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    Okay, could someone please do

    d/dx 1/2 ln [ (1+x)/(1-x) ]

    so I can check my answer? The logic keeps going astray, but I think you should get

    1 / (1-x^2)

    Can you show the method steps, so I know where I keep messing up?
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    so do you want to differentiate that or integrate it??
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    Thats correct
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    d/dx 1/2 ln [ (1+x)/(1-x) ]

    let w = (1+x)/(1-x) = u/v

    dw/dx = [v.u' - u,v']/v²

    dw/dx = [(1-x).(1) - (1+x).(-1)]/(1-x)²
    dw/dx = [1-x + 1 + x]/(1-x)²
    dw/dx = 2/(1-x)²

    d/dx (½ln[w])
    d/dw (½ln[w]).dw/dx
    ½.(1/w).(2/(1-x)²)
    ½.(1-x)/(1+x).(2)/(1-x)²
    1/[(1+x)(1-x)]
    1/(1-x²) + C
    ========
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    (Original post by JamesF)
    Thats correct
    Sorry, i thought you were checking your answer.
    Btw, you are meant to differentiate.
    If you didnt know,
    d/dx (f(x)) = f '(x)

    So, d/dx 1/2(ln(1+x/1-x) = d/dx 1/2 (ln|1+x| - ln|1-x|)

    Then use the chain rule.
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    Okay, new question:

    ∫tanh^-1 x

    given that tanh^-1 x = 1/2 ln [ (1+x)/(1-x) ]

    The answer to the first question is on the question paper, so can someone show me how to do it, as I keep getting 1/4-4x^2 instead of 1/1-x^2
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    Sorry, folks, for Q1 I mean differntiate ! My error.

    So, d/dx 1/2(ln(1+x/1-x) = d/dx 1/2 (ln|1+x| - ln|1-x|)

    Then use the chain rule.
    I got that far, but what's the chain rule? I haven't done P3 so stuff like this is difficult for me... :rolleyes:
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    why haven't you done p3?
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    (Original post by Mat666)
    Sorry, folks, for Q1 I mean differntiate ! My error.



    I got that far, but what's the chain rule? I haven't done P3 so stuff like this is difficult for me... :rolleyes:
    If you have a function inside a function, say y = f(g(x)), you let u = g(x):

    So y = f(u)

    dy/dx = df/du * du/dx
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    Way the school organises it. They decided to teach us p4 and p6 for further maths in yr 12, but wuthout teaching us p2 or p3. So whenever we came up against anything that needed p3 formulae we were taught it.
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    (Original post by Mat666)
    Way the school organises it. They decided to teach us p4 and p6 for further maths in yr 12, but wuthout teaching us p2 or p3. So whenever we came up against anything that needed p3 formulae we were taught it.
    Wow, thats particularly stupid.

    Have you done integration by parts, cos you need that for the next question?
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    Thats weird. Must be quite difficult doing it like that seeing as the modules are synoptic. I thought hyperbolic functions were introduced in P5?
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    Yeah, but not at the kind of depth. We've done things like 'integrate x sin x', but not huge chunks of natural log stuff like this

    This is AQA - I think hyperbolic stuff may be p5 Edexcel...
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    (Original post by Nylex)
    If you have a function inside a function, say y = f(g(x)), you let u = g(x):

    So y = f(u)

    dy/dx = df/du * du/dx
    Without sounding particularly slow, can anyone actually do this for me so I know which parts are which?
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    (Original post by Mat666)
    Okay, new question:

    ?tanh^-1 x

    given that tanh^-1 x = 1/2 ln [ (1+x)/(1-x) ]

    The answer to the first question is on the question paper, so can someone show me how to do it, as I keep getting 1/4-4x^2 instead of 1/1-x^2
    I
    = (int) tanh^-1(x) dx
    = (1/2) (int) [ln(1 + x) - ln(1 - x)] dx.

    (int) ln(1 + x) dx
    = x ln(1 + x) - (int) x/(1 + x) dx
    = x ln(1 + x) - (int) [1 - 1/(1 + x)] dx
    = x ln(1 + x) - x + ln(1 + x) + c.

    (int) ln(1 - x) dx
    = x ln(1 - x) + (int) x/(1 - x) dx
    = x ln(1 - x) + (int) [-1 + 1/(1 - x)] dx
    = x ln(1 - x) - x - ln(1 - x) + c.

    So

    I
    = (1/2) [x ln(1 + x) - x + ln(1 + x) - x ln(1 - x) + x + ln(1 - x)] + c
    = (1/2) [x ln[(1 + x)/(1 -x)] + ln(1 - x^2)] + c
    = x tanh^-1(x) + (1/2) ln(1 - x^2)] + c.

    OR ... integrate by parts using the "ln(x) = 1*ln(x)" method:

    I
    = x tanh^-1(x) - (int) x / (1 - x^2) dx
    = x tanh^-1(x) + (1/2) ln(1 - x^2) + c.
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    (Original post by Jonny W)
    I
    = (int) tanh^-1(x) dx
    = (1/2) (int) [ln(1 + x) - ln(1 - x)] dx.

    (int) ln(1 + x) dx
    = x ln(1 + x) - (int) x/(1 + x) dx
    = x ln(1 + x) - (int) [1 - 1/(1 + x)] dx
    = x ln(1 + x) - x + ln(1 + x) + c.

    (int) ln(1 - x) dx
    = (int) ln(1 + u) du . . . substituting u = -x
    = u ln(1 + u) - u + ln(1 + u) + c
    = -x ln(1 - x) + x + ln(1 - x) + c.

    So

    I
    = (1/2) [x ln(1 + x) - x + ln(1 + x) - x ln(1 - x) + x + ln(1 - x)] + c
    = (1/2) [x ln[(1 + x)/(1 -x)] + ln(1 - x^2)] + c
    = x tanh^-1(x) + (1/2) ln(1 - x^2)] + c.

    OR ... integrate by parts using the "ln(x) = 1*ln(x)" method:

    I
    = x tanh^-1(x) - (int) x / (1 - x^2) dx
    = x tanh^-1(x) + (1/2) ln(1 - x^2) + c.
    Cheers mate. Seems a lot for the four marks they give it on the question paper, but still...
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    If it's any reassurance, my school does things in the same way too. I think that because very few people do maths and further maths, they don't complement each other very well. I've had to learn all sorts in order to understadn P4. P1 to P4 is a huuge jump!
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    Arggh P4 in two days time, I bet'll it'll be solid just like P6
 
 
 
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