高数偏导数问题设u=e.那道

首先,　　　dz = sinydx+xcosydy,则　　　du = [e^(x²+y²+z²)](2xdx+2ydy+2zdz)　= 2[e^(x²+y²+z²)][xdx+ydy+z(sinydx+xcosydy)]　= 2[e^(x²+y²+z²)][(x+zsiny)dx+(y+xzcosy)dy],可知　　　Du/Dx = 2[e^(x²+y²+z²)](x+zsiny),　　　Du/Dy = 2[e^(x²+y²+z²)](y+xzcosy),于是,……