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    Differentiate 3e^x -0.5ln(x-2)

    I know how to differentiate exponentials and natural logs
    but i cant get the right answer.
    I dont know what to do with the negative 0.5 in front of the natural log.
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    Bring the -0.5 to the front of the expression

     \frac{-3}{2}e^x ln(x-2)

    and differentiate that using product rule...

    EDIT

    is the starting expression this

     3e^x \times \frac{-1}{2}ln(x-2) or  3e^x - \frac{1}{2}ln(x-2)

    If its the second, then as Pheylan suggests...
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    \frac {1}{2} (6e^x - ln(x-2))

    can you do it now?
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    how does that work?
    how can you take the -0.5 to the front?
    why would it then be multiplied by the ln(x-2) shouldnt it be added
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    I don't think the product rule is necessary.
    first differentiate 3x^x which gives us 3e^x
    then differentiate -0.5ln(x-2) which will give us (-0.5/(x-2)) or rather : -1/(2x-4)
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    (Original post by New Username)
    I don't think the product rule is necessary.
    first differentiate 3x^x which gives us 3e^x
    then differentiate -0.5ln(x-2) which will give us (-0.5/(x-2))

    thats what i did and its wrong
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    (Original post by ks1234567)
    thats what i did and its wrong
    incidently what is the answer your suppose to get?
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    (Original post by spread_logic_not_hate)
    Bring the -0.5 to the front of the expression

     \frac{-3}{2}e^x ln(x-2)

    and differentiate that using product rule...

    EDIT

    is the starting expression this

     3e^x \times \frac{-1}{2}ln(x-2) or  3e^x - \frac{1}{2}ln(x-2)

    If its the second, then as Pheylan suggests...
    it's the second
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    (Original post by New Username)
    incidently what is the answer your suppose to get?
    3e^x - 1/2x
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    (Original post by Pheylan)
    \frac {1}{2} (6e^x - ln(x-2))

    can you do it now?
    how do you get to that?
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    (Original post by ks1234567)
    how do you get to that?
    both terms have a factor of 0.5, so factorise it
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    (Original post by Pheylan)
    both terms have a factor of 0.5, so factorise it
    still a bit stuck though
    so far this is what i've got

    0.5(6e^x - 1/x-2)

    0.5 * 1/x-2 doesnt give -1/2x which is the answer
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    (Original post by Pheylan)
    \frac {1}{2} (6e^x - ln(x-2))

    can you do it now?

    Don't mean to hijack this thread, but since I've got a maths exam on wed I thought I'd ask...
    From the starting equation  3e^x - \frac {1}{2} ln(x-2)

    Did you divide both sides by 1/2 to get \frac {1}{2} (6e^x - ln(x-2)) ?

    Spoiler:
    Show
    (Which would yield the answer  3e^x - \frac {1}{2(x-2)} )


    Is that right?
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    (Original post by ks1234567)
    still a bit stuck though
    so far this is what i've got

    0.5(6e^x - 1/x-2)

    0.5 * 1/x-2 doesnt give -1/2x which is the answer
    Your book is wrong, as you rightly said, it should be  0.5(6e^x - \frac{1}{x-2} )
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    (Original post by blueJ-)
    Your book is wrong, as you rightly said, it should be  0.5(6e^x - \frac{1}{x-2} )
    yeah thats what i've got
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    (Original post by ks1234567)
    still a bit stuck though
    so far this is what i've got

    0.5(6e^x - 1/x-2)

    0.5 * 1/x-2 doesnt give -1/2x which is the answer
    hm..is this from a text book?

    the standard result is that:
    ln(x) = \frac{1}{x}
    ln(f(x)) = \frac{f'(x)}{f(x)}

    Perhaps they are using the first rule giving them 1/x and then multiplying that by 1/2 which would explain the 1/2x - however that answer isn't valid
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    (Original post by blueJ-)
    Your book is wrong, as you rightly said, it should be  0.5(6e^x - \frac{1}{x-2} )
    its an edexcel mark scheme
    C3 june 2005
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    (Original post by ks1234567)
    its an edexcel mark scheme
    C3 june 2005
    I doubt it, are you sure it's not a school-produced mark scheme?

    If that's what it does say, it's not the latest version, because that's an error.
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    (Original post by ks1234567)
    its an edexcel mark scheme
    C3 june 2005
    If it's june 2005, then the question is actually meant to be:

     3e^x - \frac {1}{2} ln(x) - 2

    The -2 is not in the brackets!!
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    (Original post by .:excel4100%:.)
    If it's june 2005, then the question is actually meant to be:

     3e^x - \frac {1}{2} ln(x) - 2

    The -2 is not in the brackets!!
    ahh that clears up things.. thanks a lot mate
 
 
 
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