1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Derivatives and the relation to limits

  1. Oct 2, 2014 #1
    I'm in calc 1 and want to make sure I'm understanding the reason that we find derivatives. From what I understand, a derivative is simply an equation for the rate of change at any given point on the original function. Is that correct? And the tangent line at point (x,y) is obtained by using the x value in the derivative equation to find the slope at that point, and then use the y value along with that slope to create the equation for the tangent line, right?

    I'm a bit confused on whether or not the derivative is also an equation for instantaneous velocity at any given point in the function's domain. It seems like it is, but in the previous chapter we were using limits with the [itex]\frac{f(a + h) - f(a)}{a^2}[/itex] formula (I don't know if that's the right word, isn't it less of a formula and more of the definition of a limit?) to find the average and instantaneous velocity. So is the derivative another way to do that? I guess I'm not sure as to what the relationship is between a derivative and a limit, as it seems like last chapter we were focused on limits and now have dropped them for derivatives.. I've got to be missing some key point here.
    Last edited: Oct 2, 2014
  2. jcsd
  3. Oct 2, 2014 #2


    Staff: Mentor

    The derivative is not an equation; it's a function. It might appear in an equation, such as f'(x) = 2x + 3, for example. Here the function is one that maps (pairs) a number x with the number 2x + 3.
    If you start with a function that represents the position of an object at time t, then the derivative gives the instantaneous velocity at any time t.

    The derivative of a function is defined as a limit; more specifically the limit of the difference quotient. For a function f, the derivative at x = a is defined as
    $$f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$$
    For the derivative at an arbitrary x, just replace a with x everywhere above.

    After you go through the exercise of finding the derivative by the definition (i.e., using the limit above or one similar), the usual progression is to present some shortcut rules, such as the sum rule, product rule, quotient rule, and chain rule.
  4. Oct 2, 2014 #3


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The main reason you study limits and indeterminate forms in the first place in calculus is so you can work with expressions like$$
    \frac {f(x+h)-f(x)} h$$which is an indeterminate form ##0/0## if ##h= 0##. When the limit exists as ##h\to 0##, we denote it by ##f'(x)##, or ##\frac{dy}{dx}##. Once you learn how to calculate derivatives, you learn their formulas and use them so you don't have to do the work of calculating limits forever.

    You are correct that if you are thinking of the graph of ##y=f(x)##, the derivative ##f'(x)## represents the slope at ##(x,f(x))## on the graph. As ##x## varies, the slope at the point ##(x,f(x))## varies unless the curve is a straight line. You can think of the derivative as the instantaneous rate of change of ##f(x)## with respect to ##x##.

    Since functions can represent many things, the interpretation of what the derivative represents does also. So if you are looking at an equation describing the position ##s## at time ##t## such as ##s = f(t)##, the instantaneous rate if change of ##s## with respect to ##r## would be the instantaneous velocity.

    In a business setting you might have the cost ##C## of producing quantity ##Q## given by ##C=f(Q)##. In this case ##f'(Q)## is the marginal cost, which can be interpreted as the cost of producing the next unit.

    My point is that the mathematics is the same. The interpretation of what a derivative represents depends on the setting.

    Finally, derivatives are not "another way" to calculate these things. They are the only way.
  5. Oct 3, 2014 #4


    User Avatar
    Science Advisor

    "Velocity" is physics, not mathematics. IF the function happens to be describing the "position" of some object, then the derivative, the "function's instantaneous rate of change" can be interpreted as the object's instantaneous speed.

    No, it's a formula for the "difference quotient" or "averagea rate of change". You haven't yet taken the limit to find the derivative (the instantaneous rate of change).
    (And the difference quotient is [itex]\frac{f(a+ h)- f(a)}{h}[/itex], NOT "[itex]\frac{f(a+h)- f(a)}{a^2}[/itex]"!)

    There are many different uses for a limit. The derivative is one of them. The derivative is the limit of the difference quotient. If you think of the given function as describing the position of some object (again, that's a physics application of the mathematics, not the mathematics itself) then you can think of the difference quotient as the "average velocity" and the derivative as the "instantaneous velocity".
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted