Instantaneous Velocity
Key Questions

Provided that the graph is of distance as a function of time, the slope of the line tangent to the function at a given point represents the instantaneous velocity at that point.
In order to get an idea of this slope, one must use limits. For an example, suppose one is given a distance function
#x = f(t)# , and one wishes to find the instantaneous velocity, or rate of change of distance, at the point#p_0 = (t_0, f(t_0))# , it helps to first examine another nearby point,#p_1 = (t_0+a, f(t_0+a))# , where#a# is some arbitrarily small constant. The slope of the secant line passing through the graph at these points is:#[f(t_0+a)f(t_0)]/a# As
#p_1# approaches#p_0# (which will occur as our#a# decreases), our above#difference quotient# will approach a limit, here designated#L# , which is the slope of the tangent line at the given point. At that point, a pointslope equation using our above points can provide a more exact equation.If instead one is familiar with differentiation, and the function is both continuous and differentiable at the given value of
#t# , then we can simply differentiate the function. Given that most distance functions are polynomial functions, of the form#x = f(t) = at^n + bt^(n1) + ct^(n2) + ... + yt + z,# these can be differentiated using the power rule which states that for a function#f(t) = at^n, (df)/dt# (or#f'(t)# ) =#(n)at^(n1)# .Thus for our general polynomial function above,
#x' = f'(t) = (n)at^(n1) + (n1)bt^(n2) + (n2)ct^(n3) + ... + y# (Note that since#t = t^1# (as any number raised to the first power equals itself), reducing the power by 1 leaves us with#t^0 = 1# , hence why the final term is simply#y# . Note also that our#z# term, being a constant, did not change with respect to#t# and thus was discarded in differentiation).This
#f'(t)# is the derivative of the distance function with respect to time; thus, it measures the rate of change of distance with respect to time, which is simply the velocity. 
Instantaneous velocity is the velocity at which an object is travelling at exactly the instant that is specified.
If I travel north at exactly 10m/s for exactly ten seconds, then turn west and travel exactly 5m/s for another ten seconds exactly, my average velocity is roughly 5.59m/s in a (roughly) northbynorthwest direction. However, my instantaneous velocity is my velocity at any given point: at exactly five seconds into my trip, my instantaneous velocity is 10m/s north; at exactly fifteen seconds in, it's 5m/s west.
Questions
Derivatives

Tangent Line to a Curve

Normal Line to a Tangent

Slope of a Curve at a Point

Average Velocity

Instantaneous Velocity

Limit Definition of Derivative

First Principles Example 1: x²

First Principles Example 2: x³

First Principles Example 3: square root of x

Standard Notation and Terminology

Differentiable vs. Nondifferentiable Functions

Rate of Change of a Function

Average Rate of Change Over an Interval

Instantaneous Rate of Change at a Point