INCREMENT NOTATION - The Derivative of a Function - The Calculus Primer

The Calculus Primer (2011)

Part II. The Derivative of a Function


2—1. A Convenient Symbolism. When discussing average and instantaneous rates in Chapter One, we frequently had occasion to refer to small “intervals.” We shall now introduce a convenient notation for small intervals, or for small changes in the value of a variable or a function. When a variable changes from one numerical value to another, we refer to the difference of the values as the increment of the variable; for example, if x changes from x1 to x2, then the increment equals x2x1. The increment is represented by the symbol Δx, read “delta x”; thus Δx = x2x1. The reader should note carefully that the symbol Δx does not mean “delta times x”; the “Δ” is not a number or a quantity. But Δx, taken as a single symbol, is a quantity, with the meaning we have just given it.

An increment is always equal to the new value less the original value, never the other way about. If the variable is increasing (x2 > x1), then Δx is positive; if the variable is decreasing (x2 < x1), then Δx is negative.

If in the function y = f(x) we let the independent variable x take on an increment Δx, then the corresponding increment in the function f(x), that is, the corresponding increment in y, is called Δy. Thus Δy = y2y1, or Δy = f(x2) − f(x1), where the initial values of x and y are x1 and y1, respectively, and the final values of x and y are x2 and y2, respectively. For example, if y = x3 + 5, and we start with x1 = 2, then y1 = 8 + 5 = 13; now if x takes on an increment from x1 = 2 to x2 = 3, then Δx = x2x1 = 3 − 2 = 1; also, y2 = (3)3 + 5 = 32, and hence Δy = y2y1 = 32 − 13 = 19. In other words, an increment of 1 unit in the value of x from 2 to 3 gives a corresponding increment in y of 19 units, from 13 to 32.

Thus the quantity Δy is always computed from a specific initial value of y which corresponds to the particular fixed initial value taken for x when selecting the increment Δx.

2—2. Using the Increment Notation. We shall now show how the idea of increments may serve a useful purpose.

EXAMPLE 1.Consider the function A = s2, where A is the area of a square of side s. Now let s take on a small positive increment, say Δs; the area will then take on a corresponding increment, say ΔA, and we have:

A = s2.(1)

A + ΔA = (s + Δs)2.


A + ΔA = s2 + 2s·Δs + (Δs)2.(2)

Subtracting (1) from (2):

ΔA = 2s·Δs+ (Δs)2.(3)

Dividing both sides of (3) by Δs:

images = 2s + Δs(4)

Suppose that the initial value of s = 5; then the initial area of the square is 25. By considering successively smaller values for Δs, beginning with Δs = 1, we obtain, from equation (4), the table shown here. It will be seen from the table that as we assign smaller and smaller values to Δs, the nearer the quantity images approaches the value 10. We thus see that the average rate of change in A is represented by the quantity images ; and the smaller we take the interval, or increment, Δs, the closer the average rate of change images approaches the limiting value 10. Hence the limit 10 represents the instantaneous rate of change in A at the instant when s = 5; when the side of the square equals 5, the area of the square is changing 10 times as fast as a side. In other words, the instantaneous rate of change of the area of a square is 2s times as fast as the rate of change of a side s.


images (= 2s + Δs)













EXAMPLE 2.Let us consider another illustration. Suppose an object thrown vertically upward, and the height in feet which it attains after t seconds is given by the function

h = 150t − 16t2.

We propose the question, how fast is it moving 4 seconds after its motion began, that is, when t = 4? Using the increment notation, and considering a small interval Δt beginning at t1 = 4, we have:

At t1 = 4,h1 = 150(4) − 16 (4)2 = 344.

At t2 = 4 + Δt,h2 = 150(4 + Δt) − 16(4 + Δt)2

= 344 + 22Δt − 16 (Δt)2.

Now the difference between the heights h2 and h1 is the distance that the object rose during the interval of Δt seconds; or

h2h1 = Δh = 22Δt − 16(Δt)2.(1)

Dividing equation (1) by Δt, we get:

images = 22 − 16 (Δt).

In other words, the average rate of change of the height, h, during the interval of Δt seconds, equals 22 − 16(Δt) feet.

But the rate of change in height (distance) per unit of time is the speed; hence the average speed of the object during the interval Δt is 22 — 16(Δt). For example, if

Δt = images sec.,

then images = 22 − 16 images = 20.4 ft. per sec.;

if Δt = images sec.,

thenimages = 22 − 16(.01) = 21.84 ft. per sec.

Now as Δt approaches zero, the limiting value approached by the average speed 22 − 16 (Δt) is exactly equal to 22, at the instant when t = 4. In short, the instantaneous speed at t = 4 is exactly equal to 22 ft. per sec.