第3节 多元复合函数的求导法则

一、基础题

1. 设 $z=u^2+uv+v^2$，而 $u=x+y,\ v=x-y$，求 $\frac{\partial z}{\partial x} , \frac{\partial z}{\partial y}$.

利用多元复合函数求导法则（链式法则）：

$\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial z}{\partial v}\frac{\partial v}{\partial x} = (2u+v) \cdot 1 + (u+2v) \cdot 1 = 3u + 3v$

代入 $u, v$ 的表达式：$3(x+y) + 3(x-y) = 6x$

$\frac{\partial z}{\partial y} = \frac{\partial z}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial z}{\partial v}\frac{\partial v}{\partial y} = (2u+v) \cdot 1 + (u+2v) \cdot (-1) = 2u+v-u-2v = u-v$

代入 $u, v$ 的表达式：$(x+y) - (x-y) = 2y$

(注：也可直接代入化简 $z=(x+y)^2+(x+y)(x-y)+(x-y)^2 = 3x^2+y^2$，然后直接求偏导，结果一致)

2. 设 $z=\ln(x^2+y)$，而 $x=\text{e}^{s^2},\ y=t^2+s$，求 $\frac{\partial z}{\partial t} , \frac{\partial z}{\partial s}$.

$\frac{\partial z}{\partial t} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial t} = \frac {2x}{x^2+y} \cdot 0 + \frac{1}{x^2+y} \cdot (2t) = \frac{2t}{x^2+y} = \frac{2t}{\text{e}^{2s^2}+t^2+s}$

$\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial s} = \frac {2x}{x^2+y} \cdot (\text{e}^{s^2} \cdot 2s) + \frac{1}{x^2+y} \cdot 1 = \frac{4s x\text{e}^{s^2} + 1}{x^2+y} = \frac{4s\text{e}^{2s^2}+1}{\text{e}^{2s^2}+t^2+s}$

3. 设 $z=x^y$，而 $x=\sin t,\ y=\cos t$，求 $\frac{\text{d}z}{\text{d}t}$.

$\frac{\text{d}z}{\text{d}t} = \frac{\partial z}{\partial x}\frac{\text{d}x}{\text{d}t} + \frac{\partial z}{\partial y}\frac{\text{d}y}{\text{d}t} = y x^{y-1} \cdot \cos t + x^y \ln x \cdot (-\sin t)$

代入 $x, y$ 的表达式：

$\frac{\text{d}z}{\text{d}t} = (\cos t)(\sin t)^{\cos t - 1} \cos t - (\sin t)^{\cos t} \ln(\sin t) \sin t$

$= (\sin t)^{\cos t} \left[ \frac{\cos^2 t}{\sin t} - \sin t \ln(\sin t) \right]$ 或写成 $(\sin t)^{\cos t} [ \cos t \cot t - \sin t \ln(\sin t) ]$

4. 设 $z=\arcsin(x-y)$，而 $x=3t,\ y=4t^3$，求 $\frac{\text{d}z}{\text{d}t}$.

$\frac{\text{d}z}{\text{d}t} = \frac{\partial z}{\partial x}\frac{\text{d}x}{\text{d}t} + \frac{\partial z}{\partial y}\frac{\text{d}y}{\text{d}t} = \frac{1}{\sqrt {1-(x-y)^2}} \cdot 3 + \frac{-1}{\sqrt{1-(x-y)^2}} \cdot 12t^2 = \frac{3-12t^2}{\sqrt{1-(3t-4t^3)^2}}$

化简根号内的表达式：$1-(3t-4t^3)^2 = 1 - 9t^2 + 24t^4 - 16t^6 = (1-t^2)(1-4t^2)^2$

因此，$\frac{\text{d}z}{\text{d}t} = \frac{3(1-4t^2)}{\sqrt{(1-t^2)(1-4t^2)^2}} = \frac{3(1-4t^2)}{|1-4t^2|\sqrt{1-t^2}}$

5. 设 $z=\frac{y}{x}$，而 $y=\sqrt{1-x^2}$，求 $\frac{\text{d}z}{\text{d}x}$.

