设z=u^2 v^2,而u=x y,v=x-y,求-搜答案-拍题作业网

由z=u²v²,其中u=x-y,v=x+y,题型：求复合函数的偏导数：z=（x-y）²（x+y）²,dz/dx=（x-y）²×2（x+y）+2（x-y

∂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′

z=f(x,u),u=xy,求z对x的二阶偏导数∂z/∂x=∂f/∂x+(∂f/∂u)(∂u/∂x)=&

(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+

v'y=2x,因此u'x=v'y=2x,积分得u=x^2+g(y),又由于u'y=-v'x,所以g'(y)=-2y,g(y)=-y^2+c,故u=x^2-y^2+c,f(z)=x^2-y^2+c+2i

最容易理解的办法,代进去有z=x+y+xy那么对x偏导数有那个偏导数=1+y

其实就是求z的导数,cost^2求导为2cost*(-sint),t^6求导是6t^5,cost*t^3求导是-sint*t^3+cost*3*t^2,综合起来就是2cost*(-sint)+6t^5

∫∫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+

见下图

dy/dx=dy/du*du/dx+dy/dv*dv/dx=v*e^(x+y)+u*y/x=ln(xy)*e^(x+y)+e^(x+y)*y/x=e^(x+y)[ln(xy)+y/x]所以dy=e^(

2(x+y),2(x-y).下次弄个难点的

dz/dx=dz/du*(du/dx)=2u*1=2udz/dy=dz/du*(du/dy)=2u*1=2u和v没关系

说明：eu应该是e的x次幂,dz/dx,dz/dy应该是偏导数.∵v=xy,u=x2-y2∴du/dx=2x,du/dy=-2y,dv/dx=y,dv/dy=x∵z=ln(e^u+v),∴dz/dx=

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)

①偏z/偏x=偏z/偏u偏u/偏x+偏z/偏v偏v/偏x=(2uv-v^2)siny+(2uv-v^2)cosy=(2x^2sinycosy-x^2(cosy)^2)siny+(2x^2sinycos

grad(u)=(∂u/∂x,∂u/∂y,∂u/∂z)=(y^2,2xy,3z^2),所以div(grad(u))=div(y^