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Using logarithmic differentiation

  1. Oct 28, 2007 #1
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    using logarithmic differentiation I can get it to a point where I have...

    ln(y)=20[ln(x+sqrt(x))-ln(x^2-2^x)]

    I do not know what to do at that point
     
  2. jcsd
  3. Oct 28, 2007 #2
    just take the derivative as you normally would

    also you will have [tex]\frac{y'}{y}= ...[/tex]

    bring Y to the other side and plug in what Y is. the rest is straight forward differentiation.
     
  4. Oct 28, 2007 #3
    yeah but then you have a term that is:

    [1/(x^2-2^x)]*(2x-d/dy2^x)

    thats where I get stuck
     
  5. Oct 28, 2007 #4
    i'm not really following

    you mean on the [tex] x^{2}-2^{x}[/tex] part? how would you take it's derivative?
     
  6. Oct 28, 2007 #5
    [tex] y=((x+sqrt(x))/x^{2}-2^{x})^{20} [/tex]

    [tex] lny=20(ln(x+sqrt(x))-ln(x^{2}-2^{x})) [/tex]

    [tex] y'/y=(20/(x+sqrt(x))*(1+1/2(x)^{-1/2}))-(20/x^{2}-2^{x})*(2x-derivative of 2^x) [/tex]
     
  7. Oct 28, 2007 #6
    yes that part sorry
     
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