If we knowf(x)andg(x), then what are the derivatives off (x)+g(x),f (x)−g(x), f (x)ãg(x), andf (x)/g(x)? You will probably guess the first two correctly, but are less likely to be right about the last two (unless you have already learned the answers).
Sums and Differences
Supposef andgare both defined in a setAof real numbers.
D I F F E R E N T I A T I O N O F S U M S A N D D I F F E R E N C E S
If bothf andgare differentiable at x, then the sumf +gand the difference f −gare both differentiable atx, and
F (x)=f (x)±g(x) ⇒ F(x)=f(x)±g(x)
(1)
In Leibniz’s notation:
d dx
f (x)±g(x)
= d
dxf (x)± d dxg(x) Proof for the caseF(x)=f(x)+g(x): The Newton quotient ofFis
F (x+h)−F (x)
h = (f (x+h)+g(x+h))−(f (x)+g(x)) h
= f (x+h)−f (x)
h +g(x+h)−g(x) h
Whenh→0, the last two fractions tend tof(x)andg(x), respectively, and thus the sum of the fractions tends tof(x)+g(x). Hence,
F(x)=lim
h→0
F (x+h)−F (x)
h =f(x)+g(x)
The proof of the other case is similar—only some of the signs change in an obvious way.
E X A M P L E 1 Compute d dx
3x8+x100/100 . Solution: dxd
3x8+x100/100
=dxd (3x8)+dxd
x100/100
=24x7+x99, where we used (1) and the results from Example 6.6.2.
E X A M P L E 2 In Example 6.4.3,C(x)denoted the cost of producingx units of some commodity in a given period. IfR(x)is the revenue from selling thosex units, then the profit function is the difference between the revenues and the costs,π(x)=R(x)−C(x). According to (1), π(x)=R(x)−C(x). In particular,π(x)=0 whenR(x)=C(x). In words:Marginal profit is 0 when marginal revenue is equal to marginal cost.
S E C T I O N 6 . 7 / S U M S , P R O D U C T S , A N D Q U O T I E N T S 179
Rule (1) can be extended to sums of an arbitrary number of terms. For example, d
dx(f (x)−g(x)+h(x))=f(x)−g(x)+h(x)
which we see by writingf (x)−g(x)+h(x)as(f (x)−g(x))+h(x), and then using (1).
Using the rules above makes it easy to differentiate any polynomial.
Products
Supposef (x) =x andg(x)= x2, then(f ãg)(x) = x3. Heref(x)= 1,g(x)=2x, and(fãg)(x)=3x2. Hence, the derivative of(fãg)(x)isnotequal tof(x)ãg(x)=2x.
The correct rule for differentiating a product is a little more complicated.
T H E D E R I V A T I V E O F A P R O D U C T
If bothf andgare differentiable at the pointx, then so isF =f ãg, and F (x)=f (x)ãg(x)⇒F(x)=f(x)ãg(x)+f (x)ãg(x)
(2)
Briefly formulated:The derivative of the product of two functions is equal to the derivative of the first times the second, plus the first times the derivative of the second. The formula, however, is much easier to digest than these words.
In Leibniz’s notation, the product rule is expressed as:
d dx
f (x)ãg(x)
= d dxf (x)
ãg(x)+f (x)ã d dxg(x) Before demonstrating why (2) is valid, here are two examples:
E X A M P L E 3 Use (2) to findh(x)whenh(x)=(x3−x)ã(5x4+x2).
Solution: We see thath(x)=f (x)ãg(x)withf (x)=x3−xandg(x)=5x4+x2.Here f(x)=3x2−1 andg(x)=20x3+2x.Thus,
h(x)=f(x)ãg(x)+f (x)ãg(x)=(3x2−1)ã(5x4+x2)+(x3−x)ã(20x3+2x) Usually we simplify the answer by expanding to obtain just one polynomial. Simple com- putation gives
h(x)=35x6−20x4−3x2
Alternatively, we can begin by expanding the expression forh(x). Do so and verify that you get the same expression forh(x)as before.
E X A M P L E 4 We illustrate the product rule for differentiation in a simple economic setting. LetD(P ) denote the demand function for a product. By selling D(P ) units at price P per unit, the producer earns revenueR(P )given by R(P ) =P D(P ). UsuallyD(P )is negative because demand goes down when the price increases. According to the product rule for differentiation,
R(P )=D(P )+P D(P ) (∗)
For an economic interpretation, suppose P increases by one dollar. The revenue R(P ) changes for two reasons. First,R(P )increases by 1ãD(P ), because each of theD(P )units brings in one dollar more. But a one dollar increase in the price per unit causes demand to change byD(P+1)−D(P )units, which is approximatelyD(P ). The (positive) loss due to a one dollar increase in the price per unit is then−P D(P ), which must be subtracted fromD(P )to obtainR(P ), as in equation (*). The resulting expression merely expresses the simple fact thatR(P ), the total rate of change ofR(P ), is what you gain minus what you lose.
