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    How do I differentiate y=0.5ln(1+sinx)

    I used the chain rule and got it as far as dy/dx = (cosx)/2(1+sinx)

    the markscheme got (cosx)/2e^2y

    how do I do this last bit please?
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    0.5 is a constant.

    ln(1+Sinx)

    t = 1 + Sinx
    dt/dx = ???

    ln(t)
    d/dt = ???

    d/dx = d/dt * dt/dx * 0.5 (constant)

    I'm getting:

    Cosx/(2+2Sinx)

    As well
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    (Original post by jumblehunter)
    How do I differentiate y=0.5ln(1+sinx)

    I used the chain rule and got it as far as dy/dx = (cosx)/2(1+sinx)

    the markscheme got (cosx)/2e^2y

    how do I do this last bit please?
    y=0.5ln(1+sinx)

    gives 2y = ln(1+sinx)

    gives 1+sinx = ????
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    (Original post by L'Evil Fish)


    As well
    see post 3
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    (Original post by TenOfThem)
    see post 3
    Aaah I see, I didnt know they'd ask that
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    (Original post by L'Evil Fish)
    Aaah I see, I didnt know they'd ask that
    No idea what board this is

    But if the question just asked for the differential then jumblehunter and your answer would gain full marks


    I cannot see any reason for the version they have given unless that format were specified in the question
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    (Original post by TenOfThem)
    No idea what board this is

    But if the question just asked for the differential then jumblehunter and your answer would gain full marks


    I cannot see any reason for the version they have given unless that format were specified in the question
    I was going to say, unless it said:

    Prove that it can be written in the form. Bla bla.

    I wouldn't have done anything else
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    My bad all. I missed a snippet of information from a previous question ( this is one of those ridiculously many parted MEI ones) that pretty much confirmed my answer was the same as the one they wanted me to find. Cheers for your help.
 
 
 
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