#d/dx (x^(e^(x^2)))#?

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Answer 1 · 2017-09-16T08:48:08

#dy/dx = x^(e^(x^2)-1).e^(x^2)(2x^2lnx+1)#

#y = x^(e^(x^2))#

#ln y = e^(x^2)lnx#

Apply implicit diferentiation

#1/ydy/dx = d/dx(e^(x^2)lnx)#

#= e^(x^2).d/dx lnx + d/dx e^(x^2) lnx# [Product rule]

#= e^(x^2)1/x + e^(x^2) . 2x.lnx# [Standard differential and Chain rule]

#= e^(x^2)/x (2x^2lnx +1)#

#:. dy/dx = x^(e^(x^2)) xx e^(x^2)/x (2x^2lnx +1)# [Replace #y#]

#= x^(e^(x^2))/x e^(x^2) (2x^2lnx +1)#

#=x^(e^(x^2)-1).e^(x^2)(2x^2lnx+1)#