📝 题目
5.2.5 求下列复合函数的偏导数:
(1) $u = f\left( \frac{xz}{y}\right)$ ; (2) $u = f\left( {x + y,z}\right)$ ;
(3) $u = f\left( {x,{xy},{xyz}}\right)$ ; (4) $u = f\left( {x + y + z,{x}^{2} + {y}^{2} + {z}^{2}}\right)$ ;
(5) $u = f\left( {\frac{x}{y},\frac{y}{z}}\right)$ ; (6) $u = f\left( {{x}^{2} + {y}^{2},{x}^{2} - {y}^{2},{2xy}}\right)$ .
💡 答案与解析
### 题目 5.2.4
设 $$ f(x,y) = \begin{cases} \frac{\sin(xy)}{x}, & x \neq 0, \\ y, & x = 0. \end{cases} $$ 证明 $f(x,y)$ 在平面上可微。
**证明：**
1. **连续性** 当 $x \neq 0$ 时，函数由初等函数定义，连续。 当 $x=0$ 时，需验证 $\displaystyle \lim_{(x,y)\to(0,y_0)} f(x,y) = y_0$。 若 $x \to 0$，则 $$ \frac{\sin(xy)}{x} = y \cdot \frac{\sin(xy)}{xy} \to y_0 \cdot 1 = y_0, $$ 所以处处连续。
2. **偏导数存在性** - 当 $x \neq 0$ 时， $$ f_x = \frac{xy\cos(xy) - \sin(xy)}{x^2}, \quad f_y = \cos(xy). $$ - 在 $(0,y_0)$ 处： $$ f_x(0,y_0) = \lim_{h\to 0} \frac{f(h,y_0)-f(0,y_0)}{h} = \lim_{h\to 0} \frac{\frac{\sin(hy_0)}{h} - y_0}{h} = \lim_{h\to 0} \frac{\sin(hy_0) - hy_0}{h^2}. $$ 利用 $\sin t = t - \frac{t^3}{6} + o(t^3)$，得分子为 $-\frac{(hy_0)^3}{6} + o(h^3)$，除以 $h^2$ 后极限为 0。所以 $f_x(0,y_0)=0$。 $$ f_y(0,y_0) = \lim_{k\to 0} \frac{f(0,y_0+k)-f(0,y_0)}{k} = \lim_{k\to 0} \frac{(y_0+k)-y_0}{k} = 1. $$
3. **可微性** 在任意点 $(x_0,y_0)$，考虑增量 $$ \Delta f = f(x_0+h, y_0+k) - f(x_0,y_0). $$ 若 $x_0 \neq 0$，由 $f_x,f_y$ 连续（易验证）知可微。 若 $x_0=0$，则 $$ \Delta f = f(h, y_0+k) - y_0. $$ 当 $h=0$ 时，$\Delta f = k$，线性部分为 $0\cdot h + 1\cdot k$，余项为0。 当 $h\neq 0$ 时， $$ \Delta f = \frac{\sin(h(y_0+k))}{h} - y_0. $$ 展开 $\sin(h(y_0+k)) = h(y_0+k) - \frac{h^3(y_0+k)^3}{6} + \cdots$，得 $$ \Delta f = (y_0+k) - y_0 - \frac{h^2(y_0+k)^3}{6} + \cdots = k - \frac{h^2(y_0+k)^3}{6} + \cdots. $$ 于是 $$ \Delta f - (0\cdot h + 1\cdot k) = -\frac{h^2(y_0+k)^3}{6} + \cdots. $$ 除以 $\sqrt{h^2+k^2}$，当 $(h,k)\to(0,0)$ 时趋于0，因此可微。
综上，$f$ 在平面上处处可微。 **证毕。**
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### 题目 5.2.5
求下列复合函数的偏导数（设 $f$ 可微）：
