Question

一道求偏导数的题目，画圈的第三小题？

Answer 1

方法如下图所示，请作参考，祝学习愉快：

Answer 2

u'x = (y/z)lnx + y/z + φ + x[φ'1 (-y/x^2) + φ'2 (-z/x^2) , where "1" = y/x, "2" = z/x
u'y = (x/z)lnx + xφ'1 (1/x)
u'z = (-xy/z^2)lnx + xφ'2 (1/x)