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Ln derivative

  1. Apr 22, 2006 #1
    The question is to compute the derivative of
    ln (7x+1)^1/2 (3X^2+x)^5
    (x^2-3)^3 e^2x
    the ln is for both the numerator and denominator

    I tried the chain rule and came up with this

    1
    (7x+1)^1/2 (3x^2+x)^5
    (x^2-3)^3 e^2x

    and finally 6x(x^2-3)^2 2e^2x
    7/2(7x+1)^-1/2 30x(3x^2+x)^4
     
  2. jcsd
  3. Apr 22, 2006 #2
    You may want to try to apply some logarithm properties to that to simplify it.

    For example, lna/b is the same thing as lna - lnb. lna^b = blna. lna*b = lna + lnb.....
     
  4. Apr 22, 2006 #3

    nrqed

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    Are you sayingthat your final expression is your final result for the derivative? That is not correct.

    As Moose said, use first properties of the ln to write this as a *SUM* of 4 terms and bring down the exponents using ln(a^n)= n ln(a). Also, use [itex] ln ( e^{2x}) = 2x [/itex]. Only after doing all that you should take the derivative and it will be quite simple.
     
  5. Apr 23, 2006 #4
    ok I used the ln property and here is what I got.

    ln(7x+1)^1/2+ln(3x^2+x)^5-ln(x^2-3)^3+lne^2x

    1/2ln(7x+1) +5ln(3x^2+x)-3ln(x^2-3)+2x

    and finally 1/2*7/7x+1+5*6x/3x^2+x -3*2x/x^2-3 + 2

    am I right?
     
  6. Apr 23, 2006 #5
    Apart from the '+1' missing from numerator of the third term(in the final answer), everything else seems to be right.
     
  7. Apr 23, 2006 #6

    HallsofIvy

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    It's -2x, not +2x (e2x was in the denominator of the fraction.)
    And, of course, you'll want to simplify those fractions.
     
  8. Apr 23, 2006 #7
    Thanks everyone
     
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