Problem
微积分每日一题11.22
微积分每日一题
大学生数学竞赛教程
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$$ { 设u=f\left( x,y,xyz \right) ,函数z=z\left( x,y \right) 由方程\int_{xy}^z{g\left( xy+z-t \right)}\mathrm{d}t=\mathrm{e}^{xyz}确定,其中f可微,g连续,求x\frac{\partial u}{\partial x}} $$

Tips: $$ { 分析:首先对变限积分换元,简化积分;再利用隐函数存在定理求解z_x,z_y;最后求u_x,u_y和题目所给结构.} $$
Answer:

$$ { 解:令u=xy+z-t,则t=xy+z-u,dt=\mathrm{d}\left( xy+z-u \right) =-\mathrm{d}u;} \\ { 积分上下限为:u_{上}=xy+z-t_{上}=xy+z-z=xy;u_{下}=xy+z-t_{下}=xy+z-xy=z;} \\ { \int_{xy}^z{g\left( xy+z-t \right)}\mathrm{d}t=\int_z^{xy}{g\left( u \right)}\left( -\mathrm{d}u \right) =\int_{xy}^z{g\left( u \right)}\mathrm{d}u=\int_{xy}^z{g\left( t \right)}\mathrm{d}t;} \\ { 令G\left( x,y,z \right) =\int_{xy}^z{g\left( t \right)}\mathrm{d}t-\mathrm{e}^{xyz},则有:\begin{cases} G_x=-yg\left( xy \right) -yz\mathrm{e}^{xyz}\\ G_y=-xg\left( xy \right) -xz\mathrm{e}^{xyz}\\ G_z=g\left( xy \right) -z\mathrm{e}^{xyz}\\ \end{cases};} \\ { 根据隐函数存在定理求解\frac{\partial z}{\partial x}和\frac{\partial z}{\partial y}:} \\ { \begin{cases} \frac{\partial z}{\partial x}=-\frac{G_x}{G_z}=-\frac{-yg\left( xy \right) -yz\mathrm{e}^{xyz}}{g\left( xy \right) -z\mathrm{e}^{xyz}}=\frac{yg\left( xy \right) +yz\mathrm{e}^{xyz}}{g\left( xy \right) -z\mathrm{e}^{xyz}}\\ \frac{\partial z}{\partial y}=-\frac{G_y}{G_z}=-\frac{-xg\left( xy \right) -xz\mathrm{e}^{xyz}}{g\left( xy \right) -z\mathrm{e}^{xyz}}=\frac{xg\left( xy \right) +xz\mathrm{e}^{xyz}}{g\left( xy \right) -z\mathrm{e}^{xyz}}\\ \end{cases};} \\ { 下面求解\frac{\partial u}{\partial x}和\frac{\partial u}{\partial y}:} \\ { \begin{cases} \frac{\partial u}{\partial x}=f_{1}^{\prime}+f_{3}^{\prime}\cdot \left( yz+\frac{\partial z}{\partial x} \right) =f_{1}^{\prime}+f_{3}^{\prime}\cdot \left[ yz+\frac{yg\left( xy \right) +yz\mathrm{e}^{xyz}}{g\left( xy \right) -z\mathrm{e}^{xyz}} \right]\\ \frac{\partial u}{\partial y}=f_{2}^{\prime}+f_{3}^{\prime}\cdot \left( xz+\frac{\partial z}{\partial y} \right) =f_{2}^{\prime}+f_{3}^{\prime}\cdot \left[ xz+\frac{xg\left( xy \right) +xz\mathrm{e}^{xyz}}{g\left( xy \right) -z\mathrm{e}^{xyz}} \right]\\ \end{cases};} \\ { 最后求解x\frac{\partial u}{\partial x}-y\frac{\partial u}{\partial y}:} \\ { x\frac{\partial u}{\partial x}-y\frac{\partial u}{\partial y}\begin{array}{l} =x\cdot \left\{ f_{1}^{\prime}+f_{3}^{\prime}\cdot \left[ yz+\frac{yg\left( xy \right) +yz\mathrm{e}^{xyz}}{g\left( xy \right) -z\mathrm{e}^{xyz}} \right] \right\} -y\cdot \left[ f_{2}^{\prime}+f_{3}^{\prime}\cdot \left[ xz+\frac{xg\left( xy \right) +xz\mathrm{e}^{xyz}}{g\left( xy \right) -z\mathrm{e}^{xyz}} \right] \right]\\ =xf_{1}^{\prime}-yf_{2}^{\prime}+f_{3}^{\prime}\left[ xyz+\frac{xyg\left( xy \right) +xyz\mathrm{e}^{xyz}}{g\left( xy \right) -z\mathrm{e}^{xyz}} \right] -f_{3}^{\prime}\left[ xyz+\frac{xyg\left( xy \right) +xyz\mathrm{e}^{xyz}}{g\left( xy \right) -z\mathrm{e}^{xyz}} \right]\\ =xf_{1}^{\prime}-yf_{2}^{\prime}.\\ \end{array}} \\ $$

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