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    I think I know how to do it, but mine doesn't look like the answers

    It's either:

    a) -2(sec^2)2x
    b) 2sec2xtan2x
    c) sec2xtan2x
    d) -2secxtan2x

    Thanksssss
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    I reckon it's B, but my core 4 is rusty nowadays.
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    Core 3
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    I wanna say b) but someone more sure will probably clarify.
    Sec(x) definately differentiates to sec(x)tan(x) and I assume chain rule still applies.
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    (Original post by Adam92)
    Core 3
    Differentiating trig was core 4 for OCR
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    i'm pretty sure its b
    you differentiate the X and put it at the front then keep the X with the sec and tan

    2sec2xtan2x
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    using sub u=2x gives b...
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    http://calc101.com/webMathematica/derivatives.jsp

    It's b.
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    (Original post by Bilkos_Ices)
    Differentiating trig was core 4 for OCR
    Ah I see, but I'm on AQA.
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    b

    isnt secx in the formula book (well it is for aqa) so just 2x is going to leave you with b
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    Thanks guys
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    (Original post by Adam92)
    Thanks guys
    By the way, you know 'd' is wrong because you've got a term in 'x' rather than '2x' (or it's a typo) so you should discard it immediately.
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    (Original post by jayshah31)
    By the way, you know 'd' is wrong because you've got a term in 'x' rather than '2x' (or it's a typo) so you should discard it immediately.
    It wasn't a typo. I thought it was b), I just needed a bit of clarification :]
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    I got (2tan2x)(sec2x)

    which is equiv to B.


    I can explain if you want?
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    (Original post by maxfire)
    I got (2tan2x)(sec2x)

    which is equiv to B.


    I can explain if you want?
    YES please.
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    You can find the derivative by writing \sec 2x as \dfrac{1}{\cos 2x}. Using the quotient rule, differentiating gives you \dfrac{(\cos 2x) \times 0 - 1 \times (-2\sin 2x)}{\cos^2 2x}, which can be re-written as 2 \times \dfrac{1}{\cos 2x}\times \dfrac{\sin 2x}{\cos 2x} = 2\sec 2x \tan 2x. HTH
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    sec(2x)

    1/cos(2x)

    [cos(2x)]^-1

    -1[cos(2x)]^-2.-sin(2x).2

    2[cos(2x)]^-2.sin(2x)


    2sin(2x)/cos(2x)^2

    2tan(2x).sex(2x)
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    (Original post by Adam92)
    YES please.
    I hope you realise that if you remove the bracket's it's exactly the same as b) :p:

    (Original post by nuodai)
    You can find the derivative by writing \sec 2x as \dfrac{1}{\cos 2x}. Using the quotient rule, differentiating gives you \dfrac{(\cos 2x) \times 0 - 1 \times (-2\sin 2x)}{\cos^2 2x}, which can be re-written as 2 \times \dfrac{1}{\cos 2x}\times \dfrac{\sin 2x}{\cos 2x} = 2\sec 2x \tan 2x. HTH
    \text{Chain rule} > \text{Quotient rule}

    Less prone to mistakes
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    (Original post by nuodai)
    You can find the derivative by writing \sec 2x as \dfrac{1}{\cos 2x}. Using the quotient rule, differentiating gives you \dfrac{(\cos 2x) \times 0 - 1 \times (-2\sin 2x)}{\cos^2 2x}, which can be re-written as 2 \times \dfrac{1}{\cos 2x}\times \dfrac{\sin 2x}{\cos 2x} = 2\sec 2x \tan 2x. HTH
    Don't judge me here, I've been self-teaching this module so I don't have a full understanding yet, but what's wrong with writing \dfrac{1}{\cos 2x} as (cos2x)^-1? Then using the chain rule to differentiate?

    Edit: nevermind, many people have already posted the solution :p:
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    (Original post by illy123)
    sec(2x)

    1/cos(2x)

    [cos(2x)]^-1

    -1[cos(2x)]^-2.-sin(2x).2

    2[cos(2x)]^-2.sin(2x)


    2sin(2x)/cos(2x)^2

    2tan(2x).sex(2x)
    Seems you drifted away a little on the last line there :p:
 
 
 
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