We have now seen how to differentiate products of two functions. What about products of more than two functions? For example, suppose that
y =f (x)g(x)h(x)
What isy? We extend the same technique shown earlier and put y =
f (x)g(x) h(x).
Then the product rule gives
y=[f (x)g(x)]h(x)+[f (x)g(x)]h(x)
=
f(x)g(x)+f (x)g(x)
h(x)+f (x)g(x)h(x)
=f(x)g(x)h(x)+f (x)g(x)h(x)+f (x)g(x)h(x)
If none of the three functions is equal to 0, we can write the result in the following way:4 (f gh)
f gh =f f +g
g +h h
By analogy, it is easy to write down the corresponding result for a product ofnfunctions.
In words, the relative rate of growth of the product is the sum of the relative rates at which each factor is changing.
Proof of (2): Supposef andgare differentiable atx, so that the two Newton quotients f (x+h)−f (x)
h and g(x+h)−g(x)
h
tend to the limitsf(x)andg(x), respectively, ashtends to 0. We must show that the Newton quotient ofFalso tends to a limit, which is given byf(x)g(x)+f (x)g(x). The Newton quotient ofFis
F (x+h)−F (x)
h =f (x+h)g(x+h)−f (x)g(x)
h (∗)
4 If all the variables are positive, this result is easier to show using logarithmic differentiation. See Section 6.11.
S E C T I O N 6 . 7 / S U M S , P R O D U C T S , A N D Q U O T I E N T S 181 To proceed further we must somehow transform the right-hand side (RHS) to involve the Newton quotients off andg. We use a trick: The numerator of the RHS is unchanged if we both subtract and add the numberf (x)g(x+h). Hence, with a suitable regrouping of terms, we have
F (x+h)−F (x)
h =f (x+h)g(x+h)−f (x)g(x+h)+f (x)g(x+h)−f (x)g(x) h
=f (x+h)−f (x)
h g(x+h)+f (x)g(x+h)−g(x) h
Ashtends to 0, the two Newton quotients tend tof(x)andg(x), respectively. Now we can write g(x+h)forh=0 as
g(x+h)=
g(x+h)−g(x) h
h+g(x)
By the product rule for limits and the definition ofg(x), this tends tog(x)ã0+g(x)=g(x)as htends to 0. It follows that the Newton quotient ofFtends tof(x)g(x)+f (x)g(x)ashtends to 0.
Quotients
SupposeF (x)=f (x)/g(x), wherefandgare differentiable inxwithg(x)=0. Bearing in mind the complications in the formula for the derivative of a product, one should be somewhat reluctant to make a quick guess as to the correct formula forF(x).
In fact, it is quite easy to find the formula forF(x)if weassumethatF (x)isdiffer- entiable. FromF (x) = f (x)/g(x)it follows thatf (x) = F (x)g(x). Thus, the product rule givesf(x)=F(x)ãg(x)+F (x)ãg(x). Solving forF(x)yieldsF(x)ãg(x)= f(x)−F (x)ãg(x), and so
F(x)=f(x)−F (x)g(x)
g(x) = f(x)−[f (x)/g(x)]g(x) g(x)
Multiplying both numerator and denominator of the last fraction byg(x)gives the following important formula.
T H E D E R I V A T I V E O F A Q U O T I E N T
Iff andgare differentiable atxandg(x)=0, thenF =f/gis differentiable atx, and
F (x)=f (x)
g(x) ⇒ F(x)= f(x)ãg(x)−f (x)ãg(x) g(x)2
(3)
In words:The derivative ofaquotient is equal to the derivative of the numerator times the denominator minus the numerator times the derivative of the denominator, this difference then beingdivided bythe square of the denominator. In simpler notation, we have
f g
=fg−f g g2
NOTE 1 In the product rule formula, the two functions appear symmetrically, so that it is easy to remember. In the formula for the derivative of a quotient, the expressions in the numerator must be in the right order. Here is how you check that you have the order right.
Write down the formula you believe is correct. Putg≡1. Theng ≡0, and your formula ought to reduce tof. If you get−f, then your signs are wrong.
E X A M P L E 5 ComputeF(x)andF(4)whenF (x)=3x−5 x−2 .
Solution: We apply (3) withf (x)=3x−5,g(x)=x−2. Thenf(x)=3 andg(x)=1.
