u=xy,v=u求z关于x,y的二阶偏到-搜答案-智慧作业帮

这道题运用链式法则,先求出对y偏导,然后求对x偏导,因为中间变量u,v都含有x,那么他们的二元函数f(u,v)的偏导f1,f2也是含有x的,所以对(f1+xf2)对x求偏导就是最后的结果,这里注意f1

令u=x/y,v=y/x,偏导z/x=fu(u,v)du/dx+fv(u,v)dv/dx=fu(u,v)1/y-fv(u,v)y/x^2偏导z/y=fu(u,v)du/dy+fv(u,v)dv/dy=

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∂z/∂x=(∂f(u,v)/∂u)*(∂u/∂x)+(∂f(u,v)/∂v)*(∂v/&#

偏z/偏x=(偏z/偏f)*f'x=偏z/偏f*1=偏z/偏f;偏z/偏u=(偏z/偏f)*(偏f/偏u)+偏g/偏u+偏h/偏u.

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

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

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

symsuv;d=[-5:0.5:5];[uv]=meshgrid(d);x=u.*sin(v),y=u.*cos(v),z=u;surf(x,y,z)

见下图

z=f(u,v),u=xy,v=x^2-y^2du/dx=y,du/dy=xdv/dx=2x,dv/dy=-2ydz/dx=dz/du*du/dx+dz/dv*dv/dx=df/du*y+df/dv*

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^(

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

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)

∂z／∂x=∂z／∂u*du/dx+∂z／∂v*dv/dx=1/(u^2+v)*2u+1/(u^2+v)*2xy∂z