For a functiony=f (x)of one variable, the derivativef(x)is a number which measures the function’s rate of change asxchanges. For functions of two variables, such asz=f (x, y), we also want to examine how quickly the value of the function changes w.r.t. changes in the values of the independent variables. For example, iff (x, y)is a firm’s profit when it uses quantitiesxandyof two different inputs, we want to know whether and by how much profit can increase as eitherxoryis varied.
Consider the function
z=x3+2y2 (∗)
Suppose first thatyis held constant. Then 2y2is constant. Really there is only one variable now. Of course, the rate of change ofzw.r.t.xis given by
dz dx =3x2
On the other hand, we can keepxfixed in(∗)and examine howzvaries asyvaries. This involves taking the derivative ofzw.r.t.ywhile keepingxconstant. The result is
dz dy =4y
Obviously, there are many other variations we could study. For example,xandycould vary simultaneously. But in this section, we restrict our attention to variations ineitherxory.
When we consider functions of two variables, mathematicians (and economists) usually write∂z/∂x instead ofdz/dx for the derivative ofz w.r.t.x wheny is held fixed. This slight change of notation, replacingd by∂, is intended to remind the reader that only one independent variable is changing, with the other(s) held fixed. In the same way, we write
∂z/∂yinstead ofdz/dywhenyvaries andxis held fixed. Hence, we have z=x3+2y2 ⇒ ∂z
∂x =3x2 and ∂z
∂y =4y In general, we introduce the following definitions:
P A R T I A L D E R I V A T I V E S
Ifz=f (x, y), then
(i) ∂z/∂xdenotes the derivative off (x, y)w.r.t.xwhenyis held constant;
(ii) ∂z/∂ydenotes the derivative off (x, y)w.r.t.ywhenxis held constant.
(1)
Whenz= f (x, y), we also denote the derivative∂z/∂x by∂f/∂x, and this is called the partial derivativeofz(orf) w.r.t.x. Often∂z/∂xis pronounced “partialdzbydx”. In the same way,∂z/∂y=∂f/∂yis thepartial derivativeofz(orf) w.r.t.y. Note that∂f/∂x is the rate of change off (x, y)w.r.t.xwhenyis constant, and correspondingly for∂f/∂y.
Of course, because there are two variables, there are two partial derivatives.
It is usually easy to find the partial derivatives of a functionz=f (x, y). To find∂f/∂x, just think ofyas a constant and differentiatef (x, y)w.r.t.x as iff were a function only ofx. The rules for finding derivatives of functions of one variable can all be used when we want to compute∂f/∂x. The same is true for∂f/∂y. Let us look at some further examples.
E X A M P L E 1 Find the partial derivatives of the following functions:
(a) f (x, y)=x3y+x2y2+x+y2 (b) f (x, y)= xy x2+y2 Solution:
(a) We find
∂f
∂x =3x2y+2xy2+1 (holdingyconstant)
∂f
∂y =x3+2x2y+2y (holdingxconstant) (b) For this function the quotient rule gives
∂f
∂x =y(x2+y2)−2xxy
(x2+y2)2 = y3−x2y
(x2+y2)2 , ∂f
∂y = x3−y2x (x2+y2)2
Observe that the function in (b) is symmetric inxandy, in the sense that the function value is unchanged if we interchangexandy. By interchangingxandyin the formula for∂f/∂x, therefore, we will find the correct formula for∂f/∂y. (Find∂f/∂y in the usual way and check that the foregoing answer is correct.)
Several other forms of notation are often used to indicate the partial derivatives of z = f (x, y). Some of the most common are
∂f
∂x = ∂z
∂x =zx =fx(x, y)=f1(x, y)= ∂f (x, y)
∂x
∂f
∂y = ∂z
∂y =zy =fy(x, y)=f2(x, y)=∂f (x, y)
∂y
S E C T I O N 1 1 . 2 / P A R T I A L D E R I V A T I V E S W I T H T W O V A R I A B L E S 383 Among these,f1(x, y)andf2(x, y)are the most satisfactory. Here the numerical subscript refers to the position of the argument in the function. Thus,f1indicates the partial derivative w.r.t. the first variable, andf2w.r.t. the second variable. This notation also reminds us that the partial derivatives themselves are functions ofx andy. Finally, f1(a, b)andf2(a, b) are suitable designations of the values of the partial derivatives at point(a, b)instead of at (x, y). For example, given the functionf (x, y)=x3y+x2y2+x+y2in Example 1(a), one has
f1(x, y)=3x2y+2xy2+1, f1(a, b)=3a2b+2ab2+1 In particular,f1(0,0)=1 andf1(−1,2)=3(−1)22+2(−1)22+1= −1.
