百科知识

设z=ln(x^2+y),求一阶偏导数

2012-11-21 15:43:243***
设z=ln(x^2+y), 求一阶偏导数设z=ln(x^2+y),求一阶偏导数:dz/dx=[1/(x²+y)]×2x=2x/(x²+y) dz/dy=[1/(x²?

最佳回答

  • dz/dx=[1/(x²+y)]×2x=2x/(x²+y) dz/dy=[1/(x²+y)]×1=1/(x²+y)
    2012-11-21 15:59:09
  • 很赞哦! (82)