Chapter 3 Differentiation. Historical notes. Some history. LC Abueg: mathematical economics. chapter 3: differentiation 1

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1 Chapter 3 Differentiation Lectures in Mathematical Economics L Cagandahan Abueg De La Salle University School of Economics Historical notes Some history Prior to the seventeenth century [1600s], a curve was generally described as a locus of points satisfying some geometric condition (eamples are construction of circles and ellipses) Also, a tangent line is obtained from these curves through geometric construction. chapter 3: differentiation 1

2 Some history With the birth of analytic geometry [the study of geometry involving linear algebra], the viewpoints on tangent lines and curves changed drastically. From analytic geometry followed the discovery of finding maima and minima of functions; and computation of areas under the curve (i.e., integration). The derivative The derivative Definition [Newton (1668)]. Let f be a function defined on an open interval (a,b) and let + (a,b). The derivative of f at is given as f( + h) f( ) f ( ) = lim h 0 h if this limit eists. We then say that f is differentiable at, and this process is called differentiation. chapter 3: differentiation 2

3 Differentiation notations dy/ d D f yɺ f ( ) Gottfried Wilhelm Leibniz [ ] Leonhard Euler [ ] Sir Isaac Newton [ ] Joseph-Louis Lagrange [ ] Geometric interpretation Remark. Consider a differentiable function of y = f(). Define the following loci of points: i. a secant is a curve passing through at least two points on the graph of f. ii. a chord is a line segment connecting two points on f. Geometric interpretation y y = f() chord secant a 1 chapter 3: differentiation 3

4 Geometric interpretation y y = f() tangent line at point a a Differentiability & continuity Theorem 3.1. If f is differentiable at a point c, then f is continuous at c. Differentiability & continuity Eample. The absolute value function f( ) =, R is continuous everywhere but whose derivative does not eist at = 0. chapter 3: differentiation 4

5 Differentiability & continuity Karl Weierstrass [ ] He showed that f 1 ( n ) = cos( b π ) n n= 0 a is continuous but whose derivative does not eist everywhere: a nowhere differentiable function. Theorems on differentiation Theorem 3.2. [Rules of differentiation] Let f and g be differentiable functions and let c be a constant. Then d n n i = n n Z d ( ) ( ) 1, *, 0 d d a a ( ii) ( ) = a 1, a R, > 0 Theorems on differentiation d d ( iii) ( c) = 0 d iv cf( ) cf( ) d ( ) ( ) = d v f( ) g( ) f( ) g( ) d ( ) ( + ) = + chapter 3: differentiation 5

6 Theorems on differentiation Eample. Obtain the derivative of the following functions: ( i) y = ( ii) y = ( ) iii y 5 4 = Theorems on differentiation Theorem 3.3. [Product rule (G. W. Leibniz)] Let f and g be differentiable functions. Then d ( f ( ) g ( )) d = f ( ) g( ) + g ( )( f ) Theorems on differentiation Eample. Obtain the derivative of the following functions: ( i) y = ( 3 + 1)( 3 + 2) ii y = iii y = (3 )( 7 ) 5 ( ) ( ) 2 1 chapter 3: differentiation 6

7 Theorems on differentiation Eercise. If f 1, f 2, f n are differentiable, then n n n d fj( ) fj( ) fi( ) d = = 1 j= 1 j i j Theorems on differentiation Theorem 3.4. [Quotient rule] Let f and g be differentiable functions and let Then g( ) 0, domg d f( ) g( ) f ( ) f( ) g ( ) = 2 d g( ) [ g( )] Theorems on differentiation Eample. Obtain the derivatives of the following functions ( i) ( ii) y = y = chapter 3: differentiation 7

8 Theorems on differentiation Eercise. Consider the function y = f( ) = a + b 2 2 for some constants a and b. Find the derivatives of y, y, y/, and y 1. Theorems on differentiation Theorem 3.5. [Chain rule] If the function g is differentiable at and the function f is differentiable at g(), then the composite function f g is differentiable at, and d ( f ( g ( ))) = f ( g ( )) g ( ) d Theorems on differentiation Eample. Obtain the derivatives of the following functions ( ) = + i y 4( 3 ) 1/3 100 ( ii) y = ( iii) y = ( ) 2 chapter 3: differentiation 8

9 Theorems on differentiation Remark. Consider the function f( ) h( ) = = f( ) g( ) g( ) 1 Theorems on differentiation Theorem 3.6. [Derivatives of inverse functions] Let f be a 1-1 differentiable function. Then f 1 is also differentiable and d ( f ) f d 1 ( ) = ( ) 1 provided that f ( ) 0, dom f Theorems on differentiation Eample. Consider the function y = f( ) = 3 chapter 3: differentiation 9

10 Theorems on differentiation Theorem 3.7. [Derivatives of eponential functions] If e is the natural number and if k is a positive constant not equal to 1), then d i e = e d ( ) ( ) d ii k k k d ( ) ( ) = ln Theorems on differentiation Eample. Find the derivative of y = e e Theorems on differentiation Theorem 3.8. [Derivatives of logarithmic functions] If ln is the natural logarithm, log b is the logarithm with respect to base b, then d 1 ( i) ( ln) = d d 1 b = d lnb ( ii) ( log ) chapter 3: differentiation 10

11 Theorems on differentiation Eample. Find the derivative of 3 y = ln(2 + 5) + 4 e Theorems on differentiation Eercise. Show that 2 + ( e ) d 2 ln( 1) + 3 ln3 ln 3 = d 3 1 Theorems on differentiation Remark. If f() is differentiable, then by Theorem 3.5 (chain rule), we obtain the following: given a positive constant c, then dy f( ) f( ( i) c = c ) (ln c) f ( ) d dy 1 ii log c f( ) f( ) d f( )lnc ( ) = chapter 3: differentiation 11

12 Theorems on differentiation Eample. The hyperbolic sine of, denoted sinh, is given by e sinh := e 2 and the hyperbolic cosine of, denoted cosh, is given by e cosh := + e 2 Theorems on differentiation Eercise. Find the derivatives of ( i) y = eln( ln( ) ) ( ) = + e ( iii) y = ( iv) y = 2 ln( + 1) 2 ep( + 2 4) ( v) y = π ii y ln( e e) 2 cosh(ln ) Logarithmic differentiation chapter 3: differentiation 12

13 Logarithmic differentiation Definition. The process of obtaining dy/d of a function y = f( ) such that it is necessary to perform a logarithmic transformation lny = ln f( ) = g( ) where g is a differentiable function of using elementary theorems is called logarithmic differentiation. Logarithmic differentiation Eample. Find the derivative of y =, > 0 Logarithmic differentiation Eample. Find the derivative of y = k, k > 0 chapter 3: differentiation 13

14 Logarithmic differentiation Eample. Find the derivative of y = log, b > 0 b Logarithmic differentiation Eercise. Find the derivative of the following functions: 1 i lny = 2 + e 2 ( ) ( ) = ln ii y e + 3 e iii y = ln( 1), a > 0 a ( ) Logarithmic differentiation Remark. Suppose u and v are functions of with u( ) > 0, R If y = u v, then it can be shown that dy v dv u du u = lnu d + d v d chapter 3: differentiation 14

15 Implicit differentiation Implicit differentiation Definition. Aneplicit function is an epression of the form f(, y ) = 0 which can be rewritten as y = g( ) or = h( y) If the above is not possible given f, we then say f is an implicit function. Implicit differentiation The process of obtaining dy/d given an implicit function f(, y ) = 0 is called implicit differentiation. Remark. Given an implicit function f(,y) = 0, we assume that y = g() to obtain dy/d (for some g), given f. chapter 3: differentiation 15

16 Implicit differentiation Eample. Let 2 2 0,,, = a + by + cy a b c R Find dy/d. Implicit differentiation Eample. [Niels Hendrik Abel (1827)] Consider the lemniscate ( ) ( ) y = a y 2, a R a a Find dy/d and d/dy. Implicit differentiation Eample. Find dy/d if y = y chapter 3: differentiation 16

17 Higher-order differentiation Higher- order differentiation Definition. If f is a differentiable function, we call dy f ( ) = d the first derivative of f at. If f () is again differentiable with respect to, then we can take again its derivative, denoted Higher- order differentiation 2 d dy d y f ( ) = = 2 d d d and we call the above the second derivative of f at. Continuing in this manner (assuming that the derivative at every step of differentiation is still a differentiable function), we have chapter 3: differentiation 17

18 Higher- order differentiation n ( n d y f ) ( ) = n d called the n th derivative of f at. Remark. Not all functions possess higher-order derivatives. The following set of terms are important in characterizing functions used in economic theory. Higher- order differentiation Definition. If f is a differentiable function (i.e., f eists), we say that f isdifferentiable. If f is a continuous function, we say that f iscontinuously differentiable. If f is again differentiable (i.e., f eists), we say that f is twice differentiable. If f is a continuous function, we say that f is twice continuously differentiable. Higher- order differentiation Eample. Consider the function y = 4 chapter 3: differentiation 18

19 Higher- order differentiation Eample. Consider the general quadratic polynomial 2 y a b c a = + +, R* Higher-order differentiation Eercise. If n n + y = m for some nonzero real number m, show that d y m( n 1) n = 2 2n 1 d y 2 2 Higher- order differentiation Eample. Consider the natural eponential function y = e, R chapter 3: differentiation 19

