Rule 6 Derivative of the Sum of an Arbitrary but Finite Number of Functions hx gx

This result also applies to the case where some or all of the operations involve subtraction rather than addition.

Rule 6 is a straightforward generalization of rule 5. That is. since :he derivative of the sum of two functions is simply the sum of the derivatives of the functions taken separately, then doing this iteratively allows one to establish rule 6. Thus, for example, if h(x) = v4 + 8.v2 + 2a. we can treat h(x) as the sum of two functions, fix) = (a-4 + 8.v-) and g(x) = 2a, and write li'ix) = /'(a) + g'ix) = fix) + 2 using rule 5. Apply rule 5 again to the function / (a) to obtain fix) = 4.v'+ I for. Substitution gives h'ix) = 4a3 + I 6a+ 2. Of course, if we treat each term a4. 8a3. and 2a as separate functions, then we can simply apply rule 6 to establish directly that h'ix) = 4x} -f- 16a + 2.

Rule 6 offers a useful notation for writing out the derivative for a general polynomial function. That is, for the function n h(x) = ci() + iiyx + tii.r2 + Uj.r5 H-----1- a„x" = ^ ay.t1

we can use rule 6 to write n

Rule 7 Derivative of the Product of Two Functions lfh(x) = fix) ■ g(x), then h'(x) = ,f'(x)g(x) + f(x)g'ix).

Example 5.9 Find the derivative of lux) = (6a4 + 2a- )(5.v - 10a3 + 18.x5 - 4)

Solution

we get

= (24aj + bx2)i5x - 10a"2 + 18-r5 - 4) + (6a:4 - lr3)(5 - 20.r + 90.v4|

This example shows the usefulness of this rule; that is. it is often much simpler to use rule 7 in this manner than it is to expand the original expression for h(x) and find the derivatives term by term. Moreover, if we are faced with an expression of the form h(x) = /(x)g(x) but only know certain properties of functions fix) and ), we can sometimes gain some insights by using this technique. The following example illustrates this point.

Marginal Revenue Function for a Competitive Firm and a Monopoly Firm

The total revenue of either a competitive firm or a simple monopolist is the unit price of its output times Ihe quantity it produces/sells. Thus we write TRuy) = pq. A competitive firm treats the price as a constant value, equal to the market price, p. Thus we can write TR(q) - pq for total revenue, and so marginal revenue is MR(</) = dTR(q)/dq - p (recall example 5.4). A monopolist, however, is the only firm in ihe industry and so recognizes that the amount it can sell is determined by the price it sets according to the market demand function, q - D(p). Writing the demand function in its inverse form, p — D {(q) = p(q). we see that the monopolist's total revenue function is

Thus the monopolist's total revenue function is the product of two functions. p{q) and q, and so we need to use the product rule to find MRu/ ). We then have

MR(^) = p'(q)q + p where p'(q) is the derivative of the inverse demand function. If We consider the usual case of price being negatively related to quantity sold. p'(q)<(). then it follows that the term p'(q Iq is negative, and so MR(<y ) < p\ that is, the marginal revenue of increased sales is less than the unit price being charged at any q > 0.

To understand the result above, work through the following steps, which arc illustrated in figure 5.24. A firm sells output q at the price p - p(q). To sell an additional bit of output A q. it must reduce price on all the units it sells by A p.

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