## The Calculus Primer (2011)

### Part II. The Derivative of a Function

### Chapter 5. INCREMENT NOTATION

**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 *x*_{1} to *x*_{2}, then the increment equals *x*_{2} − *x*_{1}. The increment is represented by the symbol Δ*x*, read “delta *x*”; thus Δ*x* = *x*_{2} − *x*_{1}. The reader should note carefully that the symbol Δ*x* does *not* mean “delta times *x*”; the “Δ” is not a number or a quantity. But Δ*x*, 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 (*x*_{2} > *x*_{1}), then Δ*x* is positive; if the variable is decreasing (*x*_{2} < *x*_{1}), then Δ*x* is negative.

If in the function *y* = *f*(*x*) we let the independent variable *x* take on an increment Δ*x*, then the corresponding increment in the function *f*(*x*), that is, the corresponding increment in *y*, is called Δ*y*. Thus Δ*y* = *y*_{2} − *y*_{1}, or Δ*y* = *f*(*x*_{2}) − *f*(*x*_{1}), where the initial values of *x* and *y* are *x*_{1} and *y*_{1}, respectively, and the final values of *x* and *y* are *x*_{2} and *y*_{2}, respectively. For example, if *y* = *x*^{3} + 5, and we start with *x*_{1} = 2, then *y*_{1} = 8 + 5 = 13; now if *x* takes on an increment from *x*_{1} = 2 to *x*_{2} = 3, then Δ*x* = *x*_{2} − *x*_{1} = 3 − 2 = 1; also, *y*_{2} = (3)^{3} + 5 = 32, and hence Δ*y* = *y*_{2} − *y*_{1} = 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 Δ*y* 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 Δ*x*.

**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* = *s*^{2}, where *A* is the area of a square of side *s*. Now let *s* take on a small positive increment, say Δ*s*; the area will then take on a corresponding increment, say Δ*A*, and we have:

*A* = *s*^{2}.(1)

*A* + Δ*A* = (*s* + Δ*s*)^{2}.

Expanding:

*A* + Δ*A* = *s*^{2} + 2*s*·Δ*s* + (Δ*s*)^{2}.(2)

Subtracting (1) from (2):

Δ*A* = 2*s*·Δ*s*+ (Δ*s*)^{2}.(3)

Dividing both sides of (3) by Δ*s*:

= 2*s* + Δ*s*(4)

Suppose that the initial value of *s* = 5; then the initial area of the square is 25. By considering successively smaller values for Δ*s*, beginning with Δ*s* = 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 Δ*s*, the nearer the quantity approaches the value 10. We thus see that the average rate of change in *A* is represented by the quantity ; and the smaller we take the interval, or increment, Δ*s*, the closer the average rate of change 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 2*s* times as fast as the rate of change of a side *s*.

Δ |
(= 2s + Δ |

1.0 |
11.00 |

.5 |
10.50 |

.2 |
10.20 |

.1 |
10.10 |

.01 |
10.01 |

.001 |
10.001 |

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* = 150*t* − 16*t*^{2}.

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 Δ*t* beginning at *t*_{1} = 4, we have:

At *t*_{1} = 4,*h*_{1} = 150(4) − 16 (4)^{2} = 344.

At *t*_{2} = 4 + Δ*t*,*h*_{2} = 150(4 + Δ*t*) − 16(4 + Δ*t*)^{2}

= 344 + 22Δ*t* − 16 (Δ*t*)^{2}.

Now the difference between the heights *h*_{2} and *h*_{1} is the distance that the object rose during the interval of Δ*t* seconds; or

*h*_{2} − *h*_{1} = Δ*h* = 22Δ*t* − 16(Δ*t*)^{2}.(1)

Dividing equation (1) by Δ*t*, we get:

= 22 − 16 (Δ*t*).

In other words, the average rate of change of the height, *h*, during the interval of Δ*t* seconds, equals 22 − 16(Δ*t*) 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 Δ*t* is 22 — 16(Δ*t*). For example, if

Δ*t* = sec.,

then = 22 − 16 = 20.4 ft. per sec.;

if Δ*t* = sec.,

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

Now as Δ*t* approaches zero, the *limiting value* approached by the average speed 22 − 16 (Δ*t*) 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.