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Math Help - quick Q about differentiating

  1. #1
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    Exclamation quick Q about differentiating

    Hi, this is kinda urgent so please respond quickly if you can.

    My "root" function (dunno if it's called that in English, I'm from Austria) is:

    f(x) = 2x lnx

    My first "derivative" (?) is:

    f'(x) = 2 lnx + 2

    Now my question is what the second derivative would be. I don't know how to apply the rules here since I have a multiplication and an addittion there.

    Is this correct:

    f''(x) = 2/x + lnx

    Thanks...
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  2. #2
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    Quote Originally Posted by temhawk View Post
    My first "derivative" (?) is:

    f'(x) = 2 lnx + 2

    Now my question is what the second derivative would be. I don't know how to apply the rules here since I have a multiplication and an addittion there.

    Is this correct:

    f''(x) = 2/x + lnx
    No, the second derivative is 2/x. (Both of the 2s in 2 lnx + 2 are constants, so their derivative is 0.)
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  3. #3
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    Hi, thanks for your reply. I don't understand why it is 2/x. The rules for differentiating are:

    (Products)
    f(x) = u • v ........ f'(x) = u' • v + u • v'

    (Sums)
    f(x) = u + v ........ f'(x) = u' + v'

    My comprehension problem is: how are these two rules combined when you have something like:

    f(x) = u • v + s

    (Like I have^^)

    I would think:

    f'(x) = u' • v + u • v' + s'

    That would give me:

    f'(x) = lnx + 2 • (1/x) + 0 = 2/x + lnx

    What am I missing??? (I have the feeling it's something to do with constants vs variables but I can't figure it out.)

    Thanks...

    Edit: Actually, I think I know how you did it. In my formula reference there is a rule like this:

    y = k • f(x) ........ y' = k • f'(x) CONSTANT FACTOR IS KEPT

    That works when you consider the first "2" (from f(x) = 2 • lnx + 2) to be "k" and the rest ("lnx + 2") f(x). But there is another formula right above that one in my reference book that goes like this:

    y = f(x) + k ........ y' = f'(x) CONSTANT SUMMAND IS OMITTED

    And "2 • lnx" could be considered as "f(x)" and the last "2" as the summand ("k"), then you would get what I got. So is this a matter of which rule takes precedence? If so, when do I use the rule, with which I would get my (wrong) result in my current function here?

    Thanks...
    Last edited by temhawk; November 5th 2009 at 12:26 AM. Reason: new idea
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  4. #4
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    Quote Originally Posted by temhawk View Post
    Hi, thanks for your reply. I don't understand why it is 2/x. The rules for differentiating are:

    (Products)
    f(x) = u v ........ f'(x) = u' v + u v'

    (Sums)
    f(x) = u + v ........ f'(x) = u' + v'

    My comprehension problem is: how are these two rules combined when you have something like:

    f(x) = u v + s

    (Like I have^^)
    Okay, you're trying to differentiate 2 ln x + 2. So u = 2, v = ln x and s = 2.

    Quote Originally Posted by temhawk View Post
    I would think:

    f'(x) = u' v + u v' + s'

    That would give me:

    f'(x) = lnx + 2 (1/x) + 0 = 2/x + lnx
    If u = 2 then u' = 0. So f'(x) = 0 ln x + 2 (1/x) + 0 = 2/x.
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  5. #5
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    Ah, it's clear to me now! So I can use both formulae I found in my reference interchangeably as long as I don't forget that differentiated constants are zero and not just nothing =P

    And my own combo-formula works too!

    Thanks!!
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