题目
设z=(x+2y)x,则在点(1,0)处的全微分dz|(1,0)=______.正确答案:dx+2dy
设z=(x+2y)x,则在点(1,0)处的全微分dz|(1,0)=______.
正确答案:dx+2dy
题目解答
答案
解析: ∵dz=
,dz=
(设z=μν,μ=x+2y,ν=x)=(νμν-1.1+μνlnμ.1)dx+(νμν-1.2+μνlnμ.0)dy=[x(x+2y)x-1+(x+2y)xln(x+2y)]dx+2x(x+2y)x-1dy,∴dz|(1,0)=dx+2dy.
解析
步骤 1:确定函数z的表达式
给定函数为z=(x+2y)x,这是一个复合函数,其中x和y是自变量,z是因变量。
步骤 2:计算全微分dz
全微分dz可以表示为dz=$\dfrac {\partial z}{\partial x}dx+\dfrac {\partial z}{\partial y}dy$,其中$\dfrac {\partial z}{\partial x}$和$\dfrac {\partial z}{\partial y}$分别是z对x和y的偏导数。
步骤 3:计算偏导数$\dfrac {\partial z}{\partial x}$和$\dfrac {\partial z}{\partial y}$
首先,计算$\dfrac {\partial z}{\partial x}$,即对x求偏导数。由于z=(x+2y)x,可以将其视为z=μν,其中μ=x+2y,ν=x。根据链式法则,$\dfrac {\partial z}{\partial x}=\dfrac {\partial z}{\partial u}\cdot \dfrac {\partial u}{\partial x}+\dfrac {\partial z}{\partial v}\cdot \dfrac {\partial v}{\partial x}$。计算得$\dfrac {\partial z}{\partial x}=(νμν-1.1+μνlnμ.1)=[x(x+2y)x-1+(x+2y)xln(x+2y)]$。
然后,计算$\dfrac {\partial z}{\partial y}$,即对y求偏导数。同样地,$\dfrac {\partial z}{\partial y}=\dfrac {\partial z}{\partial u}\cdot \dfrac {\partial u}{\partial y}+\dfrac {\partial z}{\partial v}\cdot \dfrac {\partial v}{\partial y}$。计算得$\dfrac {\partial z}{\partial y}=(νμν-1.2+μνlnμ.0)=2x(x+2y)x-1$。
步骤 4:代入点(1,0)计算dz
将点(1,0)代入dz的表达式中,得到dz|(1,0)=[1(1+2*0)1-1+(1+2*0)1ln(1+2*0)]dx+2*1(1+2*0)1-1dy=dx+2dy。
给定函数为z=(x+2y)x,这是一个复合函数,其中x和y是自变量,z是因变量。
步骤 2:计算全微分dz
全微分dz可以表示为dz=$\dfrac {\partial z}{\partial x}dx+\dfrac {\partial z}{\partial y}dy$,其中$\dfrac {\partial z}{\partial x}$和$\dfrac {\partial z}{\partial y}$分别是z对x和y的偏导数。
步骤 3:计算偏导数$\dfrac {\partial z}{\partial x}$和$\dfrac {\partial z}{\partial y}$
首先,计算$\dfrac {\partial z}{\partial x}$,即对x求偏导数。由于z=(x+2y)x,可以将其视为z=μν,其中μ=x+2y,ν=x。根据链式法则,$\dfrac {\partial z}{\partial x}=\dfrac {\partial z}{\partial u}\cdot \dfrac {\partial u}{\partial x}+\dfrac {\partial z}{\partial v}\cdot \dfrac {\partial v}{\partial x}$。计算得$\dfrac {\partial z}{\partial x}=(νμν-1.1+μνlnμ.1)=[x(x+2y)x-1+(x+2y)xln(x+2y)]$。
然后,计算$\dfrac {\partial z}{\partial y}$,即对y求偏导数。同样地,$\dfrac {\partial z}{\partial y}=\dfrac {\partial z}{\partial u}\cdot \dfrac {\partial u}{\partial y}+\dfrac {\partial z}{\partial v}\cdot \dfrac {\partial v}{\partial y}$。计算得$\dfrac {\partial z}{\partial y}=(νμν-1.2+μνlnμ.0)=2x(x+2y)x-1$。
步骤 4:代入点(1,0)计算dz
将点(1,0)代入dz的表达式中,得到dz|(1,0)=[1(1+2*0)1-1+(1+2*0)1ln(1+2*0)]dx+2*1(1+2*0)1-1dy=dx+2dy。