这是一个全导数问题：

$\frac{\text{d}z}{\text{d}x} = \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y}\frac{\text{d}y}{\text{d}x} = -\frac{y}{x^2} + \frac{1}{x} \cdot \frac {-2x}{2\sqrt{1-x^2}} = -\frac{\sqrt{1-x^2}}{x^2} - \frac{1}{\sqrt{1-x^2}}$

通分化简：

$\frac{\text{d}z}{\text{d}x} = \frac{-(1-x^2) - x^2}{x^2\sqrt{1-x^2}} = -\frac{1}{x^2\sqrt{1-x^2}}$

6. 设 $u=\text{e}^{x+2y}\sin 3z$，而 $x=\ln t,\ y=\ln(1+t^2),\ z=t$，求 $\frac{\text{d}u}{\text{d}t}$.

$\frac{\text{d}u}{\text{d}t} = \frac{\partial u}{\partial x}\frac{\text{d}x}{\text{d}t} + \frac{\partial u}{\partial y}\frac{\text{d}y}{\text{d}t} + \frac {\partial u}{\partial z}\frac{\text{d}z}{\text{d}t}$

$= (\text{e}^{x+2y}\sin 3z) \cdot \frac{1}{t} + (2\text{e}^{x+2y}\sin 3z) \cdot \frac{2t}{1+t^2} + (3\text{e}^{x+2y}\cos 3z) \cdot 1$

注意 $\text{e}^{x+2y} = \text{e}^{\ln t + 2\ln(1+t^2)} = t(1+t^2)^2$ ，代入上式得：

$\frac{\text{d}u}{\text{d}t} = t(1+t^2)^2\sin 3t \cdot \frac{1}{t} + 2t(1+t^2)^2\sin 3t \cdot \frac{2t}{1+t^2} + 3t(1+t^2)^2\cos 3t$

$= (1+t^2)^2\sin 3t + 4t^2(1+t^2)\sin 3t + 3t(1+t^2)^2\cos 3t$

$= (1+t^2)[(1+t^2+4t^2)\sin 3t + 3t(1+t^2)\cos 3t] = (1+t^2)[(1+5t^2)\sin 3t + 3t(1+t^2)\cos 3t]$

7. 求下列函数的一阶偏导数（其中 $f$ 具有一阶连续偏导数）：

(1) $u = f(x+y, xy)$

设中间变量 $v_1 = x+y, v_2 = xy$，则 $u = f(v_1, v_2)$

$\frac{\partial u}{\partial x} = f_1' \frac{\partial v_1}{\partial x} + f_2' \frac{\partial v_2}{\partial x} = f_1' \cdot 1 + f_2' \cdot y = f_1' + yf_2'$

$\frac{\partial u}{\partial y} = f_1' \frac{\partial v_1}{\partial y} + f_2' \frac{\partial v_2}{\partial y} = f_1' \cdot 1 + f_2' \cdot x = f_1' + xf_2'$

(2) $u = f\left(\frac{x}{y}, \frac{y}{z}\right)$

设中间变量 $v_1 = \frac{x}{y}, v_2 = \frac{y}{z}$

$\frac{\partial u}{\partial x} = f_1' \cdot \frac{1}{y} + f_2' \cdot 0 = \frac{1}{y}f_1'$

$\frac{\partial u}{\partial y} = f_1' \cdot \left(-\frac{x}{y^2}\right) + f_2' \cdot \frac{1}{z} = -\frac{x}{y^2}f_1' + \frac{1}{z}f_2'$

$\frac{\partial u}{\partial z} = f_1' \cdot 0 + f_2' \cdot \left(-\frac{y}{z^2}\right) = -\frac{y}{z^2}f_2'$

(3) $u = f(x, xy, xyz)$

设中间变量 $v_1 = x, v_2 = xy, v_3 = xyz$

$\frac{\partial u}{\partial x} = f_1' \cdot 1 + f_2' \cdot y + f_3' \cdot yz = f_1' + yf_2' + yzf_3'$

$\frac{\partial u}{\partial y} = f_1' \cdot 0 + f_2' \cdot x + f_3' \cdot xz = xf_2' + xzf_3'$

$\frac{\partial u}{\partial z} = f_1' \cdot 0 + f_2' \cdot 0 + f_3' \cdot xy = xyf_3'$

二、提高题

8. 设 $z = xy + xF(u)$，而 $u = \frac{y}{x}$，$F(u)$ 为可导函数，证明：$x\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y} = xy+z$.

