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How cubed root of x is not differentiable at 0

  1. Sep 20, 2011 #1
    I know that it isn't. I just want to know how I could, step by step, prove that this function is not differentiable at x=0.
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  3. Sep 20, 2011 #2


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    [tex]f'(0) = \lim_{h\rightarrow 0}\frac{(0+h)^{\frac 1 3} - 0^{\frac 1 3}}{h}[/tex]
  4. Sep 20, 2011 #3


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    Welcome to Physics Forums.

    You can either look at the derivative and consider where it is (not) defined. Or go back to the limit definition of the derivative and show that it doesn't exist for x=0.
  5. Sep 20, 2011 #4


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    I would like to ask - is this a big deal in mathematics?

    Presumably the inverse of any continuous function is undefined at its extrema or horizontal inflection point? However ones like this still have a tangent at all points on the curve. If you just rotate the paper this inflection point is still special in one way, but no longer in that way?

    So is this exercise really in the nature of a pedantry, or is there something more profound and important (and even then, is it relevant at this stage, I imagine, of the student?)
  6. Sep 20, 2011 #5

    Stephen Tashi

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    It's a big deal that mathematical definitions mean what they say. And it's a big deal that one's private intuitions about the way things are is not the basis for mathematical definitions or proofs. Lawyers and legal wrangling aren't popular in our culture, so writers on mathematics rarely describe math as "legalistic", but it is. If you practice it any other way you find that subjective arguments are more futile than legalism.

    In single variable calculus, the derivative is used to define the tangent line. It isn't our intuition about what a tangent line should be that defines the derivative. A person might be able to develop a consistent system of mathematics by making a different definition of derivative. (If you express a curve parametrically and pretend a particle is moving along it, you can get the kind of tangent line you want.) However, I have yet to see a definition of derivative that is more useful than the one currently used.

    It's pedantry in the sense that it attempts to teach the student the correct answer. As to "this stage", I don't know what stage the student is at. If the student is at the stage where the precise definitions of derivative and limit are being taught, the question in the original post is good exercise. What's the alternative? Keep them in the dark?
  7. Sep 21, 2011 #6
    As LCKurtz writes, the derivative here is lim_{h->0}h^{-2/3} = lim_{h->0}/cuberoot{h^{-2}} = /lim_{h->0}(/frac{1}{\cuberoot{h^2}} -> /infty.
  8. Sep 21, 2011 #7


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    Really? Are you sure about that?
  9. Sep 22, 2011 #8
    Thank you LCKurtz, I don't know why I didn't think about just using the limit rule.
  10. Sep 23, 2011 #9
    It just means the tangent line is vertical.
  11. Sep 14, 2012 #10
    I'm just getting into derivatives myself and this is a problem I'm working on but the comment above concerned me. "The tangent line is a vertical line" How can a tangent line be vertical, does that not violate the definition of function? And if it were vertical wouldn't the derivative be undefined because it's essentially showing asymptotic behavior?? While a derivative can get very very close to being vertical, I was taught once it's vertical it's no longer a function and is undefined at that point. Maybe I'm just not far enough in to full understand that, but then this is Honors Calc 1.
  12. Sep 14, 2012 #11


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    Why should a "line" necessarily be a "function"?

    This is badly stated. A derivative is a function, not a line, so "being vertical" makes no sense.

    A vertical line is NOT a function but that has nothing to do with it being a "line"!

    And the point you are missing is just that everyone is saying what you are- the limit defining the derivative at 0 does not exist so [itex]\sqrt[3]{x}[/itex] is not differentiable at x= 0.
  13. Sep 14, 2012 #12
    Alright so essentially you're saying im right. Just the fact that a line is a line and not based on a function itself. I simply meant that if you're talking about derivative and mentioning vertical lines that led me to believe that someone was saying its possible to have a tangent line that's vertical while its a line in and of itself its not a viable answer to the derivative function since it causes discontinuity.
  14. Sep 14, 2012 #13
    The tangent line is vertical at x = 0. So the slope of the tangent line is undefined.

    Therefore the function that represents the slope of the tangent line is undefined at x = 0. Which is another way of saying the derivative is undefined at x = 0.

    In other words (or really the same words, repeated) the graph does have a tangent line, but the line is vertical so its slope is undefined. So the derivative is not defined at x = 0.
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