The alternative notationfx(x, y)andfy(x, y)is often used, but especially in connection with composite functions it is sometimes too ambiguous. For instance, what is the meaning offx(x2y, x−y)?
Remember that the numbersf1(x, y)andf2(x, y)measure the rate of change offw.r.t.x andy, respectively. For example, iff1(x, y) >0, then a small increase inxwill lead to an increase inf (x, y).
E X A M P L E 2 In Example 11.1.2 we studied the functionx =Ap−1.5m2.08. Find the partial derivatives ofxw.r.t.pandm, and discuss their signs.
Solution: We find that∂x/∂p = −1.5Ap−2.5m2.08and∂x/∂m=2.08Ap−1.5m1.08. Be- causeA,p, andmare positive,∂x/∂p <0 and∂x/∂m >0. These signs are in accordance with the final remarks in the example.
Formal Definitions of Partial Derivatives
So far the functions have been given by explicit formulas and we have found the partial derivatives by using the ordinary rules for differentiation. If these rules cannot be used, however, we must resort to the formal definition of partial derivative. This is derived from the definition of derivative for functions of one variable in the following rather obvious way.
Ifz = f (x, y), then withg(x) = f (x, y)(y fixed), the partial derivative off (x, y) w.r.t.xis simplyg(x). Now, by definition,g(x)=limh→0[g(x+h)−g(x)]/ h. Because f1(x, y)=g(x), it follows that
f1(x, y)=lim
h→0
f (x+h, y)−f (x, y)
h (2)
In the same way,
f2(x, y)=lim
k→0
f (x, y+k)−f (x, y)
k (3)
If the limit in (2) (or (3)) does not exist, then we say thatf1(x, y)(orf2(x, y))does not exist, or thatz isnot differentiable w.r.t. x or y at the point. For instance, the function f (x, y)= |x| + |y|is not differentiable at the point(x, y)=(0,0).
Ifhis small in absolute value, then from (2) we obtain the approximation f1(x, y)≈f (x+h, y)−f (x, y)
h (4)
Similarly, ifkis small in absolute value,
f2(x, y)≈ f (x, y+k)−f (x, y)
k (5)
These approximations can be interpreted as follows:
(A) The partial derivative f1(x, y) is approximately equal to the change in f (x, y)per unit increase inx, holdingyconstant.
(B) The partial derivative f2(x, y) is approximately equal to the change in f (x, y)per unit increase iny, holdingxconstant.
(6)
NOTE 1 The approximations in (6) must be used with caution. Roughly speaking, they will not be too inaccurate provided that the partial derivatives do not vary too much over the actual intervals.
E X A M P L E 3 LetY =F (K, L)be the number of units produced whenKunits of capital andLunits of labour are used as inputs in a production process. What is the economic interpretation of FK(100,50)=5?
Solution: From (6),FK(100,50)= 5 means that, starting fromK = 100 and holding labour input fixed at 50, output increases by 5 units per unit increase inK.
Higher-Order Partial Derivatives
Ifz =f (x, y), then∂f/∂xand∂f/∂y are calledfirst-order partial derivatives. These partial derivatives are, in general, again functions of two variables. From∂f/∂x, provided this derivative is itself differentiable, we can generate two new functions by taking the partial derivatives w.r.t.x andy. In the same way, we can take the partial derivatives of
∂f/∂yw.r.t.xandy. The four functions we obtain by differentiating twice in this way are calledsecond-order partial derivativesoff (x, y). They are expressed as
∂
∂x ∂f
∂x
= ∂2f
∂x2 , ∂
∂x ∂f
∂y
= ∂2f
∂x∂y , ∂
∂y ∂f
∂x
= ∂2f
∂y∂x, ∂
∂y ∂f
∂y
= ∂2f
∂y2 For brevity, we sometimes refer to the first- and second-order “partials”, suppressing the word “derivatives”.
E X A M P L E 4 For the function in Example 1(a), we obtain
∂2f
∂x2 =6xy+2y2, ∂2f
∂y∂x =3x2+4xy= ∂2f
∂x∂y, ∂2f
∂y2 =2x2+2
S E C T I O N 1 1 . 2 / P A R T I A L D E R I V A T I V E S W I T H T W O V A R I A B L E S 385 As with first-order partial derivatives, several other kinds of notation are also frequently used for second-order partial derivatives. For example,∂2f/∂x2is also denoted byf11(x, y)or fxx(x, y). In the same way,∂2f/∂y∂x can also be written asf12(x, y)orfxy(x, y). Note thatf12(x, y)means that we differentiatef (x, y)first w.r.t. the first argumentx and then w.r.t. the second argumenty. To findf21(x, y), we must differentiate in the reverse order.
