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Derivative question

  1. Jun 10, 2009 #1
    This has been confusing me for awhile.

    Let's say f(x) is undefined at x=2 and thus f(x) is discontinious at x=2.

    And it's derivative is f'(x)=2X. Is the derivative still defined at x=2 or not, because f(x) is undefined at x=2?

    If it is defined, wouldn't this mean f(x) is differentiable at x=2 and thus f(x) has to be continuous at x=2 because of the theorem:

    If f(x) is differentiable at a then f(x) must be continuous at a.

    One last question. Why is (x+1)(x-1)/(x-1) still undefined when x=1? Even though the x-1 cancel out.

    Thanks in advance.
  2. jcsd
  3. Jun 11, 2009 #2
    If your function has a discontinuity then it cannot be differentiable at that point. Intuitively, your derivative gives you the rate of change of the line tangent to the function at a given point. If your function is discontinuous, then a tanget line will not exist. To put it differently, differentiability is a sufficient condition for continuity, or rather, if your derivative exists at all points, then your function is continuous.

    Please take time to soak this in though because it does NOT work the other way around. There exists many functions that are continuous but not differentiable at certain points.

    For example let's say...

    [tex]f(x) = x^{2/3}[/tex]

    The graph of this function has a cusp at [tex]x=0[/tex], which means you can draw multiple tangents lines to this function at that point. Its derivative is...

    [tex]f'(x) = \frac{2}{3}x^{-1/3}[/tex]

    And this function is undefined at [tex]x=0[/tex] as it should be. :smile:

    As for your function [tex]\frac{(x+1)(x-1)}{x-1}[/tex], it is still undefined at [tex]x=1[/tex] so it is not differentiable at that point. If you cancel the [tex](x-1)[/tex] out, then you get an entirely different function, which is now continuous at that point. :approve:
  4. Jun 11, 2009 #3
    The derivative wouldn't be defined at x = 2. It's easier to see if you try taking the derivative using the definition. You can't very well take the limit of [f(2+h) - f(2)]/h if f(2) doesn't exist.

    It's an issue of domain. Since the value x = 1 gives you zero in the denominator, it isn't in the domain of the function. Thus, the function isn't defined on x = 1. You may think this is overly anal, but it is mathematics after all.
  5. Jun 11, 2009 #4
    Everything you said is true, but I am curious why you refer to a (point where the function is not defined) as a discontinuity? In fact, the function described in the initial post is continuous and differentiable over its entire domain (Reals - {2}).

    The reason that the derivative is not defined at 2 is that the function itself is not defined at 2 i.e. the point 2 is not in the domain of the function, and the derivative of a function only exists on a (not necessarily proper) subset of the original functions domain. In particular, the definition of the derivative f' of a function f evaluated at the point x has as a premise that the point x is contained in the domain of the function f.
  6. Jun 11, 2009 #5
    IMO it is an issue of being underly 'anal', since an expression involving a variable, such as (x+1)(x-1)/(x-1), is by itself not an appropriate way to define a function at all. If anything such an expression involving a variable should be a compact way of expressing the rule that associates each point in the domain with its image in the codomain. If they said "what is the domain of the function which is defined at exactly those points for which every subexpression of (x+1)(x-1)/(x-1) is defined?" then 1 would not be in the domain because 1/(x-1) is a subexpression of the given expression and 1/(x-1) is not defined at x = 1. The problem is that by writing down [itex]a b b^{-1}[/itex] one already assumes the existence of b^{-1}, and so there is nothing wrong with applying b b^{-1} = 1. "Consider a b b^{-1}, ... oh but that's not the same as just [itex]a[/itex] since b^{-1} didn't exist at all, gotcha!" (-10 points)
  7. Jun 11, 2009 #6


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    Yes, writing down [itex]a*b*b^{-1}[/itex] assumes the existence of [itex]b^{-1}[/itex] Therefore any point where [itex]b^{-1}[/itex] doesn't exist is implicitly not in the domain of the function. It's not a gotcha deal, this can have real consequences e.g. when solving differential equations and trying to find the domain of the solution
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