[Fig. 2.1]
This chapter begins with the definition of the derivative. Two examples were in Chapter 1. When the distance is t², the velocity is 2t. When ƒ(t) = sin t we found v(t)= cos t. The velocity is now called the derivative of ƒ(t). As we move to a more formal definition and new examples, we use new symbols ƒ' and dfldt for the derivative.
2A At time t, the derivative ƒ'(t) or dƒ/dt or v(t) is ƒ'(t) = limΔt→0 (ƒ(t+Δt) − ƒ(t)) ⁄ Δt (1)
The ratio on the right is the average velocity over a short time Δt. The derivative, on the left side, is its limit as the step Δt (delta t) approaches zero.
Go slowly and look at each piece. The distance at time t + Δt is ƒ(t + Δt). The distance at time t is ƒ(t). Subtraction gives the change in distance, between those times. We often write A f for this difference: Δƒ = ƒ(t+Δt) − ƒ(t). The average velocity is the ratio Δƒ/Δt—change in distance divided by change in time.
The limit of the average velocity is the derivative, if this limit exists:
dƒ ⁄ dt = limΔt→0 Δƒ ⁄ Δt (2)
This is the neat notation that Leibniz invented: Δƒ ⁄ Δt approaches dƒ ⁄ dt. Behind the innocent word "limit" is a process that this course will help you understand.
Note that Δƒ is not Δ times ƒ! It is the change in ƒ. Similarly Δt is not Δ times t. It is the time step, positive or negative and eventually small. To have a one-letter symbol we replace Δt by h.
The right sides of (1) and (2) contain average speeds. On the graph of ƒ(t), the distance up is divided by the distance across. That gives the average slope Δƒ ⁄ Δt.
The left sides of (1) and (2)are instantaneous speeds Δƒ ⁄ Δt. They give the slope at the instant t. This is the derivative dfldt (when Δt and Δƒ shrink to zero). Look again at the calculation for ƒ(t) = t²:
Δƒ ⁄ Δt = (ƒ(t+Δt) − ƒ(t)) ⁄ Δt = (t² + 2tΔt + (Δt)² − t²) ⁄ Δt = 2t + Δt (3)
Important point: Those steps are taken before At goes to zero. If we set Δt = 0 too soon, we learn nothing. The ratio Δƒ ⁄ Δt becomes 0/0 (which is meaningless). The numbers Δƒ and Δt must approach zero together, not separately. Here their ratio is 2t + Δt, the average speed.
To repeat: Success came by writing out (t+Δt)² and subtracting t² and dividing by At. Then and only then can we approach Δt = 0. The limit is the derivative 2t.
There are several new things in formulas (1) and (2). Some are easy but important, others are more profound. The idea of a function we will come back to, and the definition of a limit. But the notations can be discussed right away. They are used constantly and you also need to know how to read them aloud:
ƒ(t) = "ƒ of t" = the value of the function ƒ at time t
Δt = "delta t" = the time step forward or backward from t
ƒ(t+Δt) = "ƒ of t plus delta t" = the value of ƒ at time t+Δt
Δƒ = "delta ƒ" = the change ƒ(t+Δt) − ƒ(t)
Δƒ ⁄ Δt = "delta ƒ over delta t" = the average velocity
ƒ'(t)= "ƒ prime of t" = the value of the derivative at time t
dƒ/dt = "dƒdt" = the same as ƒ' (the instantaneous velocity)
limΔt→0 = "limit as delta t goes to zero" = the process that starts with numbers Δƒ ⁄ Δt and produces the number dƒ/dt.
From those last words you see what lies behind the notation dƒ/dt. The symbol Δt indicates a nonzero (usually short) length of time. The symbol dt indicates an infinitesimal (even shorter) length of time. Some mathematicians work separately with df and dt, and dƒ/dt is their ratio. For us dƒ/dt is a single notation (don't cancel d and don't cancel Δ). The derivative dfldt is the limit of Δƒ ⁄ Δt. When that notation dfldt is awkward, use ƒ' or v.
Remark The notation hides one thing we should mention. The time step can be negative just as easily as positive. We can compute the average Δƒ ⁄ Δt over a time interval before the time t, instead of after. This ratio also approaches dƒ/dt.
