*In a graph of position vs. time, the instantaneous velocity at any given point #p(x,t)# on the function #x(t)# is the derivative of the function #x(t)# with respect to time at that point.*

The **derivative** of a function at any given point is simply the instantaneous rate of change of the function at that point. In the case of a graph of position (or distance) vs. time, that means that the derivative at a given point #p_0(t_0, x_0)# is the instantaneous rate of change in position (accounting for "positive" and "negative" direction) with respect to time.

As an example, consider a linear distance function (that is, one which can be represented with a line as opposed to a curve). If this were a function of #x# and #y#, with #y# as the dependent variable, then our function in slope-intercept form would take the form #y=mx+b#, where #m# is the slope and #b# is the value of #y# at #x=0#. In this case, #t# is our independent variable and #x# is our dependent, so our linear function would take the form #x(t) = mt+b#.

From algebra, we know that the slope of a line measures the number of units of change in the dependent variable for every single unit of change in the independent variable. Thus, in the line #x(t) = 2t + 5#, for every one unit by which #t# increases, #x# increases by 2 units. If we were to, for example, assign units of seconds to #t# and feet to #x#, then every second that passed (that is, every increase of one second in #t#), position (or distance) would increase by two feet (that is #x# would increase by two feet)

Since our change in distance per unit of change in time will remain the same no matter our starting point #(x_0,t_0)#, in this case we can be assured that our instantaneous velocity is the same throughout. Specifically, it is equal to #m = 2#. Differentiating the function with respect to #t# yields the same answer. Note that this is only identical to our **average velocity** throughout the function by design: for a non-linear function (such as #x(t) = t^2#) this would not be the case, and we would need to use differentiation techniques to find the derivatives of such functions.