z+x+y-e^(x+y+z)+2=0,x+y+z=0 顯然不滿足,即 x+y+z=0≠0.
兩邊對(duì) x 求偏導(dǎo)數(shù),注意 z 是 x,y 的函數(shù),得
z‘+1-(1+z')e^(x+y+z)=0,解得 z' = -1,
同理 z'= -1.得 dz=-dx-dy.cos y/x+ln xy求dz幫幫忙z=cos(y/x)+ln(xy)
z'= (y/x^2)sin(y/x)+1/x,
z'=(-1/x)sin(y/x)+1/y,
dz=[(y/x^2)sin(y/x)+1/x]dx+[(-1/x)sin(y/x)+1/y]dy.