Derivative x^x^x

1. Sep 9, 2003

silence

i can solve x^x but adding this new x just confuses me any help will do, X^x^x

2. Sep 9, 2003

Lonewolf

Do you mean (x^x)^x or x^(x^x)? I'm assuming the former.

3. Sep 9, 2003

silence

sorry i meant x^(x^x)

4. Sep 9, 2003

Lonewolf

Glad you picked that one. Get something like ln(y) = xxln(x) and use the product rule. Don't forget you already know d(xx)/dx

5. Sep 9, 2003

silence

wow i never noticed that i was doing it a long way which would have come out wrong anyways. thanks for the help

6. Feb 2, 2009

djuiceholder

ok ok

Last edited: Feb 2, 2009
7. Feb 2, 2009

arildno

1. It is not at all imposible; note that this equals: $$(x^{x})^{x}=x^{x^{2}}$$
Rewriting this as:
$$x^{x^{2}}=e^{x^{2}\ln(x)}$$
We may readily differentate this by means of the chain rule, yielding the derivative:
$$x^{x^{2}}(2x\ln(x)+x)$$