Economists often need even more general chain rules than the simple one for two variables presented in the previous section. Problem 7, for example, considers the example of a railway company whose fares for peak and off-peak fares are set by a regulatory authority.
The costs it faces for running enough trains to carry all the passengers depend on demand for both kinds of journey. These demands are obviously affected by both peak and off-peak fares because some passengers will choose when to travel based on the fare difference. The general chain rule we are about to present allows us to work out how these costs change when either fare is increased.
Consider the general problem of this kind where
z=F (x, y), x =f (t, s), y =g(t, s) In this case,zis a function of botht ands, with
z=F (f (t, s), g(t, s))
Here it makes sense to look for both partial derivatives∂z/∂tand∂z/∂s. If we keepsfixed, thenzis a function oftalone, and we can therefore use the chain rule (1) from the previous section. In the same way, by keepingtfixed, we can differentiatezw.r.t.s. The result is the following:
S E C T I O N 1 2 . 2 / C H A I N R U L E S F O R M A N Y V A R I A B L E S 417 T H E C H A I N R U L E
Ifz=F (x, y)withx =f (t, s)andy=g(t, s), then (a) ∂z
∂t =F1(x, y)∂x
∂t +F2(x, y)∂y
∂t (b) ∂z
∂s =F1(x, y)∂x
∂s +F2(x, y)∂y
∂s
(1)
E X A M P L E 1 Find∂z/∂tand∂z/∂swhenz=F (x, y)=x2+2y2, withx=t−s2andy=t s.
Solution: We obtain
F1(x, y)=2x, F2(x, y)=4y, ∂x
∂t =1, ∂x
∂s = −2s, ∂y
∂t =s, ∂y
∂s =t Formula (1) therefore gives:
∂z
∂t =2xã1+4yãs=2(t−s2)+4t ss=2t−2s2+4t s2
∂z
∂s =2xã(−2s)+4yãt =2(t−s2)(−2s)+4t st = −4t s+4s3+4t2s Check these answers by first expressingzas a function oftands, then differentiating.
E X A M P L E 2 Findzt(1,0)ifz=ex2+y2exy, withx=2t+3sandy =t2s3. Solution: We obtain
∂z
∂x =2xex2+y3exy, ∂z
∂y =2yexy+xy2exy, ∂x
∂t =2, ∂y
∂t =2t s3 Using somewhat more concise notation, formula (1) gives
zt(t, s)= ∂z
∂x
∂x
∂t + ∂z
∂y
∂y
∂t =(2xex2+y3exy)ã2+(2yexy+xy2exy)ã2t s3 Whent=1 ands=0, thenx=2 andy=0, sozt(1,0)=4e4ã2=8e4.
The General Case
In consumer demand theory economists typically assume that a household’s utility depends on the number of units of each good it is able to consume. The number of units consumed will depend in turn on the prices of these goods and on the household’s income. Thus the household’s utility is related, indirectly, to all the prices and to income. How does utility respond to an increase in one of the prices, or to an increase in income? The following general chain rule extends to this kind of problem.
Suppose that
z=F (x1, . . . , xn) with x1=f1(t1, . . . , tm), . . . , xn=fn(t1, . . . , tm) (2) Substituting for all the variablesxi as functions of the variables tj into the function F expresseszas acomposite function
z=F (f1(t1, . . . , tm), . . . , fn(t1, . . . , tm))
oft1, . . . , tm. In vector notation,z=F (x(t)). An obvious generalization of (1) is as follows:
T H E G E N E R A L C H A I N R U L E When (2) is true, then
∂z
∂tj = ∂z
∂x1
∂x1
∂tj + ∂z
∂x2
∂x2
∂tj + ã ã ã + ∂z
∂xn
∂xn
∂tj , j =1,2, . . . , m (3)
This is an important formula that every economist should understand. A small change in a basic variabletj sets off a chain reaction. First, everyxi depends ontj in general, so it changes whentjis changed. This affectszin turn. The contribution to the total derivative ofzw.r.t.tjthat results from the change inxiis(∂z/∂xi)(∂xi/∂tj). Formula (3) shows that
∂z/∂tjis the sum of all these contributions.
