Question

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

Answer 1

Since the rule:

`[f(x)]^g(x)=e^log([f(x)]^g(x))=e^(g(x)logf(x))`,

then:

`x^(x^x)=e^log(x^(x^x))=e^(x^xlogx)=e^(e^log(x^x)logx)=`

`=e^((e^(xlogx))logx)`.

So the function to derive is:

`y=e^((e^(xlogx))logx)`.

`y'=e^((e^(xlogx))logx)*[e^(xlogx)(1*logx+x*1/x)*logx+e^(xlogx)*1/x]=`

`=e^((e^(xlogx))logx)*e^(xlogx)(logx+1+1/x)=`

`=e^((e^(xlogx))logx)*e^(xlogx)(xlogx+x+1)/x`

And, if you want:

`y'=x^(x^x)*x^x(xlogx+x+1)/x`.