第4节 隐函数的求导法则
8 待校验
一、基础题
1. 设 $y=x\text{e}^y+1$ 确定隐函数 $y=y(x)$，求 $\frac{\text{d}y}{\text{d}x}$.
令 $F(x,y) = y - x\text{e}^y - 1 = 0$。
$F_x = -\text{e}^y, \quad F_y = 1 - x\text{e}^y$
根据隐函数求导公式：
$$\frac{\text{d}y}{\text{d}x} = -\frac{F_x}{F_y} = -\frac{-\text{e}^y}{1 - x\text{e}^y} = \frac{\text{e}^y}{1 - x\text{e}^y}$$2. 设 $\ln\sqrt{x^2+y^2} = \arctan\frac{y}{x}$ 确定隐函数 $y=y(x)$，求 $\frac{\text{d}y}{\text{d}x}$.
将原方程化简为 $\frac{1}{2}\ln(x^2+y^2) = \arctan\frac{y}{x}$，两边对 $x$ 求导：
$$\frac{1}{2} \cdot \frac{2x + 2yy'}{x^2+y^2} = \frac{1}{1 + (y/x)^2} \cdot \frac{y'x - y}{x^2}$$$$\frac{x + yy'}{x^2+y^2} = \frac{x^2}{x^2+y^2} \cdot \frac{xy' - y}{x^2}$$$$x + yy' = xy' - y \Rightarrow y'(x-y) = x+y$$$$\frac{\text{d}y}{\text{d}x} = \frac{x+y}{x-y}$$3. 设 $xy+\sin z+y=2z$ 确定隐函数 $z=z(x,y)$，求 $\frac{\partial z}{\partial x}$ 及 $\frac{\partial z}{\partial y}$.
令 $F(x,y,z) = xy + \sin z + y - 2z = 0$。
$F_x = y$
$F_y = x + 1$
$F_z = \cos z - 2$
$$\frac{\partial z}{\partial x} = -\frac{F_x}{F_z} = -\frac{y}{\cos z - 2} = \frac{y}{2 - \cos z}$$$$\frac{\partial z}{\partial y} = -\frac{F_y}{F_z} = -\frac{x+1}{\cos z - 2} = \frac{x+1}{2 - \cos z}$$4. 设 $\frac{y}{z} = \ln\frac{z}{x}$ 确定隐函数 $z=z(x,y)$，求 $\frac{\partial z}{\partial x}$ 及 $\frac{\partial z}{\partial y}$.
原式化为 $\frac{y}{z} = \ln z - \ln x$。令 $F(x,y,z) = \frac{y}{z} - \ln z + \ln x = 0$。
$F_x = \frac{1}{x}, \quad F_y = \frac{1}{z}, \quad F_z = -\frac{y}{z^2} - \frac{1}{z} = -\frac{y+z}{z^2}$
$$\frac{\partial z}{\partial x} = -\frac{F_x}{F_z} = -\frac{1/x}{-(y+z)/z^2} = \frac{z^2}{x(y+z)}$$$$\frac{\partial z}{\partial y} = -\frac{F_y}{F_z} = -\frac{1/z}{-(y+z)/z^2} = \frac{z}{y+z}$$5. 证明：$\frac{\partial x}{\partial y} \cdot \frac{\partial y}{\partial z} \cdot \frac{\partial z}{\partial x} = -1$.
由隐函数求导法则可知：
$$\frac{\partial x}{\partial y} = -\frac{F_y}{F_x}, \quad \frac{\partial y}{\partial z} = -\frac{F_z}{F_y}, \quad \frac{\partial z}{\partial x} = -\frac{F_x}{F_z} $$三式相乘：
$$\frac{\partial x}{\partial y} \cdot \frac{\partial y}{\partial z} \cdot \frac{\partial z}{\partial x} = \left(-\frac{F_y}{F_x}\right) \cdot \left(-\frac{F_z} {F_y}\right) \cdot \left(-\frac{F_x}{F_z}\right) = -1$$6. 证明：方程 $F\left(x+\frac{z}{y}, y+\frac{z}{x}\right)=0$ 所确定的隐函数 $z=z(x,y)$ 满足 $x\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y} = z - xy$.