证明：

求偏导数：

$\frac{\partial z}{\partial x} = y + F(u) + xF'(u) \cdot \left(-\frac{y}{x^2}\right) = y + F(u) - \frac{y}{x}F'(u)$

$\frac{\partial z}{\partial y} = x + xF'(u) \cdot \left(\frac{1}{x}\right) = x + F'(u)$

将上述偏导数代入等式左边：

$x\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y} = x\left( y + F(u) - \frac{y}{x}F'(u) \right) + y(x + F'(u))$

$= xy + xF(u) - yF'(u) + xy + yF'(u)$

$= 2xy + xF(u)$

将其拆分为 $xy + (xy + xF(u))$，又因为 $z = xy + xF(u)$，所以：

$x\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y} = xy + z$。原式得证。

9. 求下列函数的二阶偏导数（其中 $f$ 具有二阶连续偏导数）：

(1) $z = f(xy, x)$

设 $u = xy, v = x$，则 $z = f(u,v)$

一阶偏导：

$z_x = f_1' \cdot y + f_2' \cdot 1 = yf_1' + f_2'$

$z_y = f_1' \cdot x + f_2' \cdot 0 = xf_1'$

二阶偏导：

$z_{xx} = \frac{\partial}{\partial x}(yf_1' + f_2') = y(f_{11}'' \cdot y + f_{12}'' \cdot 1) + (f_{21}'' \cdot y + f_{22}'' \cdot 1) = y^2f_{11}'' + 2yf_{12}'' + f_{22}''$

$z_{yy} = \frac{\partial}{\partial y}(xf_1') = x(f_{11}'' \cdot x) = x^2f_{11}''$

$z_{xy} = \frac{\partial}{\partial y}(yf_1' + f_2') = f_1' + y(f_{11}'' \cdot x + f_{12}'' \cdot 0) + (f_{21}'' \cdot x + f_{22}'' \cdot 0) = f_1' + xyf_{11}'' + xf_{12}''$

(2) $u = f(x^2-y^2+z^2)$

设中间变量 $v = x^2-y^2+z^2$

一阶偏导：$u_x = 2xf', \ u_y = -2yf', \ u_z = 2zf'$

二阶偏导：

$u_{xx} = 2f' + 2x \cdot f''(v) \cdot 2x = 2f' + 4x^2f''$

$u_{yy} = -2f' - 2y \cdot f''(v) \cdot (-2y) = -2f' + 4y^2f''$

$u_{zz} = 2f' + 2z \cdot f''(v) \cdot 2z = 2f' + 4z^2f''$

$u_{xy} = 2x \cdot f''(v) \cdot (-2y) = -4xyf''$

$u_{yz} = -2y \cdot f''(v) \cdot 2z = -4yzf''$

$u_{zx} = 2z \cdot f''(v) \cdot 2x = 4xzf''$

(3) $z = f(\sin x, \cos y, \text{e}^{x+y})$

设 $u=\sin x, v=\cos y, w=\text{e}^{x+y}$

一阶偏导：

$z_x = f_1' \cos x + f_3' \text{e}^{x+y}$

$z_y = -f_2' \sin y + f_3' \text{e}^{x+y}$

二阶偏导：

$z_{xx} = \frac{\partial}{\partial x}(f_1' \cos x + f_3' \text{e}^{x+y}) = (f_{11}'' \cos x + f_{13}'' \text{e}^{x+y})\cos x - f_1' \sin x + (f_{31}'' \cos x + f_{33}'' \text{e}^{x+y})\text{e}^{x+y} + f_3' \text{e}^{x+y}$

