**Student Question:** Would you please interpret in “English” – ∆y/∆x? (It’s not about division – right?) Would you also interpret ∆y/∆x as compared to dy/dx?

**Garven Response: **In plain “English”, ∆y/∆x represents the *rate* at which y changes with respect to changes in x. It is a ratio, so it *is* all about division. The only difference between ∆y/∆x and dy/dx* *is that the former ratio involves *discrete* changes in the y and x variables, whereas the latter ratio involves *continuous* changes. Here, the variable y is related in some way to the variable x; i.e., y is a *function* of x, or in shorthand, y = f(x), and this ratio indicates the *slope* of f(x).

In the 1st lecture note on pp. 14-15, I show that in the case of a straight line, y = 10+5x, ∆y/∆x = dy/dx = 5 for all possible values of x. Next, let’s calculate the slope using both ∆y/∆x and dy/dx. Suppose that the initial value for x is 1, and we change x’s value to 2. Then ∆x = 2 – 1 = 1. If x = 1, then y = 10+5x = 10+5(1) = 15, and if x = 2, then y = 10+5x = 10+5(2) = 20. Thus, ∆y = 20 – 15 = 5, so ∆y/∆x = 5/1 = 5. In order to find dy/dx* *for a linear function like y = 10+5x, we use the “constant” rule (see page 21 in the 1st lecture note for a listing of the calculus rules that we typically use in ECO 5315). The constant rule indicates that if y = a + bX, then the slope dy/dx = b (note that dy/da = 0 because a is a constant). Thus, if y = 10+5x, then dy/dx = 5. Also, note that on p. 14 of the 1st lecture note, the definition shown there for dy/dx simply involves taking the *limit* of ∆y/∆x as ∆x becomes a small number. In other words, the definition of a continuous change is that it is the same as an arbitrarily small discrete change. There, ∆y = 5∆x, so ∆y/∆x = 5∆x/∆x = 5. In the case of a line, no matter how large or small ∆x is, the slope is constant. However, this is not the case for other (nonlinear) functions of x; e.g., parabolas. On pp. 16-17 of the 1st lecture note, I show how slopes change depending upon x’s value.