Economists often need to differentiate functions that are defined implicitly by an equation.
Section 7.2 considered some simple cases; it is a good idea to review those examples now.
Here we study the problem from a more general point of view.
LetF be a function of two variables, and consider the equation F (x, y)=c (cis a constant)
The equation represents a level curve for F. (See Section 11.3.) Suppose this equation definesyimplicitly as a functiony =f (x)ofxin some intervalI, as illustrated in Fig. 1.
This means that
F (x, f (x))=c for allxinI (∗)
Iff is differentiable, what is the derivative ofy =f (x)? If the graph off looks like the one given in Fig. 1, the geometric problem is to find the slope of the graph at each point likeP.
F(x, y) ⫽ c y
x P
I
Figure 1 What is the slope atP?
To find an expression for the slope, introduce the auxiliary functionudefined for allxinI byu(x)=F (x, f (x)). Thenu(x)=F1(x, f (x))ã1+F2(x, f (x))ãf(x)according to the chain rule. Now,(∗)states thatu(x)=cfor allxinI. The derivative of a constant is 0, so we haveu(x)=F1(x, f (x))+F2(x, f (x))ãf(x)=0. If we replacef (x)byy and solve forf(x)=y, we reach the conclusion:
S L O P E O F A L E V E L C U R V E
F (x, y)=c ⇒ y= −F1(x, y)
F2(x, y) (F2(x, y)=0) (1)
This is an important result. Before applying this formula fory, however, recall that the pair (x, y)must satisfy the equationF (x, y)=c.On the other hand, note that there is no need to solve the equationF (x, y)=cexplicitly fory before applying (1) in order to findy. (See Example 3.)
S E C T I O N 1 2 . 3 / I M P L I C I T D I F F E R E N T I A T I O N A L O N G A L E V E L C U R V E 421 The same argument withxandyinterchanged gives an analogous result to (1). Thus, if xis a continuously differentiable function ofy which satisfiesF (x, y)=c, then
F (x, y)=c ⇒ dx
dy = −∂F /∂y
∂F /∂x
∂F
∂x =0
(2)
E X A M P L E 1 Use (1) to findywhenxy=5.
Solution: We putF (x, y)=xy. ThenF1(x, y)=yandF2(x, y)=x. Hence (1) gives y= −F1(x, y)
F2(x, y) = −y x This confirms the result in Example 7.1.1.
E X A M P L E 2 For the curve given by
x3+x2y−2y2−10y=0
find the slope and the equation for the tangent at the point(x, y)=(2,1).
Solution: LetF (x, y)=x3+x2y−2y2−10y. Then the given equation is equivalent to F (x, y)=0, which is a level curve forF. First, we check thatF (2,1)=0, so(x, y) = (2,1)is a point on the curve. Also,F1(x, y)=3x2+2xyandF2(x, y)=x2−4y−10.
So (1) implies that
y= − 3x2+2xy x2−4y−10
Forx =2 andy = 1 in particular, one hasy =8/5. Then the point–slope formula for a line implies that the tangent at(2,1)must have the equationy−1 =(8/5)(x−2), or 5y =8x−11. See Fig. 2, in which the curve has been drawn by a computer program. Note that, for many values ofx, there is more than one corresponding value ofysuch that(x, y) lies on the curve. For instance,(2,1)and(2,−4)both lie on the curve. Findyat(2,−4).
(Answer:y=0.4.)
x3⫹ x2y ⫺ 2y2⫺ 10y ⫽ 0
⫺2
⫺3
⫺4
⫺5 ⫺1 1 2 3 4 5 3
⫺3 4
⫺4 5
⫺5 2
⫺2 1
⫺1 y
x
Figure 2 The graph of the equation in Example 2.
E X A M P L E 3 Assume that the equation
exy2−2x−4y=c
implicitly definesyas a differentiable functiony=f (x)ofx. Find a value of the constant csuch thatf (0)=1, and findyat(x, y)=(0,1).
Solution: Whenx = 0 andy = 1, the equation becomes 1−4 = c, soc = −3. Let F (x, y) =exy2 −2x−4y. ThenF1(x, y) =y2exy2−2, andF2(x, y) = 2xyexy2 −4.
Thus, from (1) we have
y= −F1(x, y)
F2(x, y) = − y2exy2−2 2xyexy2−4
Whenx =0 andy =1, we findy= −1/4. (Note that in this example it is impossible to solveexy2−2x−4y= −3 explicitly fory. Even so, we have managed to find an explicit expression for the derivative ofyw.r.t.x.)
Here is an important economic example using a function defined implicitly by an equation.
E X A M P L E 4 We generalize Example 7.2.2, and assume thatD =f (t, P )is the demand for a com- modity that depends on the priceP before tax, as well as on the sales tax per unit, denoted byt. Suppose thatS =g(P )is the supply function. At equilibrium, when supply is equal to demand, the equilibrium priceP =P (t )depends ont. Indeed,P =P (t )must satisfy the equation
f (t, P )=g(P ) (∗)
for allt in some relevant interval. Suppose that(∗)definesP implicitly as a differentiable function oft. Find an expression fordP /dt, then discuss its sign.
