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    Need help to differentiate this :

    3e^x - 1/2lnx - 2

    So differentiating the first term will give : 3e^x

    How about the second one ? -1/2lnx, how exactly do I differentiate that ?
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    y = 1/2 ln x
    dy/dx =
    1/2
    ---
    x
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    (Original post by uer23)

    How about the second one ? -1/2lnx, how exactly do I differentiate that ?
    Is it -1 / 2lnx or -0.5lnx ?
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    Use Chain Rule. And from then, by inspection.
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    differentiate 2lnx separately if that makes it any easier then flip the fraction
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    EDIT: Disregard this I am an idiot
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    (Original post by Andylol)
    y = 1/2 ln x
    dy/dx =
    1/2
    ---
    x
    Right so it's - \frac{1}{2x}
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    3e^x - 1/2x

    jst differentiate lnx then multiply by -0.5
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    (Original post by JeevesTSR)
    Is it -1 / 2lnx or -0.5lnx ?
    -0.5lnx.
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    (Original post by uer23)
    -0.5lnx.
    ok then i think its -0.5 * 1/x

    = -1 / 2x
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    (Original post by uer23)
    Right so it's - \frac{1}{2x}
    correct
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    So related to this question...

    The differential of the curve f(x) = 3e^x - \frac{1}{2} \ln x - 2

    is as stated above f'(x) = 3e^x - \frac{1}{2x}

    The curve f(x) has a turning point at p, the x co-ordinate of p is \alpha

    Show that \alpha = \frac{1}{6}e^{- \alpha}

    So far what I've done

    3e^ \alpha - \frac{1}{2 \alpha} = 0

    3e^ \alpha = \frac{1}{2 \alpha}

    6 \alpha e^ \alpha = 1

    e^ \alpha = \frac{1}{6 \alpha}

    So what do I do know take the natural log of both sides ? But then how is that going to get me closer to my target which I have to show.
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    Bump... +rep available.
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    you're basically there....just multiply by alpha, then divide by e^alpha and you've done it...
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    (Original post by uer23)
    So related to this question...

    The differential of the curve f(x) = 3e^x - \frac{1}{2} \ln x - 2

    is as stated above f'(x) = 3e^x - \frac{1}{2x}

    The curve f(x) has a turning point at p, the x co-ordinate of p is \alpha

    Show that \alpha = \frac{1}{6}e^{- \alpha}

    So far what I've done

    3e^ \alpha - \frac{1}{2 \alpha} = 0

    3e^ \alpha = \frac{1}{2 \alpha}

    6 \alpha e^ \alpha = 1

    e^ \alpha = \frac{1}{6 \alpha}

    So what do I do know take the natural log of both sides ? But then how is that going to get me closer to my target which I have to show.
    All the calculations are done, just need to rearrange alpha and e^a
 
 
 
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