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Dy/dx of ln(ax) for any a

  1. Nov 30, 2013 #1
    It's easy to show that [itex]\frac{dy}{dx}[/itex] of [itex]y = ln(ax)[/itex] where [itex]a \in ℝ, a > 0[/itex] is always [itex]\frac{1}{x}[/itex] :

    [itex]y = ln(ax)[/itex]
    [itex]y = ln(a) + ln(x)[/itex]

    [itex]ln(a)[/itex] is constant so its derivative is 0, and the derivative of [itex]ln(x)[/itex] is [itex]\frac{1}{x}[/itex].
    Hence:

    [itex]\frac{dy}{dx} = \frac{1}{x}[/itex]

    Attached is an image of y=ln(ax) with a = 1,2,3,4,5. It also shows all of their derivatives to be the same (the curve at the top). But they are clearly different curves!!!! There must be some point at which one has a larger gradient than the others?? What am I missing?
     

    Attached Files:

  2. jcsd
  3. Nov 30, 2013 #2

    mfb

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    The curves are just shifted upwards/downwards with respect to each other. This is not easy to see in the graph (as their non-vertical "distance" is not constant), but it is right. And constant offsets do not show up in the derivative.
     
  4. Nov 30, 2013 #3
    oh of course!!!! :D I feel a bit stupid now :P I guess they're shifted up by ln(a) right?
     
  5. Nov 30, 2013 #4

    Office_Shredder

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    That's right.
     
  6. Nov 30, 2013 #5

    Mark44

    Staff: Mentor

    Also, you don't "take dy/dx" of some function. dy/dx already represents the derivative of y with respect to x.
     
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