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Homework Help: Chain rule

  1. Feb 16, 2009 #1
    1. The problem statement, all variables and given/known data
    Hi.

    I want to solve:

    [tex]
    \frac{d\ln(-x)}{dx}.
    [/tex]

    When using the chain rule I get:

    [tex]
    \frac{d \ln(-x)}{dx} = \frac{d\ln(-x)}{d(-x)}\frac{d(-x)}{dx} = -\frac{d\ln(-x)}{d(-x)}.
    [/tex]

    But how do I find the last derivative? I know by experience that it is -x-1, but how is the derivation done?

    Thanks in advance.
     
  2. jcsd
  3. Feb 16, 2009 #2
    The last derivate is of the same form as:

    [tex]\frac{d}{dx}\ln x[/tex]

    Which is a simple solution, no?
     
  4. Feb 16, 2009 #3

    Mark44

    Staff: Mentor

    The last derivative works out to - 1/(-x) = 1/x, not -x^(-1) as you have.

    For the same reason that [itex]\int dx/x = ln |x| + C[/itex], d/dx(ln |x|) = 1/x.

    This means that d/dx(ln(x)) = d/dx(ln(-x)) = 1/x.
     
  5. Feb 17, 2009 #4
    If this is true, then we obtain a total of -1/x. And jazznaz is telling me the opposite of you?

    Thanks for replying.
     
  6. Feb 17, 2009 #5
    Well, no, but it probably helps to replace x with u in what Jazznaz wrote.

    To derive

    [tex]\frac{d}{du} \ln{u} = \frac{1}{u}[/tex]

    first let [tex]y = \ln{u}[/tex]. Then [tex]e^y = u[/tex].


    Now use implicit differentiation to find [tex]\frac{dy}{du}[/tex] (which is [tex]\frac{d}{du} \ln{u}[/tex] ) in terms of u.
     
  7. Feb 17, 2009 #6

    HallsofIvy

    User Avatar
    Science Advisor

    If you let u= -x, your last expresion is [tex]-\frac{d ln(u)}{du}[/tex]. Does that make more sense?
     
  8. Feb 17, 2009 #7
    Yes, that does make sense.

    I understand it now. Thanks to everybody for helping.
     
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