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Math Help - Basic doubt on differential of ln(ax) from first principles

  1. #1
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    Basic doubt on differential of ln(ax) from first principles

    Hi Guys,


    I have a fundamental doubt on differential of

    y = f(x) = ln(ax)
    where,
    ln - natual logorithm to base e,
    a - constant
    x - independent variable.

    I need to find the value of dy/dx from first principles.

    Can anybody help me on this?

    Thanks,
    Srini
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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  3. #3
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    Thanks for your reply on derivative of ln(x) friend.
    But, I need to know the derivative of ln(ax)
    it turns out to be:


    (1/x) lim (1+u/3) ^ (1/u)
    (u -> 0)

    I dont think the above limit result 'e'.
    This is where i'm stuck.
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    let u/3=t then, when u-->0 also t-->0. 1/u=1/3t.

    So, (1+u/3) ^ (1/u)=(1+t)^(1/3t)={(1+t)^(t)}^(1/3)=e^{1/3} (when t-->0)
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  5. #5
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    Quote Originally Posted by Also sprach Zarathustra View Post
    let u/3=t then, when u-->0 also t-->0. 1/u=1/3t.

    So, (1+u/3) ^ (1/u)=(1+t)^(1/3t)={(1+t)^(t)}^(1/3)=e^{1/3} (when t-->0)

    Ya, exactly.
    Therefore,
    derivative of ln(3x) is then, (1/3x) * (3) i.e., (1/x).
    Is that right?
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  6. #6
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    Yes, beause ln(3x)= ln(x)+ ln(3) and ln(3) is a constant: the derivative of ln(ax) is exactly the same as the derivative of ln(x) for any positive x.
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  7. #7
    MHF Contributor Also sprach Zarathustra's Avatar
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    Yes.(I think...)

    But there is better way of solving this...

    Recall: ln(a*b)=ln(a)+ln(b)

    Now:

    ln(ax)=ln(a)+ln(x)

    so the derivative of ln(ax) is:

    {ln(ax)}'= {ln(a)}' + {ln(x)}'

    But ln(a) is a constant! ({ln(a)}'=0)

    Hence:

    {ln(ax)}'= {ln(a)}' + {ln(x)}'= 0 +1/x=1/x
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  8. #8
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    Thanks so much :-)
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