设z=ln（xy）, 则dz=（）.
举一反三
- 设z=f(x+y,xy),则dz=
- 设z=xy，则dz=______ A: yxy-1dx+xylnxdy B: xy-1dx+ydy C: xy(dx+dy) D: xy(xdx+ydy)
- 函数\(z = {\left( {xy} \right)^x}\)的全微分为 A: \(dz = \left( { { {\left( {xy} \right)}^x} + \ln xy} \right)dx + x{\left( {xy} \right)^x}dy\) B: \(dz = \left( { { {\left( {xy} \right)}^x} + \ln xy} \right)dx + { { x { { \left( {xy} \right)}^x}} \over y}dy\) C: \(dz = {\left( {xy} \right)^x}\ln xydx + { { x { { \left( {xy} \right)}^x}} \over y}dy\) D: \(dz = {\left( {xy} \right)^x}\left( {1 + \ln xy} \right)dx + { { x { { \left( {xy} \right)}^x}} \over y}dy\)
- 设方程ez=1+xz+x2+y2确定隐函数z=z(x，y)，求dz与z"xy。
- 设z=xcosy，则dz=（ ).