Z=arctan(y/x)
zx=1/(1+(y/x)²) *（-y/x²）=-y/(x²+y²)|（1,1）=-1/2
zy=1/(1+(y/x)²) *（1/x）=x/(x²+y²)|（1,1）=1/2
dz(1,1)=-1/2dx+1/2dy
设Z=arctan(y/x)的点m（1,1）处的全微分dz(1,1)为?
Z=arctan(y/x)
zx=1/(1+(y/x)²) *（-y/x²）=-y/(x²+y²)|（1,1）=-1/2
zy=1/(1+(y/x)²) *（1/x）=x/(x²+y²)|（1,1）=1/2
dz(1,1)=-1/2dx+1/2dy