Differentiating sec

• Oct 27th 2007, 06:03 AM
curlywurlysqurly
Differentiating sec
Hi, would someone please tell me how to differentiate sec^2 x?
I tried it and got (-4sin2x)/(cos2x+1)^2
Not sure if that is right tho
xxx
• Oct 27th 2007, 07:08 AM
kalagota
Quote:

Originally Posted by curlywurlysqurly
Hi, would someone please tell me how to differentiate sec^2 x?
I tried it and got (-4sin2x)/(cos2x+1)^2
Not sure if that is right tho
xxx

differentiate?
Dx[sec^2 x] = 2(sec x)(sec x tan x) = 2(sec^2 x)(tan x)
you just nid to apply chain rule.
• Oct 28th 2007, 03:14 AM
curlywurlysqurly
sorry but i don't understand how you did that!!!
where did the tan x come from???
• Oct 28th 2007, 09:41 AM
Krizalid
Do you know what chain rule is?

$\displaystyle f(x)=\sec^2x.$

To take its derivative, we use $\displaystyle (u^n)'=nu^{n-1}\cdot u'.$

So, $\displaystyle f'(x)=2\sec x\cdot(\sec x)'.$

Does that make sense?
• Oct 28th 2007, 10:19 AM
curlywurlysqurly
yea thats all fine, but how do u differentiate sec x? is it just a standard result or is there some way to do it?
• Oct 28th 2007, 10:25 AM
Krizalid
You know that $\displaystyle \sec x=\frac1{\cos x}$

Now $\displaystyle \Big[(\cos x)^{-1}\Big]'=-(\cos x)^{-2}\cdot(\cos x)'.$

So, $\displaystyle \frac{\sin x}{\cos^2x}=\frac1{\cos x}\cdot\frac{\sin x}{\cos x}.$
• Oct 28th 2007, 11:03 AM
curlywurlysqurly
okay thanks i get it now!
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