The notation also hides another thing: The derivative might not exist. The averages Δƒ ⁄ Δt might not approach a limit (it has to be the same limit going forward and backward from time t). In that case ƒ'(t) is not defined. Δt that instant there is no clear reading on the speedometer. This will happen in Example 2.
EXAMPLE 1 (Constant velocity V=2) The distance ƒ is V times t. The distance at time t+Δ is V times t+Δt. The diference Δƒ is V times Δt:
Δƒ ⁄ Δt = VΔt ⁄ Δt = V so that limit is dƒ ⁄ dt = V.
The derivative of Vt is V. The derivative of 2t is 2. The averages Δƒ ⁄ Δt are always V = 2, in this exceptional case of a constant velocity.
EXAMPLE 2 Constant velocity 2 up to time t = 3, then stop.
For small times we still have ƒ(t) = 2t. But after the stopping time, the distance is fixed at ƒ(t) = 6. The graph is flat beyond time 3. Then ƒ(t+Δt) = ƒ(t) and Δƒ = 0 and the derivative of a constant function is zero:
t>3: ƒ'(t) = limΔt→0 (ƒ(t+Δt) − ƒ(t)) ⁄ Δt = limΔt→0 0 ⁄ Δt = 0.
In this example the derivative is not defined at the instant when t = 3. The velocity falls suddenly from 2 to zero. The ratio Δƒ ⁄ Δt depends, at that special moment, on whether Δt is positive or negative. The average velocity after time t = 3 is zero. The average velocity before that time is 2. When the graph of ƒ has a corner, the graph of v has a jump. It is a step function.
One new part of that example is the notation (dƒ/dt or ƒ' instead of v). Please look also at the third figure. It shows how the function takes t (on the left) to ƒ(t). Especially it shows Δt and Δf. Δt the start, Δƒ ⁄ Δt is 2. After the stop at t = 3, all t's go to the same ƒ(t) = 6. So Δƒ = 0 and dƒ/dt = 0.
[Fig. 2.1]
Here is a completely different slope, for the "demand function" ƒ(t) = 1/t.The demand is 1/t when the price is t. A high price t means a low demand 1/t. Increasing the price reduces the demand. The calculus question is: How quickly does 1/t change when t changes? The "marginal demand" is the slope of the demand curve.
The big thing is to find the derivative of 1/t once and for all. It is -1/t².
EXAMPLE 3 ƒ(t) = 1/t has Δƒ = 1 ⁄ (t+Δt). This equals t − (t + Δt) ⁄ (t(t + Δt)) = -Δt ⁄ (t(t + Δt))
Divide by Δt and let Δt→0: Δƒ ⁄ Δt = -1 ⁄ (t(t + Δt)) approaches dƒ/dt = -1 ⁄ t².
Line 1 is algebra, line 2 is calculus. The first step in line 1 subtracts ƒ(t) from ƒ(t+Δt). The difference is 1/(t+Δt) minus 1/t. The common denominator is t times t+Δt—this makes the algebra possible. We can't set At = 0 in line 2, until we have divided by Δt.
The average is Δƒ ⁄ Δt = -1 ⁄ (t(t+Δt)). Now set At = 0. The derivative is -1/t². Section 2.4 will discuss the first of many cases when substituting At = 0 is not possible, and the idea of a limit has to be made clearer.
[Fig. 2.2]
Check the algebra at t = 2 and t + Δt = 3. The demand 1/t drops from 1/2 to 1/3. The difference is Δƒ = -1/6, which agrees with -1/(2)(3) in line 1. As the steps Δƒ and Δt get smaller, their ratio approaches -1/(2)(2)= -1/4.
This derivative is negative. The function 1/t is decreasing, and Δƒ is below zero. The graph is going downward in Figure 2.2, and its slope is negative:
An increasing ƒ(t) has positive slope. A decreasing ƒ(t) has negative slope.
[Fig. 2.3]
The slope -1/t² is very negative for small t. A price increase severely cuts demand.
The next figure makes a small but important point. There is nothing sacred about t. Other letters can be used—especially x. A quantity can depend on position instead of time. The height changes as we go west. The area of a square changes as the side changes. Those are not affected by the passage of time, and there is no reason to use t. You will often see y = ƒ(x), with x across and y up—connected by a function ƒ.
