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Math Help - Just a simple derivation question.

  1. #1
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    Just a simple derivation question.

    h(t) = log(base3)[log(base2)(t)]
    Evaluate for h'(8).

    This is what I have so far:

    Let g(x) = log(base2)(t)
    g'(x) = 1 / (t ln2)

    Therefore.

    h'(t) = 1 / (t ln2) / (log(base 2)t)(ln3)
    h'(t) = 1 / (t)(ln2)(log(base2)t)(ln3)

    (using the rule: log(base x)(g(x)) = g'(x) / g(x)ln(b)

    The answer in the back of my textbook gives:

    h'(8) = 1 / 24(ln2)(ln3)

    How do I go from where I am to where the book is?
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Jeavus View Post
    h(t) = log(base3)[log(base2)(t)]






    Evaluate for h'(8).

    This is what I have so far:

    Let g(x) = log(base2)(t)
    g'(x) = 1 / (t ln2)

    Therefore.

    h'(t) = 1 / (t ln2) / (log(base 2)t)(ln3)
    h'(t) = 1 / (t)(ln2)(log(base2)t)(ln3)

    Solving for y' we get y'=\frac{1}{ln(2)ln(3)3^yt}

    (using the rule: log(base x)(g(x)) = g'(x) / g(x)ln(b)

    The answer in the back of my textbook gives:

    h'(8) = 1 / 24(ln2)(ln3)

    How do I go from where I am to where the book is?
    here is whta I would do y=log_3(log_2(t))

    so then 3^y=log_2(t)

    Differenatiating both sides we would get \ln(3)3^{y}y'=\frac{1}{ln(2)t}

    and since y=\log_3(\log_2(x))

    we have y'=\frac{1}{\ln(2)\ln(3)t\log_2(t)}

    Evaluating we get y'(8)=\frac{1}{\ln(2)\ln(3)\cdot{8}\cdot\log_2(2^3  )}=\frac{1}{\ln(2)\ln(3)\cdot{24}}

    which is what you asked for
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