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#1 2006-10-13 05:00:12

luca-deltodesco
Member
Registered: 2006-05-05
Posts: 1,470

Is there an error here?









The Beginning Of All Things To End.
The End Of All Things To Come.

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#2 2006-10-13 05:09:36

polylog
Member
Registered: 2006-09-28
Posts: 162

Re: Is there an error here?

The end results all are correct as far as I can tell.

I didn't check the intermediate steps of the second part, as it seems to work out simply as:

d/dx sec^n(x) = n sec ^(n-1) * sec(x) tanx  = n sec^n(x) tanx

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#3 2006-10-13 05:13:12

luca-deltodesco
Member
Registered: 2006-05-05
Posts: 1,470

Re: Is there an error here?

well i think its just a coincidence that this is true. since in general it dones hold with functions


The Beginning Of All Things To End.
The End Of All Things To Come.

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#4 2006-10-13 05:16:19

polylog
Member
Registered: 2006-09-28
Posts: 162

Re: Is there an error here?

I'm not sure what you mean, since this is certainly true:
d/dx (f(x))^n = n(f(x))^(n-1) f'(x)

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#5 2006-10-13 05:21:59

luca-deltodesco
Member
Registered: 2006-05-05
Posts: 1,470

Re: Is there an error here?

ah right, i though you just meant n(f(x))^(n-1) which ofcourse would be wrong.


The Beginning Of All Things To End.
The End Of All Things To Come.

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#6 2006-10-13 05:31:17

polylog
Member
Registered: 2006-09-28
Posts: 162

Re: Is there an error here?

indeed smile

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