What is the differentiation!!!!!!!

Question

f(x) = (e^x)^(e^x)

Answer 1

Given that f(x) = (e^x)^(e^x)

Derivative of f(x) = d/dx((e^x)^(e^x))

                           = d/dx ( e^(log(e^x)^(e^x)) )                [Express (e^x)^(e^x) as power of e]

                           = d/dx ( e^( e^xloge^x ) )

                           = e^( e^xloge^x ) d/dx ( e^xloge^x )     [Using chain rule]

                           = (e^x)^(e^x) d/dx ( e^xloge^x )           [Express e^( e^xloge^x ) as power of e^x]

                           = (e^x)^(e^x) [ e^x d/dx(loge^x) + loge^x d/dx(e^x) ]  [Since d/dx(uv) = udv/dx + vdu/dx]

                           = (e^x)^(e^x) [ e^x d/dx(xloge) + (loge^x) (e^x) ]

                           = (e^x)^(e^x) [ e^x d/dx(x*1) + (loge^x) (e^x) ]          [Since log(e) = 1]

                           = (e^x)^(e^x) [ e^x d/dx(x) + e^xloge^x ]   

                           = (e^x)^(e^x) [ e^x (1) + e^xloge^x ]                           [Since d/dx(x) = 1]

                           = (e^x)^(e^x) [ e^x + e^xloge^x ]  

Therefore d/dx((e^x)^(e^x)) = (e^x)^(e^x) [ e^x + e^xloge^x ]