已知函数y=lnx+lnlnx+lnlnlnx,求dy

Question

如题所述

Answer 1

y'=1/x+(lnx)'/lnx+(lnlnx)'/(lnlnx)
=1/x+1/(xlnx)+1/(xlnxlnlnx)
所以dy=[1/x+1/(xlnx)+1/(xlnxlnx)]dx