$= f_{11}'' \cos^2 x + f_{33}'' \text{e}^{2(x+y)} + 2f_{13}'' \text{e}^{x+y} \cos x - f_1' \sin x + f_3' \text{e}^{x+y}$

$z_{yy} = \frac{\partial}{\partial y}(-f_2' \sin y + f_3' \text{e}^{x+y}) = -(-f_{22}'' \sin y + f_{23}'' \text{e}^{x+y})\sin y - f_2' \cos y + (-f_{32}'' \sin y + f_{33}'' \text{e}^{x+y})\text{e}^{x+y} + f_3' \text{e}^{x+y}$

$= f_{22}'' \sin^2 y + f_{33}'' \text{e}^{2(x+y)} - 2f_{23}'' \text{e}^{x+y} \sin y - f_2' \cos y + f_3' \text{e}^{x+y}$

$z_{xy} = \frac{\partial}{\partial y}(f_1' \cos x + f_3' \text{e}^{x+y}) = (-f_{12}'' \sin y + f_{13}'' \text{e}^{x+y})\cos x + (-f_{32}'' \sin y + f_{33}'' \text{e}^{x+y})\text{e}^{x+y} + f_3' \text{e}^{x+y}$

$= -f_{12}'' \cos x \sin y + f_{13}'' \text{e}^{x+y} \cos x - f_{23}'' \text{e}^{x+y} \sin y + f_{33}'' \text{e}^{2(x+y)} + f_3' \text{e}^{x+y}$

10. 设 $u = x^4 - 2x^3(y+z) + 6x^2yz + f(y-x, z-x)$，其中 $f(x,y)$ 可微，求 $\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z}$.

分别求对 $x, y, z$ 的偏导数：

$u_x = 4x^3 - 6x^2(y+z) + 12xyz + f_1' \cdot (-1) + f_2' \cdot (-1) = 4x^3 - 6x^2y - 6x^2z + 12xyz - f_1' - f_2'$

$u_y = -2x^3 + 6x^2z + f_1' \cdot 1 + f_2' \cdot 0 = -2x^3 + 6x^2z + f_1'$

$u_z = -2x^3 + 6x^2y + f_1' \cdot 0 + f_2' \cdot 1 = -2x^3 + 6x^2y + f_2'$

将三式相加，含有 $f_1', f_2'$ 的项以及其余部分刚好抵消：

$u_x + u_y + u_z = (4x^3 - 2x^3 - 2x^3) + (-6x^2y + 6x^2y) + (-6x^2z + 6x^2z) + 12xyz + (-f_1' + f_1') + (-f_2' + f_2')$

$= 12xyz$

11. 已知 $f(x,x^2) = x^2\text{e}^{-x},\ f_x(x,x^2) = -x^2\text{e}^{-x}$，则 $f_y(x,x^2) =$____.

设一元函数 $g(x) = f(x, x^2) = x^2\text{e}^{-x}$。两边对 $x$ 求全导数：

$g'(x) = \frac{\text{d}}{\text{d}x}(x^2\text{e}^{-x}) = 2x\text{e}^{-x} - x^2\text{e}^{-x} = (2x-x^2)\text{e}^{-x}$

由复合函数求导法则展开 $g'(x)$ 亦有：

$g'(x) = f_x(x, x^2) \cdot 1 + f_y(x, x^2) \cdot 2x$

将已知条件 $f_x(x, x^2) = -x^2\text{e}^{-x}$ 代入上式并令两式相等：

$-x^2\text{e}^{-x} + 2x f_y(x, x^2) = 2x\text{e}^{-x} - x^2\text{e}^{-x}$

消去两边的 $-x^2\text{e}^{-x}$：

$2x f_y(x, x^2) = 2x\text{e}^{-x}$

当 $x \ne 0$ 时，两边除以 $2x$ 得到 $f_y(x, x^2) = \text{e}^{-x}$。由连续性可知对所有 $x$ 均成立。

答案选 C

三、考研真题

12. (2021102) 已知函数 $f(x,y)$ 可微，且 $f(x+1, \text{e}^x) = x(x+1)^2, f(x, x^2) = 2x^2\ln x$，则微分 $\text{d}f(x,y)|_{(1,1)} =$____.

