C3 differentiation quick question

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  1. icouldsay's Avatar
    1 29-05-2012  09:01
    • Junior Member
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    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. Wilko94's Avatar
    2 29-05-2012  09:10
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    • Location: Weymouth, Dorset
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    Show us your working. I presume you've used the chain rule on cos^{-1}(2x)
    Last edited by Wilko94; 29-05-2012 at 09:13.
  3. BabyMaths's Avatar
    3 29-05-2012  09:11
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    Re: C3 differentiation quick question
    It's the chain rule. The derivative of 2x (with respect to x) is 2.
  4. BabyMaths's Avatar
    4 29-05-2012  09:12
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    Re: C3 differentiation quick question
    (Original post by Wilko94)
    Show us your working. I presume you've used the chain rule on cos(^-1)(2x)
    It's not a great idea to call it \cos^{-1} 2x for obvious reasons.
  5. Wilko94's Avatar
    5 29-05-2012  09:14
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    • Location: Weymouth, Dorset
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    (Original post by BabyMaths)
    It's not a great idea to call it \cos^{-1} 2x for obvious reasons.
    Sorry, I'm probably being stupid, but why? I just worked it through.
  6. BabyMaths's Avatar
    6 29-05-2012  09:16
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    Re: C3 differentiation quick question
    (Original post by Wilko94)
    Sorry, I'm probably being stupid, but why? I just worked it through.
    Because \sec x=\frac{1}{\cos x} and \cos^{-1}x is the inverse cosine of x.
  7. bananarama2's Avatar
    7 29-05-2012  09:17
    • TSR Legend
    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. Wilko94's Avatar
    8 29-05-2012  09:18
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    (Original post by BabyMaths)
    Because \sec x=\frac{1}{\cos x} and \cos^{-1}x is the inverse cosine of x.
    Oh yeah - in that case I'm learning something too. :-)

    How would you write it?
  9. JulietheCat's Avatar
    9 29-05-2012  09:21
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    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. king0vdarkness's Avatar
    10 29-05-2012  09:23
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    • Posts: 123
    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. Zuzuzu's Avatar
    11 29-05-2012  11:47
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    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. Pin's Avatar
    12 29-05-2012  12:08
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    • Posts: 276
    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
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