How do I find the derivative of x^x^x?

[silence]

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

 

[Lonewolf]

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

 

[silence]

sorry i meant x^(x^x)

 

[Lonewolf]

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

 

[silence]

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

 

[djuiceholder]

ok ok

 

[arildno]

but can you do (x^x)^x ??
it seems impossible

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:
$$x^{x^{2}}(2x\ln(x)+x)$$

2. Please do not re-open nearly 6-year old threads.

 

[djuiceholder]

ok ok