设 $g(x) = f(x+1, \text{e}^x) = x(x+1)^2$。对 $x$ 求导得：

$g'(x) = f_x(x+1, \text{e}^x) \cdot 1 + f_y(x+1, \text{e}^x) \cdot \text{e}^x = (x+1)^2 + 2x(x+1)$

我们要找 $(x,y)=(1,1)$ 时的偏导数。在 $g(x)$ 中令 $x=0$，此时 $x+1=1, \text{e}^0=1$，对应点正是 $(1,1)$。

将 $x=0$ 代入 $g'(x)$ 表达式：

$f_x(1, 1) + f_y(1, 1) \cdot 1 = (0+1)^2 + 0 \implies f_x(1,1) + f_y(1,1) = 1 \quad \dots \text{①}$

设 $h(x) = f(x, x^2) = 2x^2\ln x$。对 $x$ 求导得：

$h'(x) = f_x(x, x^2) \cdot 1 + f_y(x, x^2) \cdot 2x = 4x\ln x + 2x^2 \cdot \frac{1}{x} = 4x\ln x + 2x$

在 $h(x)$ 中令 $x=1$，对应点也是 $(1,1)$。

将 $x=1$ 代入 $h'(x)$ 表达式：

$f_x(1, 1) + f_y(1, 1) \cdot 2 = 4(1)\ln 1 + 2(1) \implies f_x(1,1) + 2f_y(1,1) = 2 \quad \dots \text{②}$

联立方程组 ① 和 ②，解得：$f_y(1,1) = 1, \ f_x(1,1) = 0$。

因此，全微分 $\text{d}f(1,1) = f_x(1,1)\text{d}x + f_y(1,1)\text{d}y = 0\text{d}x + 1\text{d}y = \text{d}y$。

答案选 C

13. (2015205) 已知函数 $f(u,v)$ 可微，且 $f\left(x+y, \frac{y}{x}\right) = x^2-y^2$，则 $\left.\frac{\partial f}{\partial u}\right|_{\begin{smallmatrix}u=1\v=1\end {smallmatrix}}$ 与 $\left.\frac{\partial f}{\partial v}\right|_{\begin{smallmatrix}u=1\v=1\end{smallmatrix}}$ 分别为____.

令 $u = x+y, v = \frac{y}{x}$。当 $u=1, v=1$ 时，由 $x+y=1$ 且 $\frac{y}{x}=1$（即 $y=x$）可解得 $x=\frac{1}{2}, y=\frac{1}{2}$。

对等式 $f\left(x+y, \frac{y}{x}\right) = x^2-y^2$ 两边分别对 $x$ 和 $y$ 求偏导：

对 $x$ 求导：$f_u \cdot 1 + f_v \cdot \left(-\frac{y}{x^2}\right) = 2x$

对 $y$ 求导：$f_u \cdot 1 + f_v \cdot \left(\frac{1}{x}\right) = -2y$

将 $(x,y) = (\frac{1}{2}, \frac{1}{2})$ 以及对应的 $(u,v) = (1,1)$ 代入上面两个式子中：

$f_u(1,1) + f_v(1,1) \cdot \frac{-1/2}{(1/2)^2} = 2(\frac{1}{2}) \implies f_u - 2f_v = 1$

$f_u(1,1) + f_v(1,1) \cdot \frac{1}{1/2} = -2(\frac{1}{2}) \implies f_u + 2f_v = -1$

将两式相加得 $2f_u = 0 \implies f_u = 0$。

将 $f_u=0$ 代入得 $2f_v = -1 \implies f_v = -\frac{1}{2}$。

即 $\frac{\partial f}{\partial u} = 0, \frac{\partial f}{\partial v} = -\frac{1}{2}$。

答案选 D