Derivative x^x^x

  • Thread starter silence
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  • #1
silence
i can solve x^x but adding this new x just confuses me any help will do, X^x^x
 

Answers and Replies

  • #2
334
1
Do you mean (x^x)^x or x^(x^x)? I'm assuming the former.
 
  • #3
silence
sorry i meant x^(x^x)
 
  • #4
334
1
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 :wink:
 
  • #5
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
ok ok
 
Last edited:
  • #7
arildno
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but can you do (x^x)^x ??
it seems impossible
1. It is not at all imposible; note that this equals: [tex](x^{x})^{x}=x^{x^{2}}[/tex]
Rewriting this as:
[tex]x^{x^{2}}=e^{x^{2}\ln(x)}[/tex]
We may readily differentate this by means of the chain rule, yielding the derivative:
[tex]x^{x^{2}}(2x\ln(x)+x)[/tex]

2. Please do not re-open nearly 6-year old threads.
 
  • #8
ok ok
 
Last edited:

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