x=y^2+y,u=(x^2+2)^3/2,求dy/du 求详细过程

帮忙证明个高数题u=x^2*y^2/(x+y),求证 x*u(x)+y*u(y)=3u 函数u^2lnv,u=x/y,v=3x-2y y≥0,x^2 +y^2 =3,u=(y+1)/(x+3) ,求u的最值. 证明函数u(x,y)=f(y/x^2)*x^n满足x*（∂u/∂x)+2y*(∂u/∂y)=nu du(x,y)=2xycos(x^2y)dx+x^2cos(x^y)dy,求u(x,y) 已知x,y∈R*,x+y=xy,求u=x+2y最小值 四元一次方程的解法y+z+u=1z+u+x=2u+x+y=3x+y+z=4 已知函数y=[(x+1)^2]u(x)为方程y'-2y/(x+1)=(x+1)^3的通解,求u(x) u=x(z+y) z=sin(x+y) 求二阶偏导数σ2u／σxσy y=4^x-2^(x+1)+1变形,y=4^x-2^(x+1)+1y=u^2-2u+1u=2^x 全集u={(x,y)|x属于实数,y属于实数},集合m={(x,y)|y-3/x-2=1},集合n=[(x,y)|y不等于x+1,那[u(MUN）等于 x,y>0,且x+2y=1,u=(x+1/x)(y+1/(4y)最小值是: {(x,y)x+y=-1}u{(x,y)x-2y=8},求这连个集合的并集, 假设集合U={(x,y)|x∈R,y∈R },A={(x,y)|2x-y+m>0},B={(x,y)|x+y-n 设x+y+z=11求函数u=2x*x+3y*y+z*z的最小值 已知x^2+y^2=4,求u=3x+4y的值域 设z=u^2lnv,而u=x/y,v=3x-2y,求下偏导 设z=u^2lnv,而u=x/y,v=3x-2y,求下偏导