人邮高数 第6章 第6-3-15题

[AI解答]

[AI解答]

我们逐一用隐函数求导法（公式法或直接求偏导）求解。

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### （1）$\mathrm{e}^{x} - x y z = 0$

令 $F(x,y,z) = e^x - xyz = 0$，则 $$ \frac{\partial F}{\partial x} = e^x - yz,\quad \frac{\partial F}{\partial y} = -xz,\quad \frac{\partial F}{\partial z} = -xy $$ 由隐函数定理（当 $xy \neq 0$ 时）： $$ \frac{\partial z}{\partial x} = -\frac{F_x}{F_z} = -\frac{e^x - yz}{-xy} = \frac{e^x - yz}{xy} $$ $$ \frac{\partial z}{\partial y} = -\frac{F_y}{F_z} = -\frac{-xz}{-xy} = -\frac{xz}{xy} = -\frac{z}{y} $$

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### （2）$z^{3} - 3xyz = 0$

令 $F = z^3 - 3xyz$，则 $$ F_x = -3yz,\quad F_y = -3xz,\quad F_z = 3z^2 - 3xy $$ 所以 $$ \frac{\partial z}{\partial x} = -\frac{-3yz}{3z^2 - 3xy} = \frac{yz}{z^2 - xy} $$ $$ \frac{\partial z}{\partial y} = -\frac{-3xz}{3z^2 - 3xy} = \frac{xz}{z^2 - xy} $$

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### （3）$2xz + \ln(xyz) = 0$

令 $F = 2xz + \ln(xyz)$，注意 $\ln(xyz) = \ln x + \ln y + \ln z$，则 $$ F_x = 2z + \frac{1}{x},\quad F_y = \frac{1}{y},\quad F_z = 2x + \frac{1}{z} $$ 因此 $$ \frac{\partial z}{\partial x} = -\frac{2z + \frac{1}{x}}{2x + \frac{1}{z}} = -\frac{2xz + 1}{x(2x + \frac{1}{z})} = -\frac{2xz + 1}{2x^2 + \frac{x}{z}} $$ 可化简为 $$ \frac{\partial z}{\partial x} = -\frac{z(2xz + 1)}{x(2xz + 1)} = -\frac{z}{x} $$ （注意 $2xz+1\neq0$ 时可约） 同理 $$ \frac{\partial z}{\partial y} = -\frac{1/y}{2x + 1/z} = -\frac{z}{y(2xz+1)} $$

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### （4）$\sin(x - 2y + 3z) = x + 2y - 3z$

令 $F = \sin(x - 2y + 3z) - x - 2y + 3z = 0$，则 $$ F_x = \cos(x - 2y + 3z) - 1,\quad F_y = -2\cos(x - 2y + 3z) - 2,\quad F_z = 3\cos(x - 2y + 3z) + 3 $$ 所以 $$ \frac{\partial z}{\partial x} = -\frac{\cos u - 1}{3\cos u + 3},\quad u = x-2y+3z $$ 利用 $\cos u - 1 = -2\sin^2\frac{u}{2},\; \cos u + 1 = 2\cos^2\frac{u}{2}$，得 $$ \frac{\partial z}{\partial x} = -\frac{-2\sin^2\frac{u}{2}}{3\cdot 2\cos^2\frac{u}{2}} = \frac{1}{3}\tan^2\frac{u}{2} $$ 而 $$ \frac{\partial z}{\partial y} = -\frac{-2\cos u - 2}{3\cos u + 3} = \frac{2(\cos u + 1)}{3(\cos u + 1)} = \frac{2}{3} $$

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### （5）$x^{2} + y^{2} + 2x - 2yz = e^{z}$

令 $F = x^2 + y^2 + 2x - 2yz - e^z = 0$，则 $$ F_x = 2x + 2,\quad F_y = 2y - 2z,\quad F_z = -2y - e^z $$ 所以 $$ \frac{\partial z}{\partial x} = -\frac{2x+2}{-2y - e^z} = \frac{2(x+1)}{2y + e^z} $$ $$ \frac{\partial z}{\partial y} = -\frac{2y - 2z}{-2y - e^z} = \frac{2(y - z)}{2y + e^z} $$

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### （6）$z^{3} - 3xyz = a^{3}$

令 $F = z^3 - 3xyz - a^3$，则 $$ F_x = -3yz,\quad F_y = -3xz,\quad F_z = 3z^2 - 3xy $$ 所以 $$ \frac{\partial z}{\partial x} = -\frac{-3yz}{3z^2 - 3xy} = \frac{yz}{z^2 - xy} $$ $$ \frac{\partial z}{\partial y} = -\frac{-3xz}{3z^2 - 3xy} = \frac{xz}{z^2 - xy} $$

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**难度评级**：★★☆☆☆ （均为标准隐函数求偏导，计算量小，但需注意化简细节）