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silence
Sep9-03, 04:00 PM
i can solve x^x but adding this new x just confuses me any help will do, X^x^x

Lonewolf
Sep9-03, 04:28 PM
Do you mean (x^x)^x or x^(x^x)? I'm assuming the former.

silence
Sep9-03, 04:35 PM
sorry i meant x^(x^x)

Lonewolf
Sep9-03, 04:49 PM
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 [;)]

silence
Sep9-03, 04:58 PM
wow i never noticed that i was doing it a long way which would have come out wrong anyways. thanks for the help

djuiceholder
Feb2-09, 10:21 AM
ok ok

arildno
Feb2-09, 10:31 AM
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
Feb2-09, 09:06 PM
ok ok