第2节 偏导数与全微分
一、基础题
1. 求下列函数的偏导数：
(1) $z = x^2 y^3 - x$
$$\frac{\partial z}{\partial x} = 2xy^3 - 1$$$$\frac{\partial z}{\partial y} = 3x^2y^2$$(2) $z = x \cos y$
$$\frac{\partial z}{\partial x} = \cos y$$$$\frac{\partial z}{\partial y} = -x \sin y$$(3) $s = \frac{u^2 + v^2}{uv} = \frac{u}{v} + \frac{v}{u}$
$$\frac{\partial s}{\partial u} = \frac{1}{v} - \frac{v}{u^2} = \frac{u^2 - v^2}{u^2 v}$$$$\frac{\partial s}{\partial v} = -\frac{u}{v^2} + \frac{1}{u} = \frac{v^2 - u^2}{u v^2}$$(4) $z = \ln(x^2 + y)$
$$\frac{\partial z}{\partial x} = \frac{2x}{x^2 + y}$$$$\frac{\partial z}{\partial y} = \frac{1}{x^2 + y}$$(5) $z = \text{e}^{\sin(xy)}$
$$\frac{\partial z}{\partial x} = \text{e}^{\sin(xy)} \cdot \cos(xy) \cdot y = y \cos(xy) \text{e}^{\sin(xy)}$$$$\frac{\partial z}{\partial y} = \text{e}^{\sin(xy)} \cdot \cos(xy) \cdot x = x \cos(xy) \text{e}^{\sin(xy)}$$(6) $u = (xy)^z$
将函数改写为 $u = \text{e}^{z \ln(xy)} = x^z y^z$：
$$\frac{\partial u}{\partial x} = z(xy)^{z-1} \cdot y = z x^{z-1} y^z$$$$\frac{\partial u}{\partial y} = z(xy)^{z-1} \cdot x = z x^z y^{z-1}$$$$\frac{\partial u}{\partial z} = (xy)^z \ln(xy)$$2. 设 $z = \frac{x}{\sqrt{x^2+y^2}}$，求 $\left.\frac{\partial z}{\partial x}\right|_{(0,1)}, \left.\frac{\partial z}{\partial y}\right|_{(0,1)}$.
$$\frac{\partial z}{\partial x} = \frac{1 \cdot \sqrt{x^2+y^2} - x \cdot \frac{x}{\sqrt{x^2+y^2}}}{x^2+y^2} = \frac{x^2+y^2-x^2}{(x^2+y^2)^{3/2}} = \frac{y^2}{(x^2 +y^2)^{3/2}}$$$$\left.\frac{\partial z}{\partial x}\right|_{(0,1)} = \frac{1^2}{(0^2+1^2)^{3/2}} = 1$$$$\frac{\partial z}{\partial y} = x \left( -\frac{1}{2} \right) (x^2+y^2)^{-3/2} \cdot 2y = -\frac{xy}{(x^2+y^2)^{3/2}}$$$$\left.\frac{\partial z}{\partial y}\right|_{(0,1)} = -\frac{0 \cdot 1}{(0^2+1^2)^{3/2}} = 0$$3. 设 $f(x,y) = x + (y-1)\arcsin\sqrt{\frac{x}{y}}$，求 $f_x(x,1)$.
