设z=u^2cosv^2,u=x+y,v=xy,求dz/dx,dz/dy.
高数题
人气:396 ℃ 时间:2019-10-11 12:30:48
解答
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^2
dz/dy=2*(x+y)*cos(x^2*y^2)-2*(x+y)^2*sin(x^2*y^2)*x^2*y
推荐
- 设z=u^2+v^2,且u=x+y,v=x-y,求dz/dx,dz/dy
- z=x*arctan(xy),求(dz/dx)|(1,1),(dz/dy)|(1,1)
- 设z是x,y的函数,且 xy=xf(z)+yψ(z) ,xf'(z)+yψ'(z)≠0 .证明:[x-ψ(z)]·(dz/dx)=[y-f(z)]·(dz/dy)
- 设z=u^2v^2,而u=x-y,v=x+y,求dz/dx,dz/dy
- 设z=ln(eu+v),v=xy,u=x2-y2,求dz/dx,dz/dy.
- 1.She wentto Simon’s house without ______________(敲) on the door.
- 用以下词语以四个为一组造句
- 有一堆事水果,其中苹果占45%,在放入16个梨以后,苹果就只占25% ,那么,这堆水果中有苹果多少
猜你喜欢