac{\partial z}{\partial y}=-\frac{F_y}{F_z}=-\frac{2y-3xz}{2z-3xy}$$

例3：方程组确定的隐函数

题目：设方程组 $\{\begin{array}{l}x+y+u+v=0\\ x^2+y^2+u^2+v^2=1\end{array}$ 确定隐函数 $u=u(x,y)$，$v=v(x,y)$，求 $\frac{\partial u}{\partial x}$。

解：

设 $F=x+y+u+v$，$G=x^2+y^2+u^2+v^2-1$

$$F_u=1,F_v=1,F_x=1$$ $$G_u=2u,G_v=2v,G_x=2x$$

雅可比行列式：

$$J=\left|\begin{array}{cc}1& 1\\ 2u& 2v\end{array}\right|=2v-2u=2(v-u)$$ $$\frac{\partial u}{\partial x}=-\frac{1}{J}\left|\begin{array}{cc}1& 1\\ 2x& 2v\end{array}\right|=-\frac{2v-2x}{2(v-u)}=\frac{x-v}{v-u}$$

例4：参数方程确定的函数求导

题目：设 $\{\begin{array}{l}x=a(t-\sin t)\\ y=a(1-\cos t)\end{array}$，求 $\frac{dy}{dx}$ 和 $\frac{d^{2}y}{d{x}^2}$。

解：

$$\frac{dx}{dt}=a(1-\cos t),\frac{dy}{dt}=a\sin t$$ $$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{a\sin t}{a(1-\cos t)}=\frac{\sin t}{1-\cos t}$$

利用三角恒等式 $\sin t=2\sin \frac{t}{2}\cos \frac{t}{2}$，$1-\cos t=2{\sin }^2\frac{t}{2}$：

$$\frac{dy}{dx}=\frac{2\sin \frac{t}{2}\cos \frac{t}{2}}{2{\sin }^2\frac{t}{2}}=\cot \frac{t}{2}$$ $$\frac{d^{2}y}{d{x}^2}=\frac{d}{dt}\left(\cot \frac{t}{2}\right)\cdot \frac{dt}{dx}=-\frac{1}{2}{\csc }^2\frac{t}{2}\cdot \frac{1}{a(1-\cos t)}$$ $$=-\frac{1}{2a}\cdot \frac{1}{{\sin }^2\frac{t}{2}}\cdot \frac{1}{2{\sin }^2\frac{t}{2}}=-\frac{1}{4a{\sin }^4\frac{t}{2}}$$