So we obtain, forx=2:
F(x)= 3ã(x−2)−(3x−5)ã1
(x−2)2 =3x−6−3x+5
(x−2)2 = −1 (x−2)2
To findF(4), we putx =4 in the formula forF(x)to getF(4)= −1/(4−2)2 = −1/4.
Note thatF(x) <0 for allx =2. HenceF is strictly decreasing both forx <2 and for x >2. Note that(3x−5)/(x−2)=3+1/(x−2). The graph is shown in Fig. 5.1.7.
E X A M P L E 6 LetC(x)be the total cost of producing x units of a commodity. Then C(x)/x is the average costof producingx units. Find an expression fordxd
C(x)/x . Solution:
d dx
C(x) x
=C(x)x−C(x)
x2 = 1
x
C(x)−C(x) x
(4) Note that the marginal costC(x)exceeds the average costC(x)/xif and only if average cost increases as output increases. (In a similar way, if a basketball team recruits a new player, the average height of the team increases if and only if the new player’s height exceeds the old average height.)
The formula for the derivative of a quotient becomes more symmetric if we consider relative rates of change. By using (3), simple computation shows that
F (x)= f (x)
g(x) ⇒ F(x)
F (x) = f(x)
f (x) −g(x)
g(x) (5)
The relative rate of change of a quotient is equal to the relative rate of change of the numerator minus the relative rate of change of the denominator.
LetW (t) be the nominal wage rate andP (t )the price index at time t. Thenw(t ) = W (t )/P (t )is called thereal wage rate. According to (5),
˙ w(t )
w(t ) =W (t )˙ W (t ) −P (t )˙
P (t )
The relative rate of change of the real wage rate is equal to the difference between the relative rates of change of the nominal wage rate and the price index. Thus, if nominal wages increase at the rate of 5% per year but prices rise by 6% per year, then real wages fall by 1%. Also, if inflation leads to wages and prices increasing at the same relative rate, then the real wage rate is constant.
S E C T I O N 6 . 7 / S U M S , P R O D U C T S , A N D Q U O T I E N T S 183
P R O B L E M S F O R S E C T I O N 6 . 7
In Problems 1–4, differentiate the functions defined by the various formulas.
1. (a) x+1 (b) x+x2 (c) 3x5+2x4+5
(d) 8x4+2√
x (e) 12x−32x2+5x3 (f) 1−3x7 2. (a) 35x2−2x7+18−√
x (b) (2x2−1)(x4−1) (c)
x5+1 x
(x5+1)
⊂SM⊃3. (a) 1
x6 (b) x−1(x2+1)√
x (c) 1
√x3 (d) x+1 x−1 (e) x+1
x5 (f) 3x−5
2x+8 (g) 3x−11 (h) 3x−1
x2+x+1 4. (a)
√x−2
√x+1 (b) x2−1
x2+1 (c) x2+x+1
x2−x+1
5. Letx=f (L)be the output whenLunits of labour are used as input. Assume thatf (0)=0, f(L) >0, andf(L) <0 forL >0. Average productivity isg(L)=f (L)/L.
(a) LetL∗>0. Indicate on a figure the values off(L∗)andg(L∗). Which is the larger?
(b) How does the average productivity change when labour input increases?
⊂SM⊃6. For each of the following functions, determine the intervals where it is increasing.
(a) y=3x2−12x+13 (b) y=14(x4−6x2) (c) y= 2x
x2+2 (d) y= x2−x3 2(x+1)
⊂SM⊃7. Find the equations for the tangents to the graphs of the following functions at the specified points:
(a) y=3−x−x2 at x=1 (b) y=x2−1
x2+1 at x=1 (c) y=
1 x2+1
(x2−1) at x=2 (d) y= x4+1
(x2+1)(x+3) at x=0 8. Consider the extraction of oil from a well. Letx(t )be the rate of extraction in barrels per day
andp(t )the price in dollars per barrel at timet. ThenR(t )=p(t )x(t )is the revenue in dollars per day. Find an expression forR(t ), and give it an economic interpretation in the case when˙ p(t )andx(t )are both increasing. (Hint:R(t )increases for two reasons. . .)
⊂SM⊃9. Differentiate the following functions w.r.t.t:
(a) at+b
ct+d (b) tn
a√
t+b (c) 1
at2+bt+c 10. Iff (x)=√
x, thenf (x)ãf (x)=x. Use the product rule to find a formula forf(x). Compare this with the result in Problem 6.2.8.
11. Prove the power rule
y=xa ⇒ y=axa−1
fora= −n, wherenis a natural number, by using the relationf (x)=x−n =1/xnand the quotient rule (3).