In Example 4, these two “mixed” second-order partial derivatives (or “cross-partials”) are equal. For most functionsz=f (x, y), it will actually be the case that
∂2f
∂x∂y = ∂2f
∂y∂x (7)
Sufficient conditions for the equality in (7) are given in Theorem 11.6.1.
It is very important to note the exact meaning of the different symbols that have been introduced. If we consider (7), for example, it would be a serious mistake to believe that the two expressions are equal because∂x∂yis the same as∂y∂x. Here the expression on the left- hand side is in fact the derivative of∂f/∂yw.r.t.x, and the right-hand side is the derivative of
∂f/∂xw.r.t.y. It is a remarkable fact, and not a triviality, that the two are usually equal. As another example, we observe that∂2z/∂x2is quite different from(∂z/∂x)2. For example, ifz=x2+y2, then∂z/∂x=2x. Therefore,∂2z/∂x2 =2, whereas(∂z/∂x)2=4x2.
Analogously, we define partial derivatives of the third, fourth, and higher orders. For example, we write∂4z/∂x∂y3=z(4)yyyxwhen we first differentiatezthree times w.r.t.yand then differentiate the result once more w.r.t.x.
Here is an additional example.
E X A M P L E 5 Iff (x, y)=x3ey2,find the first- and second-order partial derivatives at(x, y)=(1,0).
Solution: To findf1(x, y), we differentiatex3ey2 w.r.t.x while treatingy as a constant.
Whenyis a constant, so isey2. Hence,
f1(x, y)=3x2ey2 and so f1(1,0)=3ã12e02=3 To findf2(x, y), we differentiatef (x, y)w.r.t.ywhile treatingxas a constant:
f2(x, y)=x32yey2 =2x3yey2 and so f2(1,0)=0
To find the second-order partialf11(x, y), we must differentiatef1(x, y)w.r.t.xonce more, while treatingyas a constant. Hence,
f11(x, y)=6xey2 and so f11(1,0)=6ã1e02 =6
To find f22(x, y), we must differentiatef2(x, y) = 2x3yey2 w.r.t. y once more, while treatingx as a constant. Becauseyey2 is a product of two functions, each involvingy, we use the product rule to obtain
f22(x, y)=(2x3)(1ãey2+y2yey2)=2x3ey2+4x3y2ey2
Evaluating this at(1,0)givesf22(1,0)=2. Moreover, f12(x, y)= ∂
∂y
f1(x, y)
= ∂
∂y(3x2ey2)=3x22yey2 =6x2yey2 and
f21(x, y)= ∂
∂x
f2(x, y)
= ∂
∂x(2x3yey2)=6x2yey2 Hence,f12(1,0)=f21(1,0)=0.
P R O B L E M S F O R S E C T I O N 1 1 . 2
1. Find∂z/∂xand∂z/∂yfor the following:
(a) z=2x+3y (b) z=x2+y3 (c) z=x3y4 (d) z=(x+y)2 2. Find∂z/∂xand∂z/∂yfor the following:
(a) z=x2+3y2 (b) z=xy (c) z=5x4y2−2xy5 (d) z=ex+y (e) z=exy (f) z=ex/y (g) z=ln(x+y) (h) z=ln(xy) 3. Findf1(x, y),f2(x, y), andf12(x, y)for the following:
(a) f (x, y)=x7−y7 (b) f (x, y)=x5lny (c) f (x, y)=(x2−2y2)5 4. Find all first- and second-order partial derivatives for the following:
(a) z=3x+4y (b) z=x3y2 (c) z=x5−3x2y+y6 (d) z=x/y (e) z=(x−y)/(x+y) (f) z=
x2+y2
⊂SM⊃5. Find all the first- and second-order partial derivatives of:
(a) z=x2+e2y (b) z=ylnx (c) z=xy2−exy (d) z=xy 6. LetF (S, E)=2.26S0.44E0.48. (This is an estimated production function for a certain lobster
fishery whereSdenotes the stock of lobsters,Ethe harvesting effort, andF (S, E)the catch.) (a) FindFS(S, E)andFE(S, E).
(b) Show thatSFS+EFE =kF for a suitable constantk.
7. Prove that ifz=(ax+by)2, thenxzx+yzy=2z.
8. Letz= 12ln(x2+y2). Show that∂2z/∂x2+∂2z/∂y2=0.
9. If a household consumesxunits of one good andyunits of a second good, its satisfaction is measured by the functions(x, y) = 2 lnx+4 lny. Suppose that the household presently consumes 20 units of the first good and 30 units of the second.
(a) What is the approximate increase in satisfaction from consuming one extra unit of the first good?
(b) What is the approximate increase in satisfaction from consuming one extra unit of the second good?
S E C T I O N 1 1 . 3 / G E O M E T R I C R E P R E S E N T A T I O N 387