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use the chain rule

d/dx f(g(x))=f'(g(x))g'(x)

f(x)=x^2
g(th)=sec th

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i also think so!

hm...
i have another answer \[\sec ^{2} \theta=\sec \theta ^{2}\]
it's equal, right?

i mean question

no, of course not.
sec^2(th)= sec th*sec th
sec(th^2)=sec(th*th)

No, it's not correct.Re-write sec^2 x as (sec x)^2, then power and chain rules.

oh - http://www.wolframalpha.com/input/?i=d%2Fdx+sec+x^2

how to get that answer?

use f(x)=sec x
g(x)=x^2
then f'(g(x))g'(x)=tan(x^2)sec(x^2)*d/dx x^2=2x tan(x^2)sec(x^2)

does that make sense?

ah ok i got it

okay - so the chain rule makes more sense to you now?

yup, thanks sooo much ^^

You're welcome :)