同济高数 第9章 第9-7-9题
📝 题目
9．设函数 $u(x, y, z), v(x, y, z)$ 的各个偏导数都存在且连续，证明： （1）$\nabla(c u)=c \nabla u$（其中 $c$ 为常数）； （2）$\nabla(u \pm v)=\nabla u \pm \nabla v$ ； （3）$\nabla(u v)=v \nabla u+u \nabla v$ ； （4）$\displaystyle \nabla\left(\frac{u}{v}\right)=\frac{v \nabla u-u \nabla v}{v^{2}} \quad(v \neq 0)$ ．
💡 答案解析
[AI解答]
[AI解答]
**证明**：设函数 $u(x,y,z), v(x,y,z)$ 的各个偏导数都存在且连续，则梯度算子定义为： $$ \nabla = \left( \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right) $$
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### （1）$\nabla(c u)=c \nabla u$（其中 $c$ 为常数）
由梯度定义： $$ \nabla(c u) = \left( \frac{\partial (c u)}{\partial x}, \frac{\partial (c u)}{\partial y}, \frac{\partial (c u)}{\partial z} \right) $$ 由于常数可提到偏导外： $$ = \left( c \frac{\partial u}{\partial x}, c \frac{\partial u}{\partial y}, c \frac{\partial u}{\partial z} \right) = c \left( \frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}, \frac{\partial u}{\partial z} \right) = c \nabla u $$ 证毕。
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### （2）$\nabla(u \pm v)=\nabla u \pm \nabla v$
由梯度定义： $$ \nabla(u \pm v) = \left( \frac{\partial (u \pm v)}{\partial x}, \frac{\partial (u \pm v)}{\partial y}, \frac{\partial (u \pm v)}{\partial z} \right) $$ 利用偏导的线性性质： $$ = \left( \frac{\partial u}{\partial x} \pm \frac{\partial v}{\partial x}, \frac{\partial u}{\partial y} \pm \frac{\partial v}{\partial y}, \frac{\partial u}{\partial z} \pm \frac{\partial v}{\partial z} \right) $$ $$ = \left( \frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}, \frac{\partial u}{\partial z} \right) \pm \left( \frac{\partial v}{\partial x}, \frac{\partial v}{\partial y}, \frac{\partial v}{\partial z} \right) = \nabla u \pm \nabla v $$ 证毕。
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### （3）$\nabla(u v)=v \nabla u+u \nabla v$
由梯度定义： $$ \nabla(u v) = \left( \frac{\partial (u v)}{\partial x}, \frac{\partial (u v)}{\partial y}, \frac{\partial (u v)}{\partial z} \right) $$ 利用乘积求导法则： $$ \frac{\partial (u v)}{\partial x} = v \frac{\partial u}{\partial x} + u \frac{\partial v}{\partial x} $$ 同理对 $y, z$ 成立，因此： $$ \nabla(u v) = \left( v \frac{\partial u}{\partial x} + u \frac{\partial v}{\partial x}, \; v \frac{\partial u}{\partial y} + u \frac{\partial v}{\partial y}, \; v \frac{\partial u}{\partial z} + u \frac{\partial v}{\partial z} \right) $$ $$ = v \left( \frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}, \frac{\partial u}{\partial z} \right) + u \left( \frac{\partial v}{\partial x}, \frac{\partial v}{\partial y}, \frac{\partial v}{\partial z} \right) = v \nabla u + u \nabla v $$ 证毕。
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### （4）$\nabla\left(\frac{u}{v}\right)=\frac{v \nabla u-u \nabla v}{v^{2}} \quad(v \neq 0)$
由梯度定义： $$ \nabla\left(\frac{u}{v}\right) = \left( \frac{\partial}{\partial x}\left(\frac{u}{v}\right), \frac{\partial}{\partial y}\left(\frac{u}{v}\right), \frac{\partial}{\partial z}\left(\frac{u}{v}\right) \right) $$ 利用商的求导法则： $$ \frac{\partial}{\partial x}\left(\frac{u}{v}\right) = \frac{v \frac{\partial u}{\partial x} - u \frac{\partial v}{\partial x}}{v^2} $$ 同理对 $y, z$ 成立，因此： $$ \nabla\left(\frac{u}{v}\right) = \left( \frac{v \frac{\partial u}{\partial x} - u \frac{\partial v}{\partial x}}{v^2}, \frac{v \frac{\partial u}{\partial y} - u \frac{\partial v}{\partial y}}{v^2}, \frac{v \frac{\partial u}{\partial z} - u \frac{\partial v}{\partial z}}{v^2} \right) $$ $$ = \frac{1}{v^2} \left( v \frac{\partial u}{\partial x} - u \frac{\partial v}{\partial x}, v \frac{\partial u}{\partial y} - u \frac{\partial v}{\partial y}, v \frac{\partial u}{\partial z} - u \frac{\partial v}{\partial z} \right) $$ $$ = \frac{v \left( \frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}, \frac{\partial u}{\partial z} \right) - u \left( \frac{\partial v}{\partial x}, \frac{\partial v}{\partial y}, \frac{\partial v}{\partial z} \right)}{v^2} = \frac{v \nabla u - u \nabla v}{v^2} $$ 证毕。
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**难度评级**：★☆☆☆☆ （仅需掌握梯度定义与基本求导法则，步骤直接，无复杂技巧。）