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Is ln(x) differentiable at negative x-axis

  1. Jan 12, 2016 #1
    Since lnx is defined for positive x only shouldnt the derivative of lnx be 1/x, where x is positive. My books does not specify that x must be positive, so is lnx differentiable for all x?
     
  2. jcsd
  3. Jan 12, 2016 #2

    Mark44

    Staff: Mentor

    No. If a function isn't defined on some interval, its derivative isn't defined there, either. The real function ln(x) is defined only for x > 0, as you are aware, so the domain for the derivative is x > 0, as well.

    As it turns out, the function y = 1/x is defined for any ##x \ne 0##, but the left-hand branch does not represent the derivative of the natural log function.
     
  4. Jan 13, 2016 #3
    I assume you are referring to real numbers (you use "x" for the ind. variable and you say that lnx is defined for positive x only) because in the complex field it's all another story...

    --
    lightarrow
     
  5. Jan 13, 2016 #4

    mathman

    User Avatar
    Science Advisor
    Gold Member

    ln(x) can be defined for x < 0, using [itex]x=-xe^{\pi i}[/itex]. Therefore [itex]ln(x)=ln(-x)+\pi i[/itex].
     
  6. Jan 14, 2016 #5

    Svein

    User Avatar
    Science Advisor

    In fact, [itex]\frac{d}{dx}ln(\lvert x \rvert)=\frac{1}{x} [/itex].
     
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