## Rules Of Differentiation Involving Two Or More Functions Of The Same Variable

The three rules presented in the preceding section are each concerned with a single given function f(x). Now suppose that we have two dijferentiable functions of the same variable x, say, /(x) and g(x), and we want to differentiate the sum, difference, product, or quotient formed with these two functions. In such circumstances, are there appropriate rules that apply? More concretely, given two functions—say, f{x) = 3x2 and g(x) = 9x'2—how do we get the derivative of, say, 3x2 + 9x'2, or the derivative of (3jc2)(9x12)?

Sum-Difference Rule

The derivative of a sum (difference) of two functions is the sum (difference) of the derivatives of the two functions:

The proof of this again involves the application of the definition of a derivative and of the various limit theorems. We shall omit the proof and, instead, merely verify its validity and illustrate its application.

Example 1 From the function y = I4x3, we can obtain the derivative dy/dx = 42x2. But 14x3 = 5x3 + 9x3, so that y may be regarded as the sum of two functions f(x) = 5x3 and g(x) = 9x3. According to the sum rule, we then have

^ = -f (5x3 + 9x3) = -f 5x3 + -j- 9x3 = 15x2 + 27x2 = 42x2

which is identical with our earlier result.

This rule, stated above in terms of two functions, can easily be extended to more functions. Thus, it is also valid to write

Example 2 The function cited in Example 1, y = 14x3, can be written as y = 2x3 + I3x3 - x3. The derivative of the latter, according to the sum-difference rule, is

^ = ¿(2*3 + 13x3 - x3) = 6x2 + 39x2 - 3x2 = 42x2 which again checks with the previous answer.

This rule is of great practical importance. With it at our disposal, it is now possible to find the derivative of any polynomial function, since the latter is nothing but a sum of power functions.

Example 4

^(7x4 + 2x3 - 3x + 37) = 28x3 + 6x2 - 3 + 0 = 28x3 + 6x2 - 3

Note that in the last two examples the constants c and 37 do not really produce any effect on the derivative, because the derivative of a constant term is zero. In contrast to the multiplicative constant, which is retained during differentiation, the additive constant drops out. This fact provides the mathematical explanation of the well-known economic principle that the fixed cost of a firm does not affect its marginal cost. Given a short-run total-cost function

the marginal-cost function (for infinitesimal output change) is the limit of the quotient AC/Ag, or the derivative of the C function:

whereas the fixed cost is represented by the additive constant 75. Since the latter drops out during the process of deriving dC/dQ, the magnitude of the fixed cost obviously cannot affect the marginal cost.

In general, if a primitive function y = f(x) represents a total function, then the derivative function dy/dx is its marginal function. Both functions can, of course, be plotted against the variable x graphically; and because of the correspondence between the derivative of a function and the slope of its curve, for each value of x the marginal function should show the slope of the total function at that value of x. In Fig. 7.1a, a linear (constant-slope) total function is seen to have a constant marginal function. On the other hand, the nonlinear (varying-slope) total function in Fig. l.\b gives rise to a curved marginal function, which lies below (above) the horizontal axis when the total function is negatively (positively) sloped. And, finally, the reader may note from Fig. 7.1c (cf. Fig. 6.5) that " nonsmoothriess" of a total function will result in a gap (discontinuity) in the marginal or derivative function. This is in sharp contrast to the everywhere-smooth total function in Fig. l.\b which gives rise to a continuous marginal function. For this reason, the smoothness of a primitive function can be linked to the continuity of its derivative function. In particular, instead of saying that a certain function is smooth (and differentiable) everywhere, we may alternatively characterize it as a function with a continuous derivative function, and refer to it as a continuously differentiable function.

### Product Rule

The derivative of the product of two (differentiable) functions is equal to the first function times the derivative of the second function plus the second function

Figure 7.1

times the derivative of the first function: (7-4) £[f(x)g(x)] =f(x)±g(x)+g(x)±f(x)

Example 5 Find the derivative of y = (2x + 3)(3x2). Let f(x) = 2x + 3 and g(x) = 3x2. Then it follows that f'(x) = 2 and g'(x) = 6x, and according to (7.4) the desired derivative is

¿[(2* + 3)(3x2)] = (2x + 3)(6x) + (3x2)(2) = 18x2 + 18*

This result can be checked by first multiplying out /(x)g(x) and then taking the derivative of the product polynomial. The product polynomial is in this case /(x)g(x) = (2x + 3)(3jc2) = 6x3 + 9x2, and direct differentiation does yield the same derivative, 18x2 + 18x.

The important point to remember is that the derivative of a product of two functions is not the simple product of the two separate derivatives. Since this differs from what intuitive generalization leads one to expect, let us produce a proof for (7.4). According to (6.13), the value of the derivative of f(x)g(x) when x = N should be

But, by adding and subtracting f(x)g(N) in the numerator (thereby leaving the original magnitude unchanged), we can transform the quotient on the right of (7.5) as follows:

Substituting this for the quotient on the right of (7.5) and taking its limit, we then get

The four limit expressions in (7.5') are easily evaluated. The first one is/(iV), and the third is g(N) (limit of a constant). The remaining two are, according to (6.13), respectively, g'(N) and f'(N). Thus (7.5') reduces to

And, since N represents any value of x, (7.5") remains valid if we replace every N symbol by x. This proves the rule.

As an extension of the rule to the case of three functions, we have

In words, the derivative of the product of three functions is equal to the product of the second and third functions times the derivative of the first, plus the product of the first and third functions times the derivative of the second, plus the product of the first and second functions times the derivative of the third. This result can be derived by the repeated application of (7.4). First treat the product g(x)h(x) as a single function, say, <p(x), so that the original product of three functions will become a product of two functions, f(x)<f>(x). To this, (7.4) is applicable. After the derivative of f(x)<j>(x) is obtained, we may reapply (7.4) to the product g(x)h(x) = <j>(x) to get <f>'(x). Then (7.6) will follow. The details are left to you as an exercise.

The validity of a rule is one thing; its serviceability is something else. Why do we need the product rule when we can resort to the alternative procedure of multiplying out the two functions f(x) and g(x) and then taking the derivative of the product directly? One answer to that question is that the alternative procedure is applicable only to specific (numerical or parametric) functions, whereas the product rule is applicable even when the functions are given in the general form. Let us illustrate with an economic example.

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