**(1)** $u = f\left(\frac{xz}{y}\right)$ 令 $t = \frac{xz}{y}$，则 $$ \frac{\partial u}{\partial x} = f'(t) \cdot \frac{z}{y},\quad \frac{\partial u}{\partial y} = f'(t) \cdot \left(-\frac{xz}{y^2}\right),\quad \frac{\partial u}{\partial z} = f'(t) \cdot \frac{x}{y}. $$
**(2)** $u = f(x+y, z)$ 令 $p = x+y, q = z$，则 $$ \frac{\partial u}{\partial x} = f_p,\quad \frac{\partial u}{\partial y} = f_p,\quad \frac{\partial u}{\partial z} = f_q. $$
**(3)** $u = f(x, xy, xyz)$ 令 $p=x, q=xy, r=xyz$，则 $$ \frac{\partial u}{\partial x} = f_p + y f_q + yz f_r, $$ $$ \frac{\partial u}{\partial y} = x f_q + xz f_r, $$ $$ \frac{\partial u}{\partial z} = xy f_r. $$
**(4)** $u = f(x+y+z, x^2+y^2+z^2)$ 令 $p = x+y+z, q = x^2+y^2+z^2$，则 $$ \frac{\partial u}{\partial x} = f_p + 2x f_q,\quad \frac{\partial u}{\partial y} = f_p + 2y f_q,\quad \frac{\partial u}{\partial z} = f_p + 2z f_q. $$
**(5)** $u = f\left(\frac{x}{y}, \frac{y}{z}\right)$ 令 $p = x/y, q = y/z$，则 $$ \frac{\partial u}{\partial x} = \frac{1}{y} f_p, $$ $$ \frac{\partial u}{\partial y} = -\frac{x}{y^2} f_p + \frac{1}{z} f_q, $$ $$ \frac{\partial u}{\partial z} = -\frac{y}{z^2} f_q. $$
**(6)** $u = f(x^2+y^2, x^2-y^2, 2xy)$ 令 $p = x^2+y^2, q = x^2-y^2, r = 2xy$，则 $$ \frac{\partial u}{\partial x} = 2x f_p + 2x f_q + 2y f_r, $$ $$ \frac{\partial u}{\partial y} = 2y f_p - 2y f_q + 2x f_r. $$
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### 题目 5.2.6
设 $u = x^n f\left(\frac{y}{x}, \frac{z}{x}\right)$，$f$ 可微。证明： $$ x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} + z\frac{\partial u}{\partial z} = n u. $$
**证明：** 令 $p = y/x, q = z/x$，则 $$ u = x^n f(p,q). $$ 计算偏导： $$ \frac{\partial u}{\partial x} = n x^{n-1} f + x^n \left( f_p \cdot \left(-\frac{y}{x^2}\right) + f_q \cdot \left(-\frac{z}{x^2}\right) \right) = n x^{n-1} f - x^{n-2} (y f_p + z f_q). $$ $$ \frac{\partial u}{\partial y} = x^n f_p \cdot \frac{1}{x} = x^{n-1} f_p, $$ $$ \frac{\partial u}{\partial z} = x^{n-1} f_q. $$ 于是 $$ x u_x + y u_y + z u_z = x\left( n x^{n-1} f - x^{n-2}(y f_p + z f_q) \right) + y x^{n-1} f_p + z x^{n-1} f_q $$ $$ = n x^n f - x^{n-1}(y f_p + z f_q) + x^{n-1}(y f_p + z f_q) = n x^n f = n u. $$ **证毕。**
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### 题目 5.2.7
证明：$f(x,y,z)$ 为 $n$ 次齐次函数的充要条件是 $$ x\frac{\partial f}{\partial x} + y\frac{\partial f}{\partial y} + z\frac{\partial f}{\partial z} = n f(x,y,z). $$
**证明：** 必要性：若 $f(tx,ty,tz) = t^n f(x,y,z)$，两边对 $t$ 求导，再令 $t=1$ 即得。