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Having trouble understanding derivatives

  1. May 23, 2009 #1
    Hello,

    I am having a bit of trouble understanding why a derivative does not exist at points where the function has a sharp jump. Is it because the slope of the tangent line as we approach the point from the left and right direction changes sign? My book does not go into detail and I looked online and ended up confusing myself even more as some sites say that you could draw infinite number of tangents through the point or some other explanation and I am unable to convince myself of the exact reason why a sharp point is a problem.

    Looking forward to your help. Many thanks,

    Luca
     
  2. jcsd
  3. May 23, 2009 #2

    Hurkyl

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    Rather than trying to feel your way through it with prior intuition, you could try doing some computations. What happens if you try to compute the derivative at 0 of the function

    [tex]f(x) = \begin{cases} 1 & x \geq 0 \\ 0 & x < 0 \end{cases}[/tex]

    ?

    What about the function

    [tex]f(x) = \begin{cases} 1 & x = 0 \\ 0 & x \neq 0 \end{cases}[/tex]

    ?
     
  4. May 23, 2009 #3
    I think what the OP is referring to is a cusp.

    The derivative at the cusp is not defined simply because it could be argued to be whatever you want.

    Take f(x) = |x|. For x > 0, f '(x) = 1, and for x < 0, f '(x) = -1. What is f '(0)?

    What does it mean for a function to have a certain derivative at a point? It means the slope of the function is the value of the derivative there. In simpler terms, that means if you draw a line that grazes the function there, it will have slope equal to that value.

    For instance: let g(x) = x^2. At x=2, the derivative g '(2) = 4. That means if you want to draw a line that touches g where x=2 (and consequently f(x)=4), it will be the line passing through the point (2,4) and having slope 4.

    If you try that with f, it works everywhere... except where x=0. If you think about it, as long as the line passes through the point (0,0) and has a slope between -1 and 1 (otherwise, it would cut f in half) you have a line. But that means there are infinitely many tangent lines. Since every tangent line corresponds to exactly one slope, and every slope means a unique derivative, it means the derivative of f at x=0 must be able to be every number between -1 and 1.

    But this is silly. We don't want the derivative operation to be able to give us infinitely many answers. So whenever differentiation gives us anything except exactly one answer, we say it's undefined. Since we can't unambiguously say what the derivative is there, by finding one and only one tangent line, we just say that "the derivative" doesn't exist.

    Alternatively, the limit definition of the derivative fails to work since the limit doesn't exist.
     
  5. May 24, 2009 #4
    Thanks for your reply and explanation.

    This is the bit that I am having trouble understanding. I thought that there are two possible tangents here (each of them pointing to one direction of the function). This is, of course, enough and the fact that left and right side limits are not the same to conclude that the function is not differentiable at x = 0. However, I am still struggling with why the tangent could have any slope between -1 and +1.

    Thanks,

    Luca
     
  6. May 24, 2009 #5

    zcd

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    It would be infinitely many tangent lines not because there are infinitely many slopes, but because the tangent lines could be centered about any point on the line and still have the same slope (e.g points (0,0); (0.1,0.1); (0.2,0.2) and so forth, and still have slope m=1). And no, the slope cannot be -1 because going either which way it's still 1/1 or -1/-1 = 1.
     
  7. May 24, 2009 #6

    Hurkyl

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    For the record, the notion of a "tangent line" to the graph of a function is only clearly defined at points where the function is differentiable; one really shouldn't talk about such things without defining what they really mean!

    AUMathTutor is probably describing the behavior of the limit
    [tex]\lim_{P \rightarrow 0, Q \rightarrow 0} \frac{f(Q) - f(P)}{Q - P}[/tex]
    which, if f is continuous, is an alternative definition for the derivative at zero. The geometric interpretation is that you are looking at the limiting behavior of slope of the chord through the points (P,f(P)) and (Q,f(Q)), as you vary the parameters P and Q.

    For f(x) = |x|, this limit doesn't exist -- the set of limit points is the entire interval [-1,1]. (a limit exists if and only if there is only one limit point) For a number to be a limit point as x approaches zero, we don't require that all "small" values of x are "near" the number, we only require that some of them are. (I can give a precise definition if you like....)

    Basically, the point is that if you vary P and Q in a specific way (rather than ranging through all allowed values), you can get the limit to apprach any number in [-1,1] that you want.

    If you define "tangent line" to mean the slope of the line just has to be a limit point of that expression, then you get lots of tangent lines. For example, using this meaning, the line defined by y=0 would be tangent to the curve defined by y=f(x) at (0,0).



    Oh, and a P.S. for zcd: the tangent line at (0.1,0.1) and the tangent line at (0.2,0.2) are the same line. And you seem to have forgotten about the tangent line at (-0.1,0.1)....
     
  8. May 24, 2009 #7
    OP:

    Hurkyl gives a very good explanation of why there are infinitely many "tangent lines" to the curve f(x) = |x| at x=0.

    In simpler terms: a tangent line is a line that touches the function at the point and is traveling in the same direction as the function... it's like what would happen if you were standing there on the curve and shot a gun straight out in front of you. If you were standing at the point (0,0) on f, then you could be looking diagonally down to the right, diagonally up to the right, or anything in between. If you were walking backwards, you could be looking diagonally down to the left, diagonally up to the left, or anything in between. The "anything in between" part comes from the fact that you have to "turn" or "pivot" as you're walking... and therefore at the point (0,0) you pivot over all of these values as you "walk" the function, so who's to say which way the function is going if somebody only gets to see a snapshot of you while you're at that location?
     
  9. May 24, 2009 #8
    I have a much better understanding of this now. Many thanks everyone for their help.

    Luca
     
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