设z=arctan(xy),而y=e^x,求dz dx-搜答案-拍题作业网

第一个：z=x^xy=e^[ln(x^xy)]=e^(xylnx)令u=xy*lnx,则z=e^u∂z/∂x=(x^u)'•u'=(e^u)•(xyln

设u=xy,v=lnx+g(xy),则x(∂z/∂x)-y(∂z/∂y)=∂f/∂v.原因如下：dz=(∂f/

e^y-e^x=xy两边求导,得e^y*y'-e^x=y+xy'(e^y-x)y'=(e^x+y)所以y'=(e^x+y)/(e^y-x)x=0时,e^y-e^0=0,则e^y=1,则y=0所以y'(

dz/dx=arctan(xy)+xy/[1+(xy)^2](dz/dx)|(1,1)=π/4+1/2(dz/dy)|(1,1)=x^2/[1+(xy)^2]=1/2

e^z-z+xy^3=0偏z/偏x：z'e^z-z'+y^3=0y^3=z'(1-e^z)z'=y^3/(1-e^z)偏z/偏y:z'e^z-z'+3xy^2=0z'=3xy^2/(1-e^z)偏z/

∂z/∂x=(∂f(u,v)/∂u)*(∂u/∂x)+(∂f(u,v)/∂v)*(∂v/&#

1.az/ax=1/2*1/√ln(xy)*1/(xy)*y=1/(2x√ln(xy))同理：az/ay=1/(2y√ln(xy))2.au/am=1/(1+(m^2n)^2)*n*2m=2mn/(1

两端对x求偏导得：-ye^(-xy)-2(z/x)+(z/x)e^z=0,所以,z/x=ye^(-xy)/(e^z-2)两端对y求偏导得：-xe^(-xy)-2(z/y)+(z/y)e^z=0,所以,

z=arctan(x*e^x)z'={1/[1+(x*e^x)^2]}*(x*e^x)'(x*e^x)'=x'*e^x+x*(e^x)'=e^x+x*e^x=(x+1)*e^x所以dz/dx=(x+1

(z对x的偏导)=y+F(u)+x[F'(u)(-y/x^2)](z对y的偏导)=x+F'(u)/x代入,左边=[xy+xF(u)-yF'(u)]+[xy+yF'(u)]=xy+xF(u)+xy=z+

点击放大,右键查看图片可以进一步放大：

z'(x)=1/[1+(x^y)]*1/2√(x^y)*yx^(y-1)=yx^(y-1)/{2√(x^y)[1+(x^y)]}z'(y)=1/[1+(x^y)]*1/2√(x^y)*lnx*x^y=

dz=[yIn(xy)+y]dx+[xIn(xy)+x]dy分开求导

dz/dx是z对x的偏导,这样把u,v都带入的话直接球偏导就好了dz/dx=y*e^(xy)*sin(x+y)+e^(xy)*cos(x+y)同理也可得到dz/dy=x*e^(xy)*sin(x+y)

y=4arctanxy'=4/(1+x^2)所以y'(1)=4/(1+1^2)=2

定义域x≠0再问：答案是x≠0,y/x的绝对值

求函数偏导：z=arctan(x-y)^z因为z=arctan(x-y)^z,所以(x-y)^z=tanz；两边取对数得zln(x-y)=ln(tanz)作函数F(x,y,z)=zln(x-y)-ln

dz=1/y/(1+x^2/y^2)*dx-x/y^2/(1+x^2/y^2)*dy

设Z=x²+2xy,求dz

z=x^2+2xy两边同时求导数,得到：dz=2xdx+2ydx+2xdy即：dz=2(x+y)dx+2xdy.