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

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Answer 1 · 2018-06-12T11:45:08

$x (2 log x + 1) {(x^{x})}^{x}$ $x(2 log x+1)(x^x)^x$

Let $y = {(x^{x})}^{x} = x^{x^{2}}$ $y= (x^x)^x = x^(x^2)$

Then

$log y = x^{2} log x \Rightarrow$ $log y = x^2 log x implies$

$\frac{1}{y} \frac{d y}{d x} = 2 x log x + x^{2} \times \frac{1}{x} = x (2 log x + 1)$ $1/y dy/dx = 2x log x+x^2times 1/x= x(2 log x+1)$

$\frac{d y}{d x} = x (2 log x + 1) {(x^{x})}^{x}$ $dy/dx = x(2 log x+1)(x^x)^x$

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 !