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