先求对 $x$ 的偏导数：
$$f_x(x,y) = 1 + (y-1) \frac{\partial}{\partial x} \left( \arcsin\sqrt{\frac{x}{y}} \right)$$将 $y=1$ 代入：
$$f_x(x,1) = 1 + (1-1) \left. \frac{\partial}{\partial x} \left( \arcsin\sqrt{\frac{x}{y}} \right) \right|_{y=1} = 1 + 0 = 1$$4. 求下列函数的二阶偏导数：
(1) $z = x^3 y^3 - 3x^2 y + xy^2 + 3$
$\frac{\partial z}{\partial x} = 3x^2 y^3 - 6xy + y^2 $$ \frac{\partial z}{\partial y} = 3x^3 y^2 - 3x^2 + 2xy $
$$\frac{\partial^2 z}{\partial x^2} = 6xy^3 - 6y$$$$\frac{\partial^2 z}{\partial x \partial y} = \frac{\partial^2 z}{\partial y \partial x} = 9x^2 y^2 - 6x + 2y$$$$\frac{\partial^2 z}{\partial y^2} = 6x^3 y + 2x$$(2) $z = xy\text{e}^{x+y}$
$\frac{\partial z}{\partial x} = y\text{e}^{x+y} + xy\text{e}^{x+y} = y(x+1)\text{e}^{x+y} $$ \frac{\partial z}{\partial y} = x\text{e}^{x+y} + xy\text{e}^{x+y} = x(y+1)\text{e}^{x+y} $
$$\frac{\partial^2 z}{\partial x^2} = y\text{e}^{x+y} + y(x+1)\text{e}^{x+y} = y(x+2)\text{e}^{x+y}$$$$\frac{\partial^2 z}{\partial x \partial y} = \frac{\partial^2 z}{\partial y \partial x} = (x+1)\text{e}^{x+y} + y(x+1)\text{e}^{x+y} = (x+1)(y+1)\text{e}^{x+y}$$$$\frac{\partial^2 z}{\partial y^2} = x\text{e}^{x+y} + x(y+1)\text{e}^{x+y} = x(y+2)\text{e}^{x+y}$$5. 设 $f(x,y,z) = xy^2 + yz^2 + zx^2$，求 $f_{xx}(0,0,1), f_{xz}(1,0,2), f_{yz}(0,-1,0)$ 及 $f_{zzx}(2,0,1)$.
$f_x = y^2 + 2zx \implies f_{xx} = 2z \implies f_{xx}(0,0,1) = 2(1) = 2$
$f_{xz} = \frac{\partial}{\partial z}(y^2+2zx) = 2x \implies f_{xz}(1,0,2) = 2(1) = 2$
$f_y = 2xy + z^2 \implies f_{yz} = 2z \implies f_{yz}(0,-1,0) = 2(0) = 0$
$f_z = 2yz + x^2 \implies f_{zz} = 2y \implies f_{zzx} = 0 \implies f_{zzx}(2,0,1) = 0$
6. 求下列函数的全微分：
(1) $z = y \sin(x+y)$
$\frac{\partial z}{\partial x} = y \cos(x+y)$
$\frac{\partial z}{\partial y} = \sin(x+y) + y \cos(x+y)$
$$\text{d}z = y \cos(x+y) \text{d}x + [\sin(x+y) + y \cos(x+y)] \text{d}y$$(2) $z = \frac{x+y}{2+y}$
$\frac{\partial z}{\partial x} = \frac{1}{2+y}$
$\frac{\partial z}{\partial y} = \frac{1\cdot(2+y) - (x+y)\cdot 1}{(2+y)^2} = \frac{2-x}{(2+y)^2}$
$$\text{d}z = \frac{1}{2+y} \text{d}x + \frac{2-x}{(2+y)^2} \text{d}y$$(3) $z = \ln\sqrt{x^2+y^2} = \frac{1}{2}\ln(x^2+y^2)$
$\frac{\partial z}{\partial x} = \frac{x}{x^2+y^2}, \quad \frac{\partial z}{\partial y} = \frac{y}{x^2+y^2}$
$$\text{d}z = \frac{x}{x^2+y^2} \text{d}x + \frac{y}{x^2+y^2} \text{d}y$$(4) $u = x^{yz}$
$\frac{\partial u}{\partial x} = yz x^{yz-1}$
$\frac{\partial u}{\partial y} = z x^{yz} \ln x$
$\frac{\partial u}{\partial z} = y x^{yz} \ln x$
$$\text{d}u = yz x^{yz-1} \text{d}x + z x^{yz} \ln x \text{d}y + y x^{yz} \ln x \text{d}z$$7. 求函数 $z = \text{e}^{\frac{y}{x}}$ 在 $x=1, y=2$ 时的全微分.
