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

2 Answers
Jun 12, 2018

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

Explanation:

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

(x^(x²))^'=(2xln(x)+x)e^(x²ln(x))

Explanation:

(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 !