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Derivative and function terminology

  1. Sep 3, 2015 #1
    In mathematical parlance, we say "take the derivative of a function f" to indicate that we are computing a new function, which maps slopes, that derives from f. However, in physics, we say "take the derivative of velocity". However, velocity is a quantity, not a function. What does it mean to take a derivative of a quantity? This goes for mathematics in general. If we have y = f(x), y is a quantity, not a function that maps between two sets. So what does it mean to "take the derivative of y"?
  2. jcsd
  3. Sep 3, 2015 #2


    Staff: Mentor

    Velocity is a quantity that changes over time so it can be considered a function whose domain is time, hence taking the derivative of velocity means taking it over time.
  4. Sep 3, 2015 #3


    Staff: Mentor

    To elaborate on why jedishrfu said, velocity is a function of time, and y in your example is a function of x. Both functions are maps between two sets.
  5. Sep 3, 2015 #4
    Aren't velocity and y both outputs to functions though? i.e. v = f(t) means f maps time to velocity. How is velocity a function in this case if it is an "output" variable?
  6. Sep 3, 2015 #5


    Staff: Mentor

    You could call the relationship between t and v by another name, but the relationship is that of a function. Many textbooks, especially physics ones will write things like v = v(t) to indicate that v is a function of t.
  7. Sep 3, 2015 #6
    It is a conventional abuse of terminology. Although variables, like y, and expressions like f(x) are rigorously meant to represent numbers, or values, and the symbol f in f(x) is meant to represent the name of a function (while the function itself is a set of ordered pairs, or a collection of arrows, neither of which are convenient to explicitly write down in their entirety), it is common for both physicists and mathematicians to conflate all three entities and speak of "the function f(x)", "the function y = x^2", or even " the function t^2 + 1". It is a convenience that avoids unnecessary verbosity in practical applications in which other considerations are of far greater immediate importance. Someone who is interested in the structure of "what is a function" may be interested in the details, but seldom is that level of detail applicable outside of foundational considerations.
  8. Sep 6, 2015 #7
    I think I need to elaborate on my original question a bit. By a value I don't mean a constant number a. By a value I mean f(x). You have to agree that y=f(x) is not a function, correct? It is the value of the function f evaluated at x; therefore, f(x) is the value of the function evaluated at an input x, just as f(x+1) would be the value of the function evaluated at an input x+1. I'm not saying the value f(x) is a constant real number or integer; I am simply saying that it is the result of applying f to the argument x, and not actually the function f. Therefore, my point is we say that d/dx is a function such that f↦f′. This is the statement I don't understand. When we perform differentiation, we need what the value of f for any x, namely f(x), in order to compute a derivative. What we do is insert the expression f(x) into the derivative operator, not purely f. Therefore, wouldn't it be more appropriate to write d/dx : f(x)↦f′(x), since f(x) is what we're actually inserting into the derivative operator?
  9. Sep 7, 2015 #8


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    Gold Member

    When you relate a quantity ## v## (the velocity) with another quantity ##t## (the time) you obtain a relation ##t\mathcal{R}v=(t,v)##. You can represent ##\mathcal{R}## with points ## (t,v(t)) ## in the plane, this is the graph of a function that can be denoted simply by ##v(t)##.
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