Similarly, ƒ is not the only possibility. Not every function is named f! That letter is useful because it stands for the word function-but we are perfectly entitled to write y(x) or y(t) instead of ƒ(x) or ƒ(t). The distance up is a function of the distance across. This relationship "y of x" is all-important to mathematics.
The slope is also a function. Calculus is about two functions, y(x) and dy/dx.
Question If we add 1 to y(x), what happens to the slope? Answer Nothing.
Question If we add 1 to the slope, what happens to the height? Answer .
The symbols t and x represent independent variables—they take any value they want to (in the domain). Once they are set, ƒ(t) and y(x) are determined. Thus ƒ and y represent dependent variables—they depend on t and x. A change Δt produces a change Δƒ. A change Δx produces Δy. The independent variable goes inside the parentheses in ƒ(t) and y(x). It is not the letter that matters, it is the idea:
independent variable t or x
dependent variable ƒ or g or y or z or u
derivative dƒ/dt or dƒ/dx or dy/dx or …
The derivative dy/dx comes from [change in y] divided by [change in x]. The time step becomes a space step, forward or backward. The slope is the rate at which y changes with x. The derivative of a function is its "rate of change."
I mention that physics books use x(t) for distance. Darn it.
To emphasize the definition of a derivative, here it is again with y and x:
Δy ⁄ Δx = Δy(x+Δx) − y(x) ⁄ Δx = distance up ⁄ distance across dy ⁄ dx = limΔt→0 dy ⁄ dx y'(x).
The notation y'(x) pins down the point x where the slope is computed. In dy/dx that extra precision is omitted. This book will try for a reasonable compromise between logical perfection and ordinary simplicity. The notation dy/dx(x)is not good; y'(x) is better; when x is understood it need not be written in parentheses.
You are allowed to say that the function is y = x² and the derivative is y' = 2x —even if the strict notation requires y(x)= x² and y'(x) = 2x. You can even say that the function is x² and its derivative is 2x and its second derivative is 2—provided everybody knows what you mean.
Here is an example. It is a little early and optional but terrific. You get excellent practice with letters and symbols, and out come new derivatives.
EXAMPLE 4 If u(x) has slope du/dx, what is the slope of ƒ(x)= (u(x))²?
From the derivative of x² this will give the derivative of x4. In that case u = x² and ƒ = x4. First point: The derivative of u² is not (du/dx)². We do not square the derivative 2x. To find the "square rule" we start as we have to—with Δƒ = ƒ(x+Δx) − ƒ(x):
Δƒ = (u(x+Δx))² − (u(x))² = [u(x+Δx) + u(x)][u(x+Δx) − u(x)].
This algebra puts Δƒ in a convenient form. We factored a² − b² into [a + b] times [a − b].Notice that we don't have (Δu)². We have Δƒ, the change in u². Now divide by Δx and take the limit:
Δy ⁄ Δx = [u(x+Δx) + u(x)][(u(x+Δx) − u(x)) ⁄ Δx] approaches 2u(x)du/dx
This is the square rule: The derivative of (u(x))² is 2u(x) times du/dx. From the derivatives of x² and 1/x and sin x (all known) the examples give new derivatives.
EXAMPLE 5 (u = x²) The derivative of x4 is 2u du/dx = 2(x²)(2x)= 4x3.
EXAMPLE 6 (u = 1/x) The derivative of 1/x² is 2u du/dx = (2/x)(-1/x²) = -2/x3.
EXAMPLE 7 (u = sin x, du/dx = cos x) The derivative of u² = sin² x is 2 sin x cos x.
Mathematics is really about ideas. The notation is created to express those ideas. Newton and Leibniz invented calculus independently, and Newton's friends spent a lot of time proving that he was first. He was, but it was Leibniz who thought of writing dy/dx—which caught on. It is the perfect way to suggest the limit of Δy/Δx. Newton was one of the great scientists of all time, and calculus was one of the great inventions of all time—but the notation must help. You now can write and speak about the derivative. What is needed is a longer list of functions and derivatives.
