We have,
(xx)x
Let,
y=(xx)x=xxx
Taking log both side and we get,
logy=log(xxx)
logy=xxlogx
Again taking log both side and we get,
loglogy=log(xxlogx)
⇒loglogy=logxx+loglogx
⇒loglogy=xlogx+loglogx
On differentiation and we get,
1logy1ydydx=xddxlogx+logxdxdx+1xlogx
⇒1ylogydydx=x×1x+logx+1xlogx
⇒dydx=ylogy(1+logx+1xlogx)
⇒dydx=xxxlogxx(1+logx+1xlogx)
Hence, this is the answer.