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Simple derivative question

  1. Aug 13, 2009 #1
    given only f ' (x) , is it possible to find the slope of the secant for a given x_1 and x_2 without integrating?
    If so how?
  2. jcsd
  3. Aug 13, 2009 #2


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    No, that is not possible.

    Note, however, that you will not need to specify f precisely, it is sufficient to determine it up to an arbitrary constant, since the secant expression will be unique.

    (The secant is only sensitive to the function values at two discrete points.

    The derivative is, however, a "neighbourhood" quality.

    This is the reason why the derivatives at two points are insufficient to determine the secant value between them.)
  4. Aug 13, 2009 #3
    I am bothered by something.
    The way i was taught it was that that the derivative of a function described its slope.
    And given its slope and how fast it changes (assume f '' (x) is known), how could it not be possible to get the secant? would it be possible if f(c) was a known value?
  5. Aug 13, 2009 #4
    The problem is that f ' (x) is the infinitesimal rate of change, at a single point. So in order to find the slope the secant is just to find the average derivative between the two points, i.e. the integral of f ' (x) from x_1 to x_2 divided by x_2 - x_1.

    However just by knowing the value of f ' (x_1) and f ' (x_2) you cannot find the secant line as you don't know what it is happening to the function inbetween. They could be, say, horizontal (local maxima and minima) and inbetween there could be a massive downslope or what-have-you. There is no way with just that information.
  6. Aug 13, 2009 #5
    but lets say you were given the general f'(x) , not a specific value. then its possible? and is there a way to do it without integration?
  7. Aug 13, 2009 #6
    Such a method would surely blow my mind and, along with it, my view of calculus. The only way I can think of finding a secant line slope while knowing f ' (x) is only with integration. The only way to get from the infinitesimal description of a function which is the derivative to the large scale description is via integration.

    Although, it should be noted, my experience with calculus is far from rigorous, I would still put a moderate sum of money on betting that what you describe is impossible.
  8. Aug 13, 2009 #7
    hypothetically speaking, given f ' (x) and f '' (x) could one not construct a "weighted average" giving the secant of f (x)?
  9. Aug 13, 2009 #8
    I don't quite follow. If you mean using f ' (x) and f '' (x) at the two given points, then you run into problems of not knowing what is going on inbetween the two points. If you then incorporate, say, the f ' (x) and f ''(x) of a point between the two points into your calculation you will have a very rough answer, which can be made more accurate by incorporating more and more points, the limit of this process is the integral, and a definite answer.
  10. Aug 13, 2009 #9
    I realized im not clear. When i say f(x), f ' (x) or f '' (x) i mean the general formula. when i do something like f(x_1) then its specific.
    so i meant that assuming that you knew what happened in the middle (as you have the whole f ' (x) not just a point).
  11. Aug 13, 2009 #10
    Yes, I understand that. However the problem is that at some point along your process you will HAVE to evaluate f ' (x) or f '' (x) at some point, correct? and, unless f(x) is linear, this will just give a rough approximation (and sometimes completely inaccurate).

    Say f ' (x) = 2x^2+1, we want to find the secant line from -1 to 1. evaluating at the two points one gets f ' (-1)= 3 and f ' (1) = 3 We take f ' (0) = 1, however this is just one small part of the larger picture of the function. so lets say my original function was f(x)=(2/3)x^3+x+c (found by integration), so the secant line has a slope of

    del(y)/del(x) = (5/3-(-5/3))/2 = 10/6

    which is CLOSE to 1, but not percise. What we could do perhaps is take f ' (-1/3) and f ' (1/3) and average them and perhaps you would get a little closer, and then divide it again and again. But the limit of this process is just the integral.

    does that help to explain anything? or am i misunderstanding your theory?
  12. Aug 14, 2009 #11
    here is an example:

    a man moving in a straight line increases his speed quadratically. in other words, his speed
    s = t^2
    one can plot the derivative (2x)
    and then get the secant from summing the derivatives - the domain change (or something like it)
    so from 0 - 10, it would be (0+2+4+6...+18) = 110, and the -10 gives the correct secant slope of 100.

    in the case of a non-linear derivative, one take the second derivative, and that becomes the increment (not 2).

    please let me know if its still unclear
    Last edited: Aug 14, 2009
  13. Aug 14, 2009 #12
    However your still taking a sum of points. It is just coincidental that it works for that particular domain for that relatively simple function. You will quickly see as you get into more complicated territory, like f ' (x) = x^5 + (1/2) x^4 + 3 x^3 +2 x^2 + 100 x + 99, you'll find any method that involves taking derivatives, or second derivates, or upteenth derivatives of all integer vales will only get an approximation. If you, say, take the values at 1/2 integer values, it will get closer, and closer still as you take the limit.
  14. Aug 14, 2009 #13
    im not sure thats what you want...
    here is what i mean.
    lets say the points are on a quadratic, summing them all up (assuming it starts at zero), is something that has a solution 1/6(1+j)(2+j) for j terms. even if the values in the middle are not used, it will always hold. for the equation you gave me, the sum of j terms is:
    [tex]\frac{1}{60} (1+j) (5940+3019 j+81 j^2+59 j^3+26 j^4+10 j^5)[/tex]
    if j started at zero (otherwise, a shift is needed, but that is trivial). i cant check now, but if the above does not work to find the secant of f(x), then try:
    [tex]\frac{1}{3} (1+j) (300+j (10+j (11+3 j (2+j))))[/tex]
    this should i think give the secant on f ' (x), but im not sure.
    does this make more sense?
  15. Aug 15, 2009 #14


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    Sure enough, and you won't need the second derivative.

    The weighted average of the derivative in the interval between x1 and x2 IS the secant value for the function between x1 and x2.

    However, to compute that weighted average, it will look like this:
    Last edited: Aug 15, 2009
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