证明： 令 $u = x+\frac{z}{y}, \ v = y+\frac{z}{x}$。方程两端分别对 $x, y$ 求偏导：
对 $x$ 求偏导：$F_1' \left(1 + \frac{z_x}{y}\right) + F_2' \left(\frac{z_x x - z}{x^2}\right) = 0$
分离出 $z_x$：$z_x \left( \frac{F_1'}{y} + \frac{F_2'}{x} \right) = \frac{z}{x^2}F_2' - F_1'$
同理对 $y$ 求偏导：$F_1' \left(\frac{z_y y - z}{y^2}\right) + F_2' \left(1 + \frac{z_y}{x}\right) = 0$
分离出 $z_y$：$z_y \left( \frac{F_1'}{y} + \frac{F_2'}{x} \right) = \frac{z}{y^2}F_1' - F_2'$
将 $z_x$ 和 $z_y$ 分别乘以 $x$ 和 $y$ 并相加：
$x z_x + y z_y = \frac{x \left( \frac{z}{x^2}F_2' - F_1' \right) + y \left( \frac{z}{y^2}F_1' - F_2' \right)}{\frac{F_1'}{y} + \frac{F_2'}{x}} = \frac{\frac{z-xy} {x}F_2' + \frac{z-xy}{y}F_1'}{\frac{F_1'}{y} + \frac{F_2'}{x}} = \frac{(z-xy) \left( \frac{F_1'}{y} + \frac{F_2'}{x} \right)}{\frac{F_1'}{y} + \frac{F_2'}{x}} = z - xy$。原式得证。
二、提高题
7. 设 $z=f(x+y+z, xyz)$ 确定隐函数 $z=z(x,y)$，求 $\frac{\partial z}{\partial x}, \frac{\partial x}{\partial y}$ 和 $\frac{\partial y}{\partial z}$.
令 $F(x,y,z) = z - f(x+y+z, xyz) = 0$。
$F_x = -f_1' - yzf_2'$
$F_y = -f_1' - xzf_2'$
$F_z = 1 - f_1' - xyf_2'$
通过隐函数求导法则分别作比：
$$\frac{\partial z}{\partial x} = -\frac{F_x}{F_z} = \frac{f_1' + yzf_2'}{1 - f_1' - xyf_2'}$$$$\frac{\partial x}{\partial y} = -\frac{F_y}{F_x} = -\frac{f_1' + xzf_2'}{f_1' + yzf_2'}$$$$\frac{\partial y}{\partial z} = -\frac{F_z}{F_y} = \frac{1 - f_1' - xyf_2'}{f_1' + xzf_2'}$$8. 设 $\text{e}^z - xyz = 0$ 确定隐函数 $z=z(x,y)$，求 $\frac{\partial^2 z}{\partial x^2}$.
原方程两边取对数化简得：$z = \ln x + \ln y + \ln z$。
两边对 $x$ 求偏导：
$z_x = \frac{1}{x} + \frac{1}{z}z_x \Rightarrow z_x \left( \frac{z-1}{z} \right) = \frac{1}{x} \Rightarrow z_x = \frac{z}{x(z-1)}$
再对 $x$ 求二阶偏导：
$$z_{xx} = \frac{\partial}{\partial x} \left( \frac{z}{x(z-1)} \right) = \frac{z_x \cdot x(z-1) - z \cdot [1 \cdot (z-1) + x \cdot z_x]}{(x(z-1))^2}$$$$= \frac{-x z_x - z(z-1)}{x^2(z-1)^2}$$将 $z_x = \frac{z}{x(z-1)}$ 代入分子：
$$z_{xx} = \frac{-\frac{z}{z-1} - z(z-1)}{x^2(z-1)^2} = \frac{-z - z(z-1)^2}{x^2(z-1)^3} = \frac{-z(z^2-2z+2)}{x^2(z-1)^3} = \frac{z(z^2-2z+2)}{x^2(1-z)^3}$$9. 设 $z=z(x,y)$ 是由方程 $\sin(x+y) + \sin(y+z) = 1$ 所确定的隐函数，求 $\frac{\partial^2 z}{\partial x \partial y}$.
对原方程两边分别求 $x, y$ 的偏导：
$\cos(x+y) + \cos(y+z) \cdot z_x = 0 \Rightarrow z_x = -\frac{\cos(x+y)}{\cos(y+z)}$
$\cos(x+y) + \cos(y+z) \cdot (1+z_y) = 0 \Rightarrow 1+z_y = -\frac{\cos(x+y)}{\cos(y+z)} \Rightarrow z_y = z_x - 1$
求混合偏导数 $z_{xy}$ 即对 $z_x$ 关于 $y$ 求导：
$$z_{xy} = \frac{\partial}{\partial y}\left( -\frac{\cos(x+y)}{\cos(y+z)} \right) = -\frac{-\sin(x+y)\cos(y+z) - \cos(x+y)[-\sin(y+z)(1+z_y)]}{\cos^2(y+z)}$$将 $(1+z_y) = -\frac{\cos(x+y)}{\cos(y+z)}$ 代入分子并化简：
$$z_{xy} = \frac{\sin(x+y)\cos^2(y+z) + \cos^2(x+y)\sin(y+z)}{\cos^3(y+z)}$$10. 设 $z=x^2+y^2$，其中 $y=f(x)$ 是由方程 $x^2-xy+y^2=1$ 所确定的隐函数，求 $\frac{\text{d}z}{\text{d}x}$.