$\frac{\partial z}{\partial x} = \text{e}^{\frac{y}{x}} \cdot \left(-\frac{y}{x^2}\right) = -\frac{y}{x^2}\text{e}^{\frac{y}{x}}$
$\frac{\partial z}{\partial y} = \text{e}^{\frac{y}{x}} \cdot \frac{1}{x} = \frac{1}{x}\text{e}^{\frac{y}{x}}$
将 $x=1, y=2$ 代入：$\left.\frac{\partial z}{\partial x}\right|_{(1,2)} = -2\text{e}^2, \quad \left.\frac{\partial z}{\partial y}\right|_{(1,2)} = \text{e}^2$
$$\text{d}z = -2\text{e}^2 \text{d}x + \text{e}^2 \text{d}y$$8. 求函数 $z = xy$ 在 $x=2, y=1, \Delta x=0.1, \Delta y=-0.2$ 时的全增量和全微分.
全增量： $\Delta z = (x+\Delta x)(y+\Delta y) - xy = (2+0.1)(1-0.2) - (2)(1) = 2.1 \times 0.8 - 2 = 1.68 - 2 = -0.32$
全微分： $\text{d}z = y\Delta x + x\Delta y = (1)(0.1) + (2)(-0.2) = 0.1 - 0.4 = -0.3$
9. 求函数 $z = x^4 + y^4 - 4x^2 y^2$ 在 $x=1, y=1, \Delta x=0.15, \Delta y=0.1$ 时的全微分.
$\frac{\partial z}{\partial x} = 4x^3 - 8xy^2 \implies \left.\frac{\partial z}{\partial x}\right|_{(1,1)} = 4 - 8 = -4$
$\frac{\partial z}{\partial y} = 4y^3 - 8x^2y \implies \left.\frac{\partial z}{\partial y}\right|_{(1,1)} = 4 - 8 = -4$
$$\text{d}z = \frac{\partial z}{\partial x}\Delta x + \frac{\partial z}{\partial y}\Delta y = -4(0.15) - 4(0.1) = -0.6 - 0.4 = -1.0$$二、提高题
10. 设 $z = x \ln(xy)$，求 $\frac{\partial^3 z}{\partial x^2 \partial y}, \frac{\partial^3 z}{\partial x \partial y^2}$.
化简函数：$z = x \ln x + x \ln y$
$\frac{\partial z}{\partial x} = \ln x + 1 + \ln y = \ln(xy) + 1 $$ \frac{\partial^2 z}{\partial x^2} = \frac{1}{x} \implies \frac{\partial^3 z}{\partial x^2 \partial y} = \frac{\partial}{\partial y}\left(\frac{1}{x}\right) = 0 $
$\frac{\partial^2 z}{\partial x \partial y} = \frac{1}{y} \implies \frac{\partial^3 z}{\partial x \partial y^2} = \frac{\partial}{\partial y}\left(\frac{1}{y} \right) = -\frac{1}{y^2}$
11. 验证：
(1) $z = \ln(\sqrt{x} + \sqrt{y})$ 满足 $x\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y} = \frac{1}{2}$
$\frac{\partial z}{\partial x} = \frac{1}{\sqrt{x}+\sqrt{y}} \cdot \frac{1}{2\sqrt{x}} $$ \frac{\partial z}{\partial y} = \frac{1}{\sqrt{x}+\sqrt{y}} \cdot \frac {1}{2\sqrt{y}} $
$$x\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y} = \frac{x}{2\sqrt{x}(\sqrt{x}+\sqrt{y})} + \frac{y}{2\sqrt{y}(\sqrt{x}+\sqrt{y})} = \frac{\sqrt {x}+\sqrt{y}}{2(\sqrt{x}+\sqrt{y})} = \frac{1}{2}$$验证成立。
(2) $z = x^m y^n$，其中 $x>0, y>0, m+n=1$，满足 $\frac{\partial^2 z}{\partial x^2}\frac{\partial^2 z}{\partial y^2} - \left(\frac{\partial^2 z}{\partial x \partial y}\right)^2 = 0$.
