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Derivatives with detailed pictures and work ! help !

  1. Nov 20, 2008 #1
    deleted by accident .. ahh
     
    Last edited: Nov 20, 2008
  2. jcsd
  3. Nov 20, 2008 #2

    Mark44

    Staff: Mentor

    Speaking only for myself, I would be more inclined to help you on a single problem rather than a half dozen all at once.
     
  4. Nov 20, 2008 #3
    good idea. i'll change it right now
     
  5. Nov 20, 2008 #4

    tiny-tim

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    Yup … that was what put me off a few hours ago!

    ok, in 1. they want you to write it in the form log[f(x)] = …, and differentiate both sides.

    So start with … what is log[f(x)]?

    And in 2., try using the chain rule … what do you get? :smile:
     
  6. Nov 20, 2008 #5
    thank you tim,

    Attempt at #1

    log[f(x)]= log[ (x^2)(x-3)^9 / (x^2+3)^5 ]
    log[f(x)] = log(x^2)(x-3)^9 - log (x^2+3)^5 ?

    try#1
    log[f(x)]= log(x^2)+log(x-3)^9-log(x^2+3)^5
    i'm not sure if it is this ===== 2log(x) + 9log(x-3) - 5log(x^2+3)[/quote]
    Yes, log(ab)= b log(a).
    Now, differentiate.

    and 2x/x^2= 2/x, of course: the derivative of 2log x is 2/x.


    Are there 2 #2s?
    No, the derivative of 5ln(u) is NOT 5 ln(u)*du/dx, it is (5/u)*du/dx.


     
    Last edited by a moderator: Nov 20, 2008
  7. Nov 20, 2008 #6

    tiny-tim

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    hmm … messy, but mostly correct …

    the lines I would pick out to string together for a proper answer are:

    log[f(x)]= log[ (x^2)(x-3)^9 / (x^2+3)^5 ]

    log[f(x)]= log(x^2)+log(x-3)^9-log(x^2+3)^5

    = 2log(x) + 9log(x-3) - 5log(x^2+3)

    then you should differentiate both sides (you only did the RHS):

    d/dx(log[f(x)]) = 2x/x^2+ 9*1/x-3 - 5* 2x/x^2+3

    and then complete that equation, and multiply both sides by … something :wink: … to get an equation with d/dx(f(x)) on the LHS on its own. :smile:
    u= 5x+7ln(x)) du/dx= 5+ 7*1/x ==== 5+7/x is correct,

    but the next line is completely wrong … try again! :smile:
     
  8. Nov 20, 2008 #7
    hey guys,

    1.

    log[f(x)]= log[ (x^2)(x-3)^9 / (x^2+3)^5 ]

    log[f(x)]= log(x^2)+log(x-3)^9-log(x^2+3)^5

    = 2log(x) + 9log(x-3) - 5log(x^2+3)


    d/dx(log[f(x)]) = 2x/x^2+ 9*1/x-3 - 5* 2x/x^2+3

    1/f(x) * f'(x) = 2/x + 9/x-3 -10x/x^2+3

    f'(x)= (2/x + 9/x-3 -10x/x^2+3) * (x^2)(x-3)^9/(x^2+5)^5

    I got f(x) from the original equation

    f'(3)= (2/3+9/0 - 30/12) * (9)(0)^9/(9+5)^5

    Um ... is f'(3) = 9 ?



    2.
    I used d/dx 5ln(u) = 5/u * du/dx

    so continuing off from there, 5/5x+7ln(x) * 5+7/x is the answer?

    f'(3)= (5/5(3))+7ln(3) * 5+7/3

    ==== 1/3 + 7.6902 * 7.3333
    final answer f'(3) = 56.7287 ?


    Thanks , I tried to make it as clear as possible this time haha :)
     
  9. Nov 20, 2008 #8

    tiny-tim

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    Much better! :biggrin:

    (though you have a misprint at the end … 5 instead of 3 :wink:)

    Well, you have 09/0 …

    wouldn't it have been better to cancel an (x - 3) before you put x = 3? :rolleyes:
    Yes … much clearer!

    But clarity isn't just for the reader's benefit … it's for your benefit also …

    so your next aim should be to use brackets more often … in fact, whenever they're needed … you've made at least one mistake because you've misread a non-existent bracket. :redface:
     
  10. Nov 20, 2008 #9
    thank you , i will do that more often !

    1. sorry i made a mistake last time , i meant f'(3) = 0 (the 9 is too close to the 0)

    (2/x + 9 - 10x/x^2+3) * ( (x^2)((x-3)^8)/ (x^2+3)^5 =

    (2/3 + 9 - 30/12) * ( (9)*(0) / (12)^5 ) = 0 because 9*0 = 0
     
    Last edited: Nov 20, 2008
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