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# C3 differentiation quick question Tweet

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1. C3 differentiation quick question
Why does sec2x differentiate to 2sec2xtan2x?
Where does the 2 come from at the front?
Could someone please explain step by step?

Thanks
2. Show us your working. I presume you've used the chain rule on
Last edited by Wilko94; 29-05-2012 at 09:13.
3. Re: C3 differentiation quick question
It's the chain rule. The derivative of 2x (with respect to x) is 2.
4. Re: C3 differentiation quick question
(Original post by Wilko94)
Show us your working. I presume you've used the chain rule on
It's not a great idea to call it for obvious reasons.
5. (Original post by BabyMaths)
It's not a great idea to call it for obvious reasons.
Sorry, I'm probably being stupid, but why? I just worked it through.
6. Re: C3 differentiation quick question
(Original post by Wilko94)
Sorry, I'm probably being stupid, but why? I just worked it through.
Because and is the inverse cosine of x.
7. Re: C3 differentiation quick question
(Original post by Wilko94)
Sorry, I'm probably being stupid, but why? I just worked it through.
That means inverse cos. NEVER use that notation to mean anything else.
8. (Original post by BabyMaths)
Because and is the inverse cosine of x.
Oh yeah - in that case I'm learning something too. :-)

How would you write it?
9. Re: C3 differentiation quick question
(Original post by Wilko94)
Oh yeah - in that case I'm learning something too. :-)

How would you write it?
(cos(x))^-1
10. Re: C3 differentiation quick question
(Original post by icouldsay)
Why does sec2x differentiate to 2sec2xtan2x?
Where does the 2 come from at the front?
Could someone please explain step by step?

Thanks
Y= sec2x. Let u= 2x, so y= sec u. differentiate u= 2. differentiate y=sec2xtan2x. using the chain rule (dy/dx= dy/du . du/dx) so dy/dx = 2sec2xtan2x. Hope this Helps!
11. Re: C3 differentiation quick question
(Original post by king0vdarkness)
Y= sec2x. Let u= 2x, so y= sec u. differentiate u= 2. differentiate y=sec2xtan2x. using the chain rule (dy/dx= dy/du . du/dx) so dy/dx = 2sec2xtan2x. Hope this Helps!
I don't think that will help.
12. Re: C3 differentiation quick question
(Original post by icouldsay)
Why does sec2x differentiate to 2sec2xtan2x?
Where does the 2 come from at the front?
Could someone please explain step by step?

Thanks
You need to use the chain rule - I'm really going to simplify it so I hope it makes sense

sec2x = (cos 2x)^-1

dy/dx = (-1)(2)(-sin2x)(cos2x)^-2

*you have to differentiate 2x from cos2x, so that's where the 2 comes fro

this will simplify to 2(sin2x)(sec2x)^2

(sec2x)^2 = 1/(cos2x)^2 so if you split it
you get sin2x/cos2x x 1/cos2x

which cancels down into tan2xsec2x

so overall you get 2tan2xsec2x