$\frac{\partial z}{\partial x} = mx^{m-1}y^n, \quad \frac{\partial z}{\partial y} = nx^my^{n-1} $$ \frac{\partial^2 z}{\partial x^2} = m(m-1)x^{m-2}y^n $
$\frac{\partial^2 z}{\partial y^2} = n(n-1)x^my^{n-2} $$ \frac{\partial^2 z}{\partial x \partial y} = mnx^{m-1}y^{n-1} $
代入等式左边：
$$\left[ m(m-1)x^{m-2}y^n \right] \left[ n(n-1)x^my^{n-2} \right] - \left( mnx^{m-1}y^{n-1} \right)^2$$$$= mn(m-1)(n-1)x^{2m-2}y^{2n-2} - m^2n^2x^{2m-2}y^{2n-2}$$$$= mn x^{2m-2}y^{2n-2} [ (m-1)(n-1) - mn ]$$因为 $(m-1)(n-1) - mn = mn - m - n + 1 - mn = 1 - (m+n)$。
由已知条件 $m+n=1$，故 $1 - (m+n) = 0$。代入得式子的值为 0，验证成立。
12. 证明函数 $f(x,y) = \begin{cases} \frac{x^2 y}{x^2+y^2}, & x^2+y^2 \ne 0 \\ 0, & x^2+y^2 = 0 \end{cases}$ 在点 $(0,0)$ 处的偏导数存在，但在点 $(0,0)$ 处不可微.
证明：
(1) 偏导数存在：
$f_x(0,0) = \lim_{\Delta x \to 0} \frac{f(\Delta x, 0) - f(0,0)}{\Delta x} = \lim_{\Delta x \to 0} \frac{0 - 0}{\Delta x} = 0$
$f_y(0,0) = \lim_{\Delta y \to 0} \frac{f(0, \Delta y) - f(0,0)}{\Delta y} = \lim_{\Delta y \to 0} \frac{0 - 0}{\Delta y} = 0$
两偏导数均存在且为 0。
(2) 不可微：
如果 $f(x,y)$ 在 $(0,0)$ 可微，那么 $\Delta z = A\Delta x + B\Delta y + o(\rho)$，这里 $A=f_x(0,0)=0, B=f_y(0,0)=0$，$\rho = \sqrt{\Delta x^2 + \Delta y^2}$。
应当有 $\lim_{\rho \to 0} \frac{\Delta z - (A\Delta x + B\Delta y)}{\rho} = \lim_{\rho \to 0} \frac{\Delta x^2 \Delta y}{(\Delta x^2 + \Delta y^2)^{3/2}} = 0$。
让动点沿直线 $\Delta y = \Delta x$ 趋向原点，有：
$$\lim_{\Delta x \to 0} \frac{\Delta x^3}{(\Delta x^2 + \Delta x^2)^{3/2}} = \lim_{\Delta x \to 0} \frac{\Delta x^3}{2\sqrt{2}|\Delta x|^3}$$当 $\Delta x > 0$ 时，极限为 $\frac{1}{2\sqrt{2}} \ne 0$。
因此极限不为 0，函数在点 $(0,0)$ 处不可微。
13. 设 $z = x^2 - 2bxy + cy^2$，且 $\left.\frac{\partial z}{\partial x}\right|_{(2,1)} = 6, \left.\frac{\partial z}{\partial y}\right|_{(2,1)} = 0$，求 $\frac {\partial^2 z}{\partial y^2}$.
$\frac{\partial z}{\partial x} = 2x - 2by \implies 2(2) - 2b(1) = 6 \implies 4 - 2b = 6 \implies b = -1$
$\frac{\partial z}{\partial y} = -2bx + 2cy \implies -2(-1)(2) + 2c(1) = 0 \implies 4 + 2c = 0 \implies c = -2$
将求得的系数代回对 $y$ 的一阶偏导：
$\frac{\partial z}{\partial y} = 2x - 4y$
$$\frac{\partial^2 z}{\partial y^2} = \frac{\partial}{\partial y}(2x - 4y) = -4$$14. 设 $z = \arctan\left(\frac{y}{x}\right)$，求 $\frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2}$.
