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    http://mathsathawthorn.pbworks.com/f/FP3Jun06.pdf
    q3a?
    This is what ive done below!
    f(x)= lnsecx
    x=0 , f(x)=0

    f'(x)=tanx
    x=0, f'(x)=0

    f''(x)=sec^2x
    x=0, f''(x)=1

    Is f'''(x)=2(sec^2xtanx)

    And what is f''''(x)
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    You're right about f'''(x).

    What have you tried to work out f''''(x)?
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    (Original post by Student403)
    You're right about f'''(x).

    What have you tried to work out f''''(x)?
    Well can i use u=sec^2x and v= 2tanx? And find du/dx and dv/dx and cross multiply?
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    (Original post by Ayaz789)
    Well can i use u=sec^2x and v= 2tanx? And find du/dx and dv/dx and cross multiply?
    cross multiply? I don't see how you could cancel the du and dv. Could you explain how?
    Spoiler:
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    I'd use the quotient rule after putting tanx = sinx/cosx and secx = 1/cosx
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    (Original post by Student403)
    You're right about f'''(x).

    What have you tried to work out f''''(x)?
    Or i can do u=2sec^2x and v=tanx as dv/dx =sec^2x but du/dx is?
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    (Original post by Student403)
    cross multiply? I don't see how you could cancel the du and dv. Could you explain how?
    Spoiler:
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    I'd use the quotient rule after putting tanx = sinx/cosx and secx = 1/cosx
    Like i do u* dv/dx and v* du/dx
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    (Original post by Ayaz789)
    Or i can do u=2sec^x and v=tanx as dv/dx =sec^x but du/dx is?
    I'm asking what would you do with du/dx and dv/dx? How do you cancel the du and dv? The substitution doesn't really make sense
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    (Original post by Student403)
    I'm asking what would you do with du/dx and dv/dx? How do you cancel the du and dv? The substitution doesn't really make sense
    I do u* dv/ dx???
    And do v* du/ dx??
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    (Original post by Ayaz789)
    I do u* dv/ dx???
    And do v* du/ dx??
    Oh you're talking about the product rule here. Yeah that's fine (sorry)

    But remember since you have 2sec2x tanx and you want d/dx (2sec2x tanx), it's better to factor out the 2 to get

    2 d/dx (sec2x tanx)
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    to find the fourth derivative it is sensible to replace sec2(x) with tan2(x) + 1

    in the third derivative
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    (Original post by Student403)
    Oh you're talking about the product rule here. Yeah that's fine

    But remember since you have 2sec2x tanx and you want d/dx (2sec2x tanx), it's better to factor out the 2 to get

    2 d/dx (sec2x tanx)
    Okay can you show me how to get f''''(x)
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    (Original post by Ayaz789)
    Okay can you show me how to get f''''(x)
    No I can't show you, sorry.

    Using the bear 's identity will work well
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    (Original post by Student403)
    No I can't show you, sorry.

    Using the bear 's identity will work well
    Okay do me a favour and differentiate 2sec^2x
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    (Original post by Ayaz789)
    Okay do me a favour and differentiate 2sec^2x
    You can do that since you said

    f''(x)=sec^2x

    f'''(x)=2(sec^2xtanx)



    So if you have 2sec2x, then what is that differentiated?
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    (Original post by Ayaz789)
    Okay do me a favour and differentiate 2sec^2x
    DIY
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    (Original post by Math12345)
    DIY
    & that means?
    (Do it yourself)
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    Is f''''(X) = 2sec^4x +tanx(tanx/cosx)^2
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    (Original post by Ayaz789)
    & that means?
    (Do it yourself)
    If you are really lazy, use wolframalpha to do it for you.
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    (Original post by Math12345)
    If you are really lazy, use wolframalpha to do it for you.
    Dw ive attempted it ^
 
 
 
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