问题描述: 高数偏导数问题设u=e.那道 1个回答 分类:数学 2014-11-27 问题解答: 我来补答 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,…… 展开全文阅读