E X A M P L E 3 Example 11.7.1 considered an agricultural production functionY =F (K, L, T ), where Yis the size of the harvest,Kis capital invested,Lis labour, andTis the area of agricultural land used to grow the crop. Suppose that K,L, andT are all functions of time. Then, according to (3), one has
dY dt = ∂F
∂K dK
dt +∂F
∂L dL
dt +∂F
∂T dT
dt
In the special case whenF is the Cobb–Douglas functionF (K, L, T )=AKaLbTc, then dY
dt =aAKa−1LbTcdK
dt +bAKaLb−1TcdL
dt +cAKaLbTc−1dT
dt (∗)
Denoting time derivatives by “dots”, and dividing each term in(∗)byY =AKaLbTc, we
get Y˙
Y =aK˙ K +bL˙
L +cT˙ T
The relative rate of change of output is, therefore, a weighted sum of the relative rates of change of capital, labour, and land. The weights are the respective powersa,b, andc.
S E C T I O N 1 2 . 2 / C H A I N R U L E S F O R M A N Y V A R I A B L E S 419
P R O B L E M S F O R S E C T I O N 1 2 . 2
1. Use (1) to find∂z/∂tand∂z/∂sfor the following cases:
(a) z=F (x, y)=x+y2, x=t−s, y=t s
(b) z=F (x, y)=2x2+3y3, x=t2−s, y=t+2s3
⊂SM⊃2. Using (1), find∂z/∂tand∂z/∂sfor the following cases:
(a) z=xy2, x=t+s2, y=t2s (b) z=x−y
x+y, x=et+s, y=et s 3. (a) Ifz=F (u, v, w)whereu=r2,v= −2s2, andw=lnr+lns, find∂z/∂rand∂z/∂s.
(b) Ifz=F (x)andx=f (t1, t2), find∂z/∂t1and∂z/∂t2. (c) Ifx=F (s, f (s), g(s, t )), find∂x/∂sand∂x/∂t.
(d) Ifz=F (u, v, w)whereu=f (x, y),v=x2h(y)andw=1/y, find∂z/∂xand∂z/∂y.
4. Use the general chain rule (3) to find∂w/∂tfor the following cases:
(a) w=xy2z3, with x=t2, y=s, z=t (b) w=x2+y2+z2, with x=√
t+s, y=et s, z=s3 5. Find expressions fordz/dtwhen
(a) z=F (t, t2, t3) (b) z=F (t, f (t ), g(t2))
6. (a) SupposeZ=G+Y2+r2, whereYandrare both functions ofG. Find∂Z/∂G.
(b) SupposeZ=G+I (Y, r), whereIis a differentiable function of two variables, andY,r are both functions ofG. Find∂Z/∂G.
7. Each week a suburban railway company has a long-run costC=aQ1+bQ2+cQ21of providing Q1passenger kilometres of service during rush hours andQ2passenger kilometres during off- peak hours. As functions of the regulated faresp1andp2per kilometre for the rush hours and off-peak hours respectively, the demands for the two kinds of service areQ1=Ap−α1 1p2β1and Q2 =Bp1α2p2−β2, where the constantsA, B, α1, α2, β1, β2are all positive. Assuming that the company runs enough trains to meet the demand, find expressions for the partial derivatives of Cw.r.t.p1andp2.
⊂SM⊃8. (a) Ifu=ln(x3+y3+z3−3xyz), show that (i) x∂u
∂x +y∂u
∂y +z∂u
∂z =3 (ii) (x+y+z) ∂u
∂x +∂u
∂y +∂u
∂z
=3 (b) Ifz=f (x2y), show thatx∂z
∂x =2y∂z
∂y.
9. (a) Find a formula for∂u/∂rwhenu=f (x, y, z, w)andx,y,z, andwall are functions of two variablesrands.
(b) Supposeu=xyzw, wherex =r+s,y =r−s,z =rs,w =r/s. Find∂u/∂rwhen (r, s)=(2,1).