Solution: LetF (t, P ) =f (t, P )−g(P ). Then equation(∗)becomesF (t, P )= 0, so formula (1) yields
dP
dt = −Ft(t, P )
FP(t, P )= − ft(t, P )
fP(t, P )−g(P ) = ft(t, P )
g(P )−fP(t, P ) (∗∗) It is reasonable to assume thatg(P ) >0 (meaning that supply increases if price increases) and thatft(t, P )andfP(t, P )are both<0 (meaning that demand decreases if either the tax or the price increases). Then(∗∗)tells us thatdP /dt < 0, implying that the pre-tax price faced by suppliers decreases as the tax increases. Thus the suppliers, as well as the consumers, are adversely affected if the tax on their product rises.
Of course, we can also derive formula(∗∗)by implicit differentiation of(∗)w.r.t.t. This procedure gives
ft(t, P )ã1+fP(t, P )dP
dt =g(P )dP dt Solving this equation fordP /dtyields(∗∗)again.
S E C T I O N 1 2 . 3 / I M P L I C I T D I F F E R E N T I A T I O N A L O N G A L E V E L C U R V E 423
A Formula for the Second Derivative
Sometimes we need to know whether a level curveF (x, y)=cis the graph of a function y =f (x)that is convex or concave. One way to find out is to calculatey, which is the derivative ofy = −F1(x, y)/F2(x, y). WriteG(x) = F1(x, y)andH (x) = F2(x, y), whereyis a function ofx. Our aim now is to differentiate the quotienty= −G(x)/H (x) w.r.t.x. According to the rule for differentiating quotients,
y= −G(x)H (x)−G(x)H(x)
[H (x)]2 (∗)
Keeping in mind thatyis a function ofx, bothG(x)andH (x)are composite functions. So we differentiate them both by using the chain rule, thereby obtaining
G(x)=F11(x, y)ã1+F12(x, y)ãy H(x)=F21(x, y)ã1+F22(x, y)ãy
Assuming thatF is aC2 function, Young’s Theorem (Theorem 11.6.1) implies thatF12 = F21. Replacey in both the preceding equations by the quotient−F1/F2, and then insert the results into(∗). After some algebraic simplification, this yields the formula
F (x, y)=c ⇒ y= − 1
(F2)3[F11(F2)2−2F12F1F2+F22(F1)2] (3) Occasionally (3) is used in theoretical arguments, but generally it is easier to findy by direct differentiation, as in the examples in Section 7.1.
E X A M P L E 5 Use (3) to findywhen
xy=5
Solution: WithF (x, y)=xywe haveF1=y,F2 =x,F11 =0,F12 =1, andF22 =0.
According to (3), we obtain
y= − 1
x3(−2ã1ãyãx)= 2y x2 which is the same result we found in Example 7.1.4.
For those who are already familiar with 3×3 determinants, which this book discusses in Section 16.2, the result in (3) can be expressed in the following more memorable form:
F (x, y)=c ⇒ y=d2y dx2 = 1
(F2)3
0 F1 F2 F1 F11 F12 F2 F21 F22
(4)
This is valid only ifF2=0, obviously.
P R O B L E M S F O R S E C T I O N 1 2 . 3
1. Use formula (1) withF (x, y) =2x2+6xy+y2 andc = 18 to findywhenyis defined implicitly by 2x2+6xy+y2=18. Compare with the result in Problem 7.1.5.
⊂SM⊃2. Use formula (1) to findyfor the following level curves. Also findyusing (3).
(a) x2y=1 (b) x−y+3xy=2 (c) y5−x6=0
⊂SM⊃3. A curve in thexy-plane is given by the equation 2x2+xy+y2−8=0.
(a) Findy,y, and the equation for the tangent at the point(2,0).
(b) Which points on the curve have a horizontal tangent?
4. The equation 3x2−3xy2+y3+3y2 = 4 definesyimplicitly as a functionh(x)ofxin a neighbourhood of the point(1,1). Findh(1).
5. Suppose the demandD(P , r)for a certain commodity (like a luxury car) depends on its price Pand the interest rater. What signs should one expect the partial derivatives ofDw.r.t.Pand rto have? Suppose the supplySis constant, so that in equilibrium,D(P , r)=S. Differentiate implicitly to finddP /dr, and comment on its sign. (Problem 7.2.3 considers a special case.) 6. LetD=f (R, P )denote the demand for a commodity when the price isPandRis advertising
expenditure. What signs should one expect the partial derivativesfR andfP to have? If the supply isS=g(P ), equilibrium in the market requires thatf (R, P )=g(P ). What isdP /dR?
Discuss its sign.
7. Letf be a differentiable function of one variable, and letaandbbe two constants. Suppose that the equationx−az=f (y−bz)defineszas a differentiable function ofxandy. Prove thatzsatisfiesazx+bzy=1.