u=xy-4(x y) 10,则u有
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x2+xy+y2=(x+y)2-xy=2,所以(x+y)2=2+xy.2|xy|+xy≤x2+xy+y2=2,所以0≤xy≤2/3.或者-2≤xy≤0u=x2-xy+y2=(x+y)2-3xy=2-2
先对x求偏导u'x=f'(x,xy,xyz)+yf'(x,xy,xyz)+yzf'(x,xy,xyz)所以u'xy=yf''(x,xy,xyz)+xzf''(x,xy,xyz)+f''(x,xy,xy
∂z/∂x=(∂f(u,v)/∂u)*(∂u/∂x)+(∂f(u,v)/∂v)*(∂v/
z=(x+y)^2*cos(x^2*y^2)dz/dx=2*(x+y)*cos(x^2*y^2)-2*(x+y)^2*sin(x^2*y^2)*x*y^2dz/dy=2*(x+y)*cos(x^2*y
∵z=f(x,xy),令u=x,v=xy∴∂z∂x=f′1+yf′2∴∂2z∂x∂y=∂∂y(f′1+yf′2)=∂f′1∂y+∂∂y(yf′2)═(∂f′1∂u∂u∂y+∂f′1∂v∂v∂y)+f′
若f(u,v,w)=(u-v)^w+w^(u+v)f(x+y,x-y,xy)=[(x+y)-(x-y)]^xy+(xy)^[(x+y)+(x-y)]=(2y)^xy+(xy)^2x
x^2+xy+y^2=2(x-y)^2+3xy=2(x-y)^2=2-3xy≥0xy≤2/3x^2+xy+y^2=2(x+y)^2-xy=2(x-y)^2=2+xy≥0xy≥-2所以-2≤xy≤2/3
U=X平方+Y平方+4X-2Y=(x+2)平方+(Y-1)平方-5,但因为4X+3Y≥0,不能同时使X=-2,Y=1,所以要另找途径,4X+3Y≥0=>Y≥(-4/3X)所以U=U=X平方+Y平方+4
我是一名高中生,也没学过什么大学课本,但我可以帮你解决这个问题,导数是什么,是k,k是什么.是(y1-y2)÷(x1-x2).那么对于一个复合函数.(z1-z2)÷(y1-y2)的值乘以(y1-y2)
∂u/∂x=[∂u/∂(xy)][d(xy)/dx]+[∂u/∂(x/y)][d(x/y)/dx]=yf₁'+(1/
x、y自变量,将式子对x偏导u²+v²-x²-y=0,对x求导2uu'+2vv'-2x=0uu'+vv'-x=0(1)-u+v-xy+1=0-u'+v'-y=0(2)联立
(x+y)^2=1+3xy(x-y)^2=1-xyu=(x+y)(x-y)|u|=√(x+y)^2√(x-y)^2=√(1+3xy)√(1-xy)=√[-3(t-1/3)^2+2/3]≤√6/3故-√
u=ln(xy+z)du=d[ln(xy+z)]/dx*dx+d[ln(xy+z)]/dy*dy+d[ln(xy+z)]/dz*dz=y/(xy+z)*dx+x/(xy+z)*dy+1/(xy+z)*
∫∫f(u,v)dudv是一个数,记为A,则f(x,y)=xy+A,两边在D上作二重积分,得∫∫f(x,y)dxdy=∫∫xydxdy+A∫∫dxdy即A=∫∫xydxdy+AσA=∫xdx∫ydy+
坐标系内一以原点为一顶点的矩形面积和对角线平方的比值.再问:可以化简么?再答:经过原点的对角线所在的直线的倾斜角其二倍余角的正弦值的二分之一。手机太烂,拍不了图给你详细论证。。。
grad(u)=(∂u/∂x,∂u/∂y,∂u/∂z)=(y^2,2xy,3z^2),所以div(grad(u))=div(y^
由x>=0,y>=0,2x+y=6可知0=
dU=xdy+ydx,可得dU/dy=x,dU/dx=y,dy/dx=(dU/dx)/(dU/dy)=y/x再问:��Ĵ���ȷ��������ǵ�һ�������ף��ɷ����һ�£�3Q再答:�
∵x>0,y>0,且x+y=xy∴y=xx−1> o,∴x>1∴u=x+4y=x+4•xx−1=(x-1)+(4x−1)+5≥9故答案为:[9,+∞)