x+y+z∧3=cosz求dz

z=arctan(x/y),求dz【方法1】z'(x)=y/(x^2+y^2),z'(y)=-x/(x^2+y^2),dz=(ydx-xdy)/(x^2+y^2).【方法2】微分形式不变性dz=[1/(x^2+y^2)]d(x/y)[1/(

设z=x^y,求dzdz=d(x^y)=d(e^(ylnx))=e^(ylnx)*d(ylnx)=x^y*(lnxdy+dy/x)=x^y*lnx*dy+x^(y-1)*dy2种方法，可以直接用定义，也可以用微分形式不变性（就是楼上那位的做

z=3x^3y^2,求dzdz=9x^2y^2dx+6x^3ydy

z=(e^3y)+(x^2)Xsiny,求dzz关于x的偏导数时2x*siny,z关于y的偏导数时3(e^3y)+x^2*cosy所以dz=2x*sinydx+3(e^3y)+x^2*cosydy我来帮你解答，你那个x^2后面那个大x是不是

z=(2y+7)^2*ln(x^3+2)求dz/dx和dz/dyz=(2y+7)^2*ln(x^3+2)dz/dx=3x^2*(2y+7)^2/(x^3+2)dz/dy=2*(2y+7)*ln(x^3+2)

若z=e^（x^2+y^3）,求dz/dx,dz/dy同上令u=x^2+y^3dz/dx=dz/duXdu/dx=e^uX2x=2xe^（x^2+y^3）dz/dy=dz/duXdu/dy=e^uX3y=3ye^（x^2+y^3）考查公式(

设z=ln(x^z×y^x),求dzz=lnx^z+lny^x=zlnx+xlnyz=xlny/(1-lnx)先关于x求偏导,把y看做常数,再对y求偏导,把x看做常数dz=0dx+x/y(1-lnx)dy（此处省略了一些计算过程,）dz=x

F(x/z,y/z)=0,求dz过程见图中

z=x^y+y^x求dz/dx两边同时取对数,有lnz=ylnx+xlny上式两边对X求导,有1/zdz/dx=y/x+lny所以dz/dx=(y/x+lny)(x^y+y^x)z=x^y+y^xdz/dx=y*x^(y-1)+(y^x)*

z=x^y+y^x求dz/dx本题是全微分与偏微分之间的关系.点击放大,  荧屏放大再放大:两边同时取对数,有lnz=ylnx+xlny上式两边对X求导,有1/zdz/dx=y/x+lny所以dz/dx=(y/x+lny

设z=(x+y)/(x-y),求dzf(x)=z=x+y/x-ydz=fxdx+fydy=[[(x-y)-(x+y)]/(x-y)^2]dx+[[(x-y)+(x+y)]/(x-y)^2]dy=-2y/(x-y)^2dx+2x/(x-y)^

已知sinx+siny+sinz=cosx+cosy+cosz=0,求tan(x+y+z)+tanxtanytanz的值注：以下pi表示圆周率由于三角函数的周期性以及x,y,z地位的对等性,不妨设03

求函数：z^x=y^z的,dz/dx,dz/dy,这类题目有两种方法,不过严格的说是一种方法,只是理解的方向不同.且说是两种方法吧.1、分别将式子对x,y求偏导数,然后整理式子就可可以得到答案了.z^x*ln(z)+x*z^(x-1)*z[

设有方程x+y^2+z^2=2z,求dz/dxdz/dy两边同时微分：dx+2ydy+2zdz=2dzdz=1/(2-2z)dx+2y/(2-2z)dydz/dx=1/(2-2z)dz/dy=2y/(2-2z)注意：这是全微分求偏导数

z=x^2-y/x+y,而y=2x-3,求dz/dx.z＝x^2-(2x-3)/x+2x-3＝x^2+3/x+2x-5dz/dx＝2x-3/x^2+2

求函数f(z)=cosz/(z-3)在0函数f(z)=cosz/(z-3)在0cosz=cos(z-3+3)=cos3*cos(z-3)-sin3*sin(z-3)展开cos(z-3),sin(z-3)代入整理即可.

z=x^2*arctan(2y/x)求dzdz=2xarctan(2y/x)+x^2(1+(1+4y^2/x^2)^0.5)(2dy/x-2ydx/x^2)本人水平有限,见谅

函数z=In(1+x+y),求dzdz=d[ln(1+x+y)]=d(1+x+y)/(1+x+y)=(dx+dy)/(1+x+y).

z=arctan（x-y）,求dzdz=arctan(x+y)/(x-y)dx+arctan(x+y)/(x-y)dy=((x-y)/(1+(x+y)^2)+(x-1)arctan(x+y))dx/(x-y)^2+((x-y)/(1+(x+

设z=(x+y)^xy,求dzlnz=xyln(x+y)两边对x求导得z'x/z=yln(x+y)+xy/(x+y)z'x=z[yln(x+y)+xy/(x+y)]两边对y求导得z'y/z=xln(x+y)+xy/(x+y)z'x=z[xl