利用全导数公式：$\frac{\text{d}z}{\text{d}x} = 2x + 2yy'$
对方程 $x^2-xy+y^2=1$ 两边求导：
$2x - y - xy' + 2yy' = 0 \Rightarrow y'(2y-x) = y-2x \Rightarrow y' = \frac{y-2x}{2y-x}$
代入得：
$$\frac{\text{d}z}{\text{d}x} = 2x + 2y \left(\frac{y-2x}{2y-x}\right) = \frac{2x(2y-x) + 2y(y-2x)}{2y-x} = \frac{2(y^2-x^2)}{2y-x}$$11. 求由下列方程组所确定的隐函数的导数或偏导数.
(1) 对方程组分别求导：
$\begin{cases} 2x + 2y y' + 2z z' = 0 \Rightarrow x + y y' + z z' = 0 \\ 1 + y' + z' = 0 \Rightarrow z' = -1 - y' \end{cases}$
将 $z'$ 代入第一个方程：$x + y y' - z(1+y') = 0 \Rightarrow y'(y-z) = z-x \Rightarrow \frac{\text{d}y}{\text{d}x} = \frac{z-x}{y-z}$
从而 $\frac{\text{d}z}{\text{d}x} = -1 - \frac{z-x}{y-z} = \frac{x-y}{y-z}$
(2) 对两方程求全微分：
$\text{d}u + \text{d}v = \text{d}x + \text{d}y \quad \cdots (i)$
$\cos u \sin v \text{d}u + \sin u \cos v \text{d}v = \frac{1}{y}\text{d}x - \frac{x}{y^2}\text{d}y \quad \cdots (ii)$
将 $(i)$ 式中 $\text{d}v = \text{d}x + \text{d}y - \text{d}u$ 代入 $(ii)$ 化简得：
$\sin(v-u)\text{d}u = \left(\frac{1}{y} - \sin u \cos v\right)\text{d}x - \left(\frac{x}{y^2} + \sin u \cos v\right)\text{d}y$
$$\frac{\partial u}{\partial x} = \frac{\frac{1}{y} - \sin u \cos v}{\sin(v-u)}, \quad \frac{\partial u}{\partial y} = -\frac{\frac{x}{y^2} + \sin u \cos v}{\sin (v-u)}$$反解 $\text{d}v$ 可得：
$$\frac{\partial v}{\partial x} = \frac{\sin v \cos u - \frac{1}{y}}{\sin(v-u)}, \quad \frac{\partial v}{\partial y} = \frac{\sin v \cos u + \frac{x}{y^2}}{\sin (v-u)}$$(3) 对方程组两边同时对 $x$ 求偏导：
$\begin{cases} u_x = f_1' \cdot (u + x u_x) + f_2' \cdot v_x \\ v_x = g_1' \cdot (u_x - 1) + g_2' \cdot (2v y v_x) \end{cases}$
整理成关于 $u_x, v_x$ 的线性方程组：
$\begin{cases} (1-xf_1')u_x - f_2'v_x = u f_1' \\ -g_1'u_x + (1-2vyg_2')v_x = -g_1' \end{cases}$
用克拉默法则解得：
$$\frac{\partial u}{\partial x} = \frac{u f_1' (1 - 2v y g_2') - f_2' g_1'}{(1 - x f_1')(1 - 2v y g_2') - f_2' g_1'}$$$$\frac{\partial v}{\partial x} = \frac{g_1' [(x+u)f_1' - 1]}{(1 - x f_1')(1 - 2v y g_2') - f_2' g_1'}$$12. 设 $y=y(x)$ 是由方程 $\text{e}^{xy}+y^3-5x=0$ 所确定的隐函数，求 $\left. \frac{\text{d}^2 y}{\text{d}x^2} \right|_{x=0}$.
当 $x=0$ 时，$\text{e}^0 + y^3 = 0 \Rightarrow y=-1$。
对方程两边求导：$\text{e}^{xy}(y+xy') + 3y^2y' - 5 = 0$。代入 $x=0, y=-1$，得 $\text{e}^0(-1) + 3(-1)^2y' - 5 = 0 \Rightarrow y'(0) = 2$。
再求一次导数：$\text{e}^{xy}(y+xy')^2 + \text{e}^{xy}(2y' + xy'') + 6y(y')^2 + 3y^2y'' = 0$。
代入 $x=0, y=-1, y'=2$：
$1 \cdot (-1)^2 + 1 \cdot (4) + 6(-1)(4) + 3(1)y'' = 0 \Rightarrow 1 + 4 - 24 + 3y'' = 0 \Rightarrow 3y'' = 19$
$$\left. \frac{\text{d}^2 y}{\text{d}x^2} \right|_{x=0} = \frac{19}{3}$$13. 设函数 $z=z(x,y)$ 由方程 $\cos^2 x + \cos^2 y + \cos^2 z = 1$ 确定，求 $\text{d}z$.
