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Logaritmic Differentiation

  1. Mar 16, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the derivative of y= x^2 (x^2 - 1)^3 (x+1)^4 using logaritmic differentiation.

    2. Relevant equations



    3. The attempt at a solution

    y= x^2 (x^2 - 1)^3 (x+1)^4 = 2 ln|x| + 3 ln |(x^2)-1|+ 4 ln |x+1|

    2/x + (6x)/(x^2-1) + 4/(x+1)

    Is this right and if so what should I do next?

    Thanks a lot in advance..
     
  2. jcsd
  3. Mar 16, 2010 #2

    Dick

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    You should start by not connecting all of the steps in your problem with an '=' sign. Do a simple example first. y=x^2. ln(y)=ln(x^2)=2*ln(x). Now differentiate. y'/y=2/x. So y'=2y/x=2x^2/x=2x. That's true. Now note that y and ln(y) are not equal. Nor is y=(ln(y))'. Take the log of both sides, then solve for y'.
     
  4. Mar 18, 2010 #3
    My problem is in the first steps, you said that y=x^2. ln(y)=ln(x^2)=2*ln(x)
    Why does ln(x^2)=2*ln(x)

    I dont remember much of my algebra classes about logarithms.

    I would appreciate some help with this.
     
  5. Mar 18, 2010 #4

    Char. Limit

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    It's part of the properties of logarithms..

    [tex]log (a^x) = x * log (a)[/tex]

    This is because of the exponent property...

    [tex](b^x)^y=b^{xy}[/tex]
     
  6. Mar 19, 2010 #5
    Still im not having the result.

    I tried this:
    y= x^2 (x^2 - 1)^3 (x+1)^4



    Lny = 2Ln(x) + 3Ln(x^2-1) + 4Ln(x+1)

    y'/y = 2/x + 6x/x^2-1 + 4/x+1

    y'= 2x (x^2 - 1)^3 (x+1)^4 + 6x^3 (x^2 - 1)^2 (x+1)^4 + 4 x^2 (x^2 - 1)^3 (x+1)^3


    Is this right?

    Thanks in advance.
     
  7. Mar 19, 2010 #6

    Dick

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    Put parentheses around your denominators, ok? I.e. write y'/y = 2/x + 6x/(x^2-1) + 4/(x+1). But, yes, it's right so far. Now I think you just need to try and simplify it.
     
  8. Mar 19, 2010 #7
    How can I simplify? I cant see a way of doing that.
     
  9. Mar 19, 2010 #8

    Dick

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    y'= 2x (x^2 - 1)^3 (x+1)^4 + 6x^3 (x^2 - 1)^2 (x+1)^4 + 4 x^2 (x^2 - 1)^3 (x+1)^3 is also right. Take out the common factors and try and simplify what's left.
     
  10. Mar 19, 2010 #9
    I guess the common factors are (x^2 - 1) and (x+1) whats left is 2x + 6x^3 + 4 x^2

    I guess I could simplify by 2x
    2x ( 1 + 3x^3 + 2x^2 )

    If this is correct what is next? Im kind of confuse I dont see where this is going.
     
  11. Mar 19, 2010 #10

    Dick

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    The point is that your answer is already right. There's nothing wrong with what you did. But you said "Still im not having the result." I assume that means you haven't simplified it enough to get the form of the answer you want. Is that right? In that case, that's where it's going. You can immediately factor out the highest power of each factor in your expression. That would be x*(x^2-1)^2*(x+1)^3. Then try and factor what's left. What is the answer you are looking for anyway?
     
  12. Mar 19, 2010 #11
    The answer im looking for is 2x(x+1)^6 (x-1)^2 (6x^2 -2x -1)
     
  13. Mar 19, 2010 #12

    Dick

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    Ok, that's the same as your answer in a different form. Notice (x^2-1)=(x-1)*(x+1). Just combine and factor.
     
  14. Mar 20, 2010 #13
    Let me see, I still dont get the result. From
    2x (x^2 - 1)^3 (x+1)^4 + 6x^3 (x^2 - 1)^2 (x+1)^4 + 4 x^2 (x^2 - 1)^3 (x+1)^3



    (x^2-1)^3 = (x-1)^1.5 (x+1)^1.5 ?

    So by adding all the (x^2 - 1) terms I get (x-1)^4 (x+1)^4


    Plus the other (x+1) terms in the answer (x+1)^11 + (x+1)^4 = (x+1)^15 ?
     
  15. Mar 20, 2010 #14

    Dick

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    Factor! Take out the common factor first. All of the terms are divisible by x*(x^2-1)^2*(x+1)^3. Factor it out.
     
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