Question

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

Answer 1

`d/(dx) x^(x^x) = x^(x^x + x - 1) (x ln^2(x) + x ln(x) + 1)`

Explanation:

Use:

`ln(a^b) = b ln(a)`

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

Find:

`x^(x^x)= e^(ln(x^(x^x)))`

`= e^(x^xln(x))`

`= e^(e^(ln(x^ x)) * ln(x))`

`= e^(e^(x ln(x)) * ln(x))`

So:

`d/(dx) x^(x^x) = d/(dx) e^(e^(x ln(x))*ln(x))`

`=e^(e^(x ln(x)) * ln(x)) * d/(dx)(e^(x ln(x)) * ln(x))`

`=x^(x^x) * d/(dx)(e^(x ln(x)) * ln(x))`

`=x^(x^x) * ((d/(dx) e^(x ln(x))) * ln(x) + e^(x ln(x)) * (d/(dx) ln(x)))`

`=x^(x^x) * (e^(x ln(x)) (d/(dx) (x ln(x))) * ln(x) + e^(x ln(x)) * 1/x)`

`=x^(x^x) * x^x * ((d/(dx) (x ln(x))) * ln(x) + 1/x)`

`=x^(x^x) * x^x * ((ln(x)+1) * ln(x) + 1/x)`

`=x^(x^x) * x^x * (ln^2(x) + ln(x) + 1/x)`

`=x^(x^x + x - 1) (x ln^2(x) + x ln(x) + 1)`