What is the derivative of (x^x)^x?

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Answer 1 · 2018-06-12T11:45:08

#x(2 log x+1)(x^x)^x#

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

Then

#log y = x^2 log x implies#

#1/y dy/dx = 2x log x+x^2times 1/x= x(2 log x+1)#

#dy/dx = x(2 log x+1)(x^x)^x#

Answer 2 · 2018-06-12T11:49:33

#(x^(x²))^'=(2xln(x)+x)e^(x²ln(x))#

#(x^x)^x#

#=e^ln((x^x)^x)#

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

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

Let's differentiate:

#e^u=u'e^u#

#((x^x)^x)^'=(2xln(x)+x)e^(x²ln(x))#

\0/ here's our answer !