$\frac{\partial z}{\partial x} = \frac{1}{1+(y/x)^2} \cdot \left(-\frac{y}{x^2}\right) = \frac{-y}{x^2+y^2} $$ \frac{\partial z}{\partial y} = \frac{1}{1+(y/x)^2} \cdot \left(\frac{1}{x}\right) = \frac{x}{x^2+y^2} $
$\frac{\partial^2 z}{\partial x^2} = \frac{\partial}{\partial x}\left(\frac{-y}{x^2+y^2}\right) = \frac{0 - (-y)(2x)}{(x^2+y^2)^2} = \frac{2xy}{(x^2+y^2)^2} $$ \frac{\partial^2 z}{\partial y^2} = \frac{\partial}{\partial y}\left(\frac{x}{x^2+y^2}\right) = \frac{0 - (x)(2y)}{(x^2+y^2)^2} = \frac{-2xy}{(x^2+y^2)^2} $
$$\frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2} = \frac{2xy}{(x^2+y^2)^2} - \frac{2xy}{(x^2+y^2)^2} = 0$$三、考研真题
15. (2016206) 已知函数 $f(x,y) = \frac{\text{e}^x}{x-y}$，则 ____.
分别求偏导数：
$f_x = \frac{\text{e}^x(x-y) - \text{e}^x \cdot 1}{(x-y)^2} = \frac{\text{e}^x(x-y-1)}{(x-y)^2} $$ f_y = \frac{0 - \text{e}^x(-1)}{(x-y)^2} = \frac{\text{e}^x} {(x-y)^2} $
计算两者之和：
$f_x + f_y = \frac{\text{e}^x(x-y-1) + \text{e}^x}{(x-y)^2} = \frac{\text{e}^x(x-y)}{(x-y)^2} = \frac{\text{e}^x}{x-y} = f(x,y)$
答案选 D. $f_x + f_y = f$
16. (2020211) 已知 $z = \arctan[xy + \sin(x+y)]$，则 $\left.\text{d}z\right|_{(0,\pi)} =$ ____.
令 $u = xy + \sin(x+y)$。在 $(0,\pi)$ 处，$u = 0 + \sin(\pi) = 0$。
全微分为 $\text{d}z = \frac{1}{1+u^2} \text{d}u$。
$\text{d}u = \frac{\partial u}{\partial x}\text{d}x + \frac{\partial u}{\partial y}\text{d}y = [y+\cos(x+y)]\text{d}x + [x+\cos(x+y)]\text{d}y$
将 $x=0, y=\pi$ 代入：
$\text{d}u = [\pi + \cos(\pi)]\text{d}x + [0 + \cos(\pi)]\text{d}y = (\pi - 1)\text{d}x - \text{d}y$
因为在 $(0,\pi)$ 处 $u=0$，所以 $\frac{1}{1+u^2} = 1$。
答案填写： $(\pi - 1)\text{d}x - \text{d}y$
17. (2020112) 设函数 $f(x,y) = \int_0^{xy} \text{e}^{xt^2} \text{d}t$，则 $\left.\frac{\partial^2 f}{\partial x \partial y}\right|_{(1,1)} =$ ____.
根据含参变量积分与变上限积分的求导法则，先对 $y$ 求偏导：
$\frac{\partial f}{\partial y} = \text{e}^{x(xy)^2} \cdot \frac{\partial (xy)}{\partial y} = x\text{e}^{x^3 y^2}$
再将此结果对 $x$ 求偏导：
$ \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial}{\partial x} \left( x\text{e}^{x^3 y^2} \right) = 1 \cdot \text{e}^{x^3 y^2} + x \cdot \text{e}^{x^3 y^2} \cdot (3x^2 y^2) = (1 + 3x^3 y^2)\text{e}^{x^3 y^2} $
将 $(1,1)$ 代入上式：
$\left.\frac{\partial^2 f}{\partial x \partial y}\right|_{(1,1)} = (1 + 3(1^3)(1^2))\text{e}^{1^3 1^2} = (1+3)\text{e}^1 = 4\text{e}$
答案填写： $4\text{e}$