what is the first derivative of

sec^2 x ...

thanks =)

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- Apr 27th 2008, 02:35 PMgo.leafs.go.89derivativeswhat is the first derivative of

sec^2 x ...

thanks =)

- Apr 27th 2008, 02:38 PMstorpotaten
- Apr 27th 2008, 02:40 PMgo.leafs.go.89so the derivative of sec x is tan x?

- Apr 27th 2008, 02:42 PMTheEmptySet
let $\displaystyle g(x)=\sec^{2}(x)$

$\displaystyle f(u)=u^2 \mbox{ and }u=\sec(x)$

$\displaystyle f'(u)=2u \mbox{ and } u'=\sec(x)\tan(x)$

then $\displaystyle g(x)=f(u)$

taking the derivative by the chain rule gives

$\displaystyle g'(x)=f'(u)\cdot u'=2u \cdot (\sec(x)\tan(x))$

but u is $\displaystyle u=\sec(x) $

so we get

$\displaystyle g'(x)=f'(u)\cdot u'=2u \cdot (\sec(x)\tan(x))=2\sec(x) \cdot (\sec(x)\tan(x))=2\sec^{2}(x)\tan(x)$ - Apr 27th 2008, 02:50 PMgo.leafs.go.89hmm

i had to use that in a question

lim tan x - x

x->0 ---------

x^2

i used l'hopital's rule to do this .. but couldn't get very far ..

any suggestions?

- Apr 27th 2008, 03:02 PMTheEmptySet
- Apr 27th 2008, 03:05 PMgo.leafs.go.89
- Apr 27th 2008, 03:51 PMTheEmptySet
- Apr 27th 2008, 03:53 PMgo.leafs.go.89
thanks so much!