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    Given that y= (4ln(x-3))/(4ln(x+3)) show that dy/dx = 24/(x(4ln(x+3)^2))

    I know you use the quotient rule, but I can't seem to get the right answer?
    Help please.
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    y= u/v then dy/dx = vu'-uv'/v^2 (u and v functions of x)

    so you need to find the derivative of a logarithm (ln).

    ie what is f(x) in the expression; ln(ax+b) = integral f(x) dx?
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    (Original post by mathperson)
    y= u/v then dy/dx = vu'-uv'/v^2 (u and v functions of x)

    so you need to find the derivative of a logarithm (ln).

    ie what is f(x) in the expression; ln(ax+b) = integral f(x) dx?
    Yeah I'm aware that dy/dx of ln(ax+b) = a/ax+b

    What I get is

    v= 4lnx-3 dv/dx= 4/x-3
    u= 4lnx+3 du/dx= 4/x+3

    ((4/x+3)4lnx-3 - (4/x-3)4lnx+4) / (4lnx-3 )^2


    I dont know where to go from there.
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    (Original post by Jin3011)
    Yeah I'm aware that dy/dx of ln(ax+b) = a/ax+b

    What I get is

    v= 4lnx-3 dv/dx= 4/x-3
    u= 4lnx+3 du/dx= 4/x+3

    ((4/x+3)4lnx-3 - (4/x-3)4lnx+4) / (4lnx-3 )^2


    I dont know where to go from there.
    bearing in mind I havn't checked anything:

    a/x + b/y = (ay+xb)/xy

    where (4/x+3)ln(x-3) =a and x=1 and rather than a + sign you have a - sign, try then simplifying it using log' rules such as ln(ab)=ln(a)+ln(b) (or whatever those rules are
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    No idea how to get the given answer, but: http://www3.wolframalpha.com/input/?...%28x%2B3%29%29
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    (Original post by mathperson)
    bearing in mind I havn't checked anything:

    a/x + b/y = (ay+xb)/xy

    where (4/x+3)ln(x-3) =a and x=1 and rather than a + sign you have a - sign, try then simplifying it using log' rules such as ln(ab)=ln(a)+ln(b) (or whatever those rules are
    In this question though when would I need to use the log rules?:confused:


    The mark scheme says this, but I dont understand what its means exactly.
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    (Original post by Jin3011)
    In this question though when would I need to use the log rules?:confused:


    The mark scheme says this, but I dont understand what its means exactly.
    Methinks the question is supposed to be, in LaTex form:

    

\dfrac{4\ln (x) -3}{4\ln (x) +3}

    In which case, you should be able to get the correct answer.
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    (Original post by Jin3011)
    In this question though when would I need to use the log rules?:confused:


    The mark scheme says this, but I dont understand what its means exactly.
    y = 4ln(x-3) / 4ln(x+3)
    let u=4ln(x-3) u'=4/(x-3)
    v=4ln(x+3) v'=4/(x+3)

    dy/dx = (vu'-uv')/v^2

    = (4/x-3)ln(x+3) - (4/x+3)ln(x-3)
    -------------------------------------
    (4ln(x+3))^2

    = 4 ((1/x-3)ln(x+3) - (1/x+3)ln(x-3))
    ------------------------------------------
    4^2 . ln(x+3)^2

    = (1/x-3)ln(x+3) - (1/x+3)ln(x-3)
    ---------------------------------------
    4ln(x+3)^2

    try using the addition of fraction rule I told you in my last post

    = (x+3)ln(x+3) -(x-3)ln(x-3)
    -----------------------------------
    4ln(x+3)^2

    = (x+3)ln(x+3) - (x-3)ln(x-3)
    --------------------------------
    12ln(x+3)

    = (x+3) - (x-3) ln(x-3 - (x+3))
    ------ -----
    12 12

    something tells me I havn't done this right, or at least the way you need it.
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    (Original post by F1Addict)
    Methinks the question is supposed to be, in LaTex form:

    

\dfrac{4\ln (x) -3}{4\ln (x) +3}

    In which case, you should be able to get the correct answer.
    Honestly, I think you're right. I was thinking that it was f(x-3) when its actually f(x)-3.
    Thank you!!!
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    (Original post by mathperson)
    y = 4ln(x-3) / 4ln(x+3)
    let u=4ln(x-3) u'=4/(x-3)
    v=4ln(x+3) v'=4/(x+3)

    dy/dx = (vu'-uv')/v^2

    = (4/x-3)ln(x+3) - (4/x+3)ln(x-3)
    -------------------------------------
    (4ln(x+3))^2

    = 4 ((1/x-3)ln(x+3) - (1/x+3)ln(x-3))
    ------------------------------------------
    4^2 . ln(x+3)^2

    = (1/x-3)ln(x+3) - (1/x+3)ln(x-3)
    ---------------------------------------
    4ln(x+3)^2

    try using the addition of fraction rule I told you in my last post

    = (x+3)ln(x+3) -(x-3)ln(x-3)
    -----------------------------------
    4ln(x+3)^2

    = (x+3)ln(x+3) - (x-3)ln(x-3)
    --------------------------------
    12ln(x+3)

    = (x+3) - (x-3) ln(x-3 - (x+3))
    ------ -----
    12 12

    something tells me I havn't done this right, or at least the way you need it.
    Thanks for trying
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    (Original post by Jin3011)
    Honestly, I think you're right. I was thinking that it was f(x-3) when its actually f(x)-3.
    Thank you!!!
    well no wonder I have done it wrong, wasn't given the right questions, won't get to the right answer :rolleyes:
 
 
 
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