z=f(yx^2,xy^2)的偏导数
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/21 17:31:13
xy+xz=8-x²yx+yz=12-y²zy+zx=-4-z²x(x+y+z)=8y(x+y+z)=12z(x+y+z)=-4(x+y+z)²=8+12-4=
∵(y^2+xy^2)dx+(x^2-yx^2)dy=0==>y²(1+x)dx+x²(1-y)dy=0==>[(y-1)/y²]dy=[(1+x)/x²]dx
由于x-y=a,z-y=10得x-z=a-10并且由x²+y²+z²-xy-yx-zx=1/2[(x-y)²+(y-z)²+(z-x)²]=
e^(-xy)-2z+e^z=0-ye^(-xy)-2z'(x)+e^zz'(x)=0z'(x)=ye^(-xy)/(e^z-2)-xe^(-xy)-2z'(y)+e^zz'(y)=0z'(y)=xe
x=6-3y &nbs
令u=xy,v=x+yz=f(u,v)az/ax=y(fu)+(fv)a^2z/axay=a(az/ax)/ay=a(y(fu)+(fv))/ay=(fu)+y(a(fu)/ay)+a(fv)/ay=
∵1-8x≥0,8x-1≥0,∴x=18,y=12,∴代数式xy+yx+2-xy+yx−2=14+4+2-14+4−2=52-32=1.故选:B.
|z-2|=3,即(x-2)2+y2=3就是以(2,0)为圆心以3为半径的圆,yx的几何意义点与原点连线的斜率,易得yx的最大值是:3故答案为:3.
=xy-3xy+2xy-xy=-xy
题目有问题,yx/(y+x)=4/3应该是yz/(y+z)=4/3xy/(x+y)=-2(x+y)/(xy)=-1/21/x+1/y=-1/2(1)yz/(y+z)=4/3(y+z)/(yz)=3/4
你只要X看成是是常数求导就行了,答案就不给你了,自己动手丰衣足食
1.z'x=3x²y²z'y=2x³y2.z'x=4x³z'y=3y³3.z'x=ye^(xy)+2xyz'y=xe^(xy)+x²4.u'
∵x-y=4xy,∴2x+3xy-2yx-2xy-y=2(x-y)+3xyx-y-2xy=8xy+3xy4xy-2xy=112.故答案为:112.
根据题意,2x2-3xy+y2=0,且xy≠0,故有(yx)2−3yx+2=0,即(yx−1)(yx−2)=0,即得yx=1或2,故xy=1或12,所以xy+yx=2或212.故选A.
3xy-3xy-xy+2yx=-xy+2xy=xy
XY=XZ+YX?那么也就是XY=X(Z+Y)咯,Y=Z+Y?无法证明的.题抄错啦~`
dy/dx=(1+y^2)/[xy(1+x^2)]y/(1+y^2)dy=dx/[x(1+x^2)]2y/(1+y^2)dy=2xdx[x^2(1+x^2)]d(y^2)/(1+y^2)=d(x^2)
-186;由x:y:z=1:2:3所以y=2xz=3x代入xy+xz+yz=66解出xyz再代入满意请采纳
即(10x+y)*(10y+x)=2268101xy+10x²+10y²=2268因为后面的10x²+10y²只可能是整十的数,所以2268中的个位8要靠101
(xyz²+4yx-1)+(-3xy+z²xy-3)-(2xyz²+xy)=xyz²+4yx-1-3xy+z²xy-3-2xyz²-xy=-