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Homework Help: Derivative with logarithms

  1. Jan 26, 2008 #1
    [SOLVED] derivative with logarithms

    Ok, so the problem problem probably isn't as bad as I'm making it, either that or its because its getting late & my brain just isn't functioning.

    Find the derivative of y with respect to r.
    y=[tex]log _2 \left( r \right)[/tex] * [tex]log _4 \left( r \right)[/tex]

    The first thing I thought to do was use the product rule which yielded
    y'=[tex]log _2 \left( r \right)[/tex]*(1/ln2)(1/r)+[tex]log _4 \left( r \right)[/tex](1/ln4)(1/r)

    I then changed the [tex]log _2 \left( r \right)[/tex] to (ln r/ln2) & did the same to the other logarithm which created a complex fraction that I condensed down to
    (ln r)/(r ln 2 * ln 2) + (ln r)/(r ln 4 * ln 4)

    Cross multiplying gave
    [(ln r)(ln 4)^2+ (ln r)(ln 2)^2]/[r (ln 2)^2 (kn 4)^2]

    Here is where I'm stuck, because the book likes (2 ln r)/(r ln 2 * ln 4).

    The book also showed different steps. To begin with it shows making the logs into ln.
    which gives
    y=(ln r)^2/(ln 2 * ln 4)
    then take the derivative, but here is where I'm a bit lost. It shows
    y'=1/(ln2 *ln 4) * (2 ln r) * (1/r)
    I can see where the (2 ln r) & the (1/r) come from the chain rule, but not so much the 1/(ln2 *ln 4) part.
  2. jcsd
  3. Jan 27, 2008 #2

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    y= (ln r/ln 2)*(ln r/ln 4) = (ln r)^2/(ln 2*ln 4), as in the book. The denominator D is a const, and just keep it like that. So,

    y' = 2(ln r)(1/r)/D.
  4. Jan 27, 2008 #3
    That'd probably eventually simplify to the correct answer, but I think your teacher wanted you to apply the change-of-base formula at the start and then use the chain rule as follows:

    [tex]\frac{dy}{dr} \left ( \frac{\ln^2 r}{\ln 2 \ln 4} \right ) = \frac{2\ln r}{r \ln 2 \ln 4} [/tex]
  5. Jan 27, 2008 #4

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    How is this any different from what I'd given?

    Also, [tex]( \ln^2 r)[/tex] is not a standard notation.
  6. Jan 27, 2008 #5
    Right, duh... :) thanks.
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