1. ## derivatives

what is the first derivative of

sec^2 x ...

thanks =)

2. Originally Posted by go.leafs.go.89
what is the first derivative of

sec^2 x ...

thanks =)
2sec^2xtanx

3. so the derivative of sec x is tan x?

4. Originally Posted by go.leafs.go.89
what is the first derivative of

sec^2 x ...

thanks =)
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

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

5. hmm
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?

6. Originally Posted by go.leafs.go.89
hmm

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?

$\displaystyle \lim_{x \to 0}\frac{\tan(x)-x}{x^2}=$

by L.H.

$\displaystyle \lim_{x \to 0}\frac{\sec^{2}(x)-1}{2x}$

this is of the from 0/0 so by L.H again

$\displaystyle \lim_{x \to 0}\frac{2\sec^2(x)\tan(x)}{2}=\frac{2(1)\cdot 0}{2}=0$

I hope this helps

7. Originally Posted by TheEmptySet
$\displaystyle \lim_{x \to 0}\frac{\tan(x)-x}{x^2}=$

by L.H.

$\displaystyle \lim_{x \to 0}\frac{\sec(x)-1}{2x}$

this is of the from 0/0 so by L.H again

$\displaystyle \lim_{x \to 0}\frac{\sec^2(x)\tan(x)}{2}=\frac{1\cdot 0}{2}=0$

I hope this helps
yes it does .. thanks a lot
however .. when you first differentiated

tanx-x
------
x^2

isn't the derivative of tanx = sec^2x and not sec x
which would lead up to having the 2nd derivative = 2sec^2xtanx

?

8. Originally Posted by go.leafs.go.89
yes it does .. thanks a lot
however .. when you first differentiated

tanx-x
------
x^2

isn't the derivative of tanx = sec^2x and not sec x
which would lead up to having the 2nd derivative = 2sec^2xtanx

?
Yes it is a typo I but the limit is still zero.

I will fix my post above

9. thanks so much!