z=ln(x+y^2)求偏导数,记为（不规范啊）dz/dx=1/(x+y^2),dz/dy=1/(x+y^2)*2y=2y/(x+y^2),dz=(dz/dx)dx+(dz/dy)dy=1/(x+y^2)dx+2y/(x+y^2)dy当x=1,y=1时,即dz|(下1,1)=1/(1+1)dx+2*1/(1+1)dy=(1/2)dx+dy...