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Math Unit III: More on the derivative and differential equations

In the last unit, we discussed how mathematicians and scientists deal with quantities that change in ways other than linearly. The key idea turns out to be the rate of change of the quantity. The lowbrow way to measure the rate of change is to compute the average rate of change over a small interval. The high-class way to talk about change is to try to compute the derivative, or instantaneous rate of change.

In real-world situations, when one deals with measured data, it is often the case that the data has been measured only for certain specific values of the independent variable. Then, one can only compute average rates of change between the data points. But in mathematics, when one deals with abstractly-defined functions, it is possible to compute derivatives. Often, mathematical models are developed using derivatives, predictions are made based upon these mathematical models, and then experimental results are compared to the predictions to see how well the models reflect reality.

1. Review and extensions

To get started, we review the formulas for average and instantaneous rates of change. Let f(x) be a function defined for all values of x on some interval. Recall that the average rate of change of f over a (short) interval [a, a+h] is given by the quantity g_av(a,h) where

g_av(a,h) = (f(a + h) - f(a))/h

Geometrically, this is just the slope of the line connecting the two points (a, f(a)) and (a + h, f(a + h)) on the graph of y = f(x). For example, suppose, f(x) = x² . Then
f(a + h) = (a + h)² = a² + 2ah + a² and so

g_av(a,h) = ((a + h)² - a²)/h = (2ah + h²)/h = 2a + h

The derivative, or instantaneous rate of change of f(x) at a value of x, say x = a, is the ``limiting value'' of the average rate of change of f over [a, a + h] as h decreases to zero. The general expression for this is

.

So for our example of f(x) = x² , we see that

           f´ (a) = 2a

Do exercise 3-1 before proceeding.

This week, we need a few rules that derivatives obey, in order to continue our work with differential equations. It turns out that approximate versions of the rules are easy to derive for average rates of change, and then it becomes apparent what happens as h approaches zero.

The sum rule.When a function is defined as the sum of two other functions (think of f(x) = x⁵ + x³, for example) we can compute the derivative of the sum in terms of the derivatives of the addends.

Suppose f(x) = u(x) + v(x). We need to consider the three average rate of change functions:

Now let's notice something:

Thus, the process of taking average rates of change is additive, i.e., it respects sums. As we let h get smaller and smaller, this additive property holds all the time, so we expect (and get) that the property holds in the limit:

          f´ (x) = u´ (x) + v´ (x)

Get some practice with the sum rule by doing exercise 3-2.

The product rule.
It's a little more complicated if f is defined as the product of two other functions. Suppose f(x) = u(x)v(x). Using the average rate of change functions we defined before, we compute:

(Did you notice the clever way we added zero in the second line there?) The product rule for average rates of change is pretty complicated:

But it becomes simpler when we let . As , the expressions become derivatives, and v(x + h) becomes simply v(x):

         f´ (x) = u´ (x)v(x) + u(x)v´ (x)

Do exercise 3-3.

One special case of the product rule is worth a special mention: If one of the factors u or v or is a constant function then the product rule simplifies. This is because the derivative of a constant function is zero (Why?). So we get what some people call the ``constant-times-a-function rule''

The ``stretch'' rule.How do the graphs of y = f(x) and y = f(2x) compare? Since the argument of the function f(x) in the second expression changes twice as fast as that of the first expression, you should expect that the graphs of the two expressions would look qualitatively the same, but the horizontal scale of the second would be compressed by a factor of two (you have seen this already in Unit 1 with the trigonometric functions). Intuitively, you might expect that the second expression will change twice as fast as the first, so the average (and instantaneous) rate of change of f(2x) should be twice as big as that of f(x). Let's verify this algebraically: Let c be a constant, and set f(x) = u(cx). Then

(notice that we multiplied by 1 in a clever way this time). This is again a complicated rule. But notice that as h -> 0, it is also true that ch -> 0. so g_av,u(cx, ch) -> u´ (cx). Thus we have the stretch rule:

            (f(cx))´ = cf´ (x)

Verify the stretch rule by specific application in exercise 3-4, then summarize this section in exercise 3-5.