Suppose that, #y=x^(x^x)#.
#:. lny=(x^x)lnx#.
#:. ln(lny)=ln{(x^x)lnx}=ln(x^x)+ln(lnx)#,
#i.e., ln(lny)=xlnx+ln(lnx)#.
Diff.ing both sides w.r.t. #x#,
#d/dx{ln(lny)}=d/dx{xlnx+ln(lnx)}..................(star)#.
Here, by the Chain Rule,
#d/dx{ln(lny)}=1/lny*d/dx{lny}=1/lny*d/dy{lny}*dy/dx#.
#;. d/dx{ln(lny)}=1/lny*1/y*dy/dx=1/(ylny)*dy/dx#.
Utilising this in #(star)#, we get,
#1/(ylny)*dy/dx=d/dx{xlnx}+d/dx{ln(lnx)}#,
#=[x*d/dx{lnx}+(lnx)*d/dx{x}]+1/lnx*d/dx{lnx}#,
#=[x*1/x+(lnx)*1]+1/lnx*1/x#,
# rArr 1/(ylny)*dy/dx=1+lnx+1/(xlnx)#.
#:. dy/dx=(ylny){1+lnx+1/(xlnx)}#.
Here, sub.ing #y=x^(x^x), lny=x^xlnx#, we get,
# dy/dx=x^(x^x)*(x^xlnx){1+lnx+1/(xlnx)}#.