对方程两端微分：
$-2\cos x\sin x \text{d}x - 2\cos y\sin y \text{d}y - 2\cos z\sin z \text{d}z = 0$
利用二倍角公式化简为 $\sin(2x)\text{d}x + \sin(2y)\text{d}y + \sin(2z)\text{d}z = 0$。
$$\text{d}z = -\frac{\sin(2x)}{\sin(2z)}\text{d}x - \frac{\sin(2y)}{\sin(2z)}\text{d}y$$14. 设函数 $z=z(x,y)$ 由方程 $\sin x + 2y - z = \text{e}^z$ 确定，求 $\frac{\partial^2 z}{\partial x^2}, \frac{\partial^2 z}{\partial x\partial y}, \frac {\partial^2 z}{\partial y^2}$.
令 $F = \sin x + 2y - z - \text{e}^z = 0$。
$z_x = \frac{\cos x}{1+\text{e}^z}, \quad z_y = \frac{2}{1+\text{e}^z}$
分别对 $x, y$ 进一步求偏导：
$$\frac{\partial^2 z}{\partial x^2} = \frac{\partial}{\partial x}\left(\frac{\cos x}{1+\text{e}^z}\right) = \frac{-\sin x(1+\text{e}^z) - \cos x \cdot \text{e}^z z_x}{(1+\text{e}^z)^2} = -\frac{\sin x}{1+\text{e}^z} - \frac{\text{e}^z\cos^2 x}{(1+\text{e}^z)^3}$$$$\frac{\partial^2 z}{\partial x\partial y} = \frac{\partial}{\partial y}\left(\frac{\cos x}{1+\text{e}^z}\right) = \frac{-\cos x \cdot \text{e}^z z_y}{(1+\text{e} ^z)^2} = -\frac{2\text{e}^z\cos x}{(1+\text{e}^z)^3}$$$$\frac{\partial^2 z}{\partial y^2} = \frac{\partial}{\partial y}\left(\frac{2}{1+\text{e}^z}\right) = \frac{-2\text{e}^z z_y}{(1+\text{e}^z)^2} = -\frac{4\text{e} ^z}{(1+\text{e}^z)^3}$$三、考研真题
15. (2021213) 设函数 $z=z(x,y)$ 由方程 $(x+1)z + y\ln z - \arctan(2xy) = 1$ 确定，则 $\left. \frac{\partial z}{\partial x} \right|_{(0,2)} =$____.
将 $x=0, y=2$ 代入原方程求 $z$：$(0+1)z + 2\ln z - 0 = 1 \Rightarrow z + 2\ln z = 1$。观察得 $z=1$。
令 $F(x,y,z) = (x+1)z + y\ln z - \arctan(2xy) - 1 = 0$。
$F_x = z - \frac{2y}{1+4x^2y^2}$，代入 $(0,2,1)$ 得 $F_x = 1 - 4 = -3$。
$F_z = (x+1) + \frac{y}{z}$，代入 $(0,2,1)$ 得 $F_z = 1 + 2 = 3$。
$$\left. \frac{\partial z}{\partial x} \right|_{(0,2)} = -\frac{F_x}{F_z} = -\frac{-3}{3} = 1$$答案填写： 1
16. (2015111) 若函数 $z=z(x,y)$ 由方程 $\text{e}^z + xyz + x + \cos x = 2$ 确定，则 $\left. \text{d}z \right|_{(0,1)} =$____.
将 $x=0, y=1$ 代入原方程求 $z$：$\text{e}^z + 0 + 0 + 1 = 2 \Rightarrow \text{e}^z = 1 \Rightarrow z=0$。
对方程两边求全微分：
$\text{e}^z \text{d}z + yz \text{d}x + xz \text{d}y + xy \text{d}z + \text{d}x - \sin x \text{d}x = 0$
将 $(x,y,z) = (0,1,0)$ 代入：
$\text{e}^0 \text{d}z + 0 + 0 + 0 + \text{d}x - 0 = 0 \Rightarrow \text{d}z + \text{d}x = 0 \Rightarrow \text{d}z = -\text{d}x$
答案填写： $-\text{d}x$