20 Higher- order differentiation Eample. Given the function k y = e, k R* Higher- order differentiation Eample. Given the function y = 1, R* Higher- order differentiation Eample. From our earlier eample, we have computed for the derivatives of sinh and cosh: d d e + e sinh = = cosh d d 2 d d e e cosh = = sinh d d 2 chapter 3: differentiation 20

21 Higher-order differentiation Eercise. [Leibniz generalization of a product] If u() and v() are differentiable up to the n th order, then n n n d ( n k) ( k) ( ) ( ) ( ) ( ) n u v u = v k= 0 k d n n! where = k ( n k)! k! Marginal functions Marginal functions Definition. Let y = f() be differentiable. The marginal function of f with respect to, denoted MF(), is MF( ) = f ( ) and the average function of f with respect to, denoted AF(), is f( ) AF( ) = chapter 3: differentiation 21

22 Marginal functions Eample. Consider the cost function C 3 2 ( ) = Denote C( ) = TC( ) i.e., the total cost function (TC) with respect to the commodity. Marginal functions Recall that TC( ) = VC( ) + FC( ) i.e., total cost comes with two components: fied cost (FC) and variable cost (VC). From the given, we have 3 2 VC( ) = FC( ) = 1 Marginal functions From the total cost we obtain TC( ) = VC( ) + FC( ) 2 MC( ) = C ( ) = C( ) 2 1 AC( ) = = chapter 3: differentiation 22

23 Marginal functions Denote AC( ) = ATC( ) i.e., we call average cost as the average total cost, also having two components ATC( ) = AVC( ) + AFC( ) the average fied cost (AFC) and the average variable cost (AVC). Marginal functions From the average total cost ATC( ) = we have AVC ( ) = AFC( ) = Marginal functions Theorem 3.9. Let y = f() be a differentiable function. Then ( i) AF ( ) > 0 MF( ) > AF( ) ( ii) AF ( ) = 0 MF( ) = AF( ) ( iii) AF ( ) < 0 MF( ) < AF( ) Remark. This theorem eplains why the graphs of the cost and production functions are such. chapter 3: differentiation 23

24 Marginal functions Remark. Let y = f() be differentiable. We have f( ) AF( ) =, MF( ) = f ( ) In this lieu, the function y = f() is called the total function, (denoted TF()): TF( ) = f( ) Point elasticity Point elasticity Definition. Let y = f() be differentiable. The point elasticity ε y of y with respect to, is ε = y Remark. dy y ε = y d dy y d f = ( ) ( f ) = MF( ) AF( ) chapter 3: differentiation 24

25 Point elasticity Theorem Given the differentiable functions u = f() and v = g(). Let > 0. Then ( i) ε = ε + ε uv, u, v, ( ii) ε = ε ε /,,, u v u v uε + vε ε u + v, u + v u,, ( iii) = v Point elasticity u,, ( iv) = εu v, uε vε u v ( v) u a R ε = * = 0 ( vi) ε = ε ε u( v), u, v v, ( vii) z = u p, p R ε = pε u v z, u, Point elasticity Eample. Given the differentiable functions 2 y = e, z = 2 ( i) w = yz ( ii) v = y + z 4 Find the elasticitiesof the following with respect to : chapter 3: differentiation 25

26 Point elasticity Eample. [Linear demand function] Let q = a bp, where a, b are positive real numbers. The point elasticity of q with respect to p is Point elasticity Definition. Let y = f() with dy/d < 0. We say that y is i. elastic, if ε y > 1 ii. unit elastic, if ε y = 1 iii.inelastic, if ε y < 1 Point elasticity Eample. Consider the demand function q = 8 4p where q is quantity demanded and p is the price. Determine for what values of p so that demand is elastic and inelastic. When is demand unit elastic? chapter 3: differentiation 26

27 Point elasticity Eercise. Consider the demand function q = a bp, a, b > 0 If the midpoint of the demand curve is at (q m,p m ), then ε ( q, p ) = 1 qp m m Point elasticity Eample. [Demand functions with constant elasticity] Consider the demand function k q = cp, c > 0 Point elasticity Eercise. Obtain the elasticity of the following functions: 4 ( i) y = 3 ( ) 5 ii log y = 8 ( iii) log 2 = y 6 ( iv) y = e ε y chapter 3: differentiation 27

28 Point elasticity Eample. [Elasticity and revenue] Recall that a firm s revenue function is given by R( ) = p, p > 0 where p is the price of each sold in the market. If the firm faces the differentiable demand function for given by = g( p) Point elasticity then, using Theorem 3.3, d R ( p) = p + dp Point elasticity Theorem Let = g(p) be a differentiable demand function (that is downward sloping) for a good with corresponding market price p. Let the firm s revenue function be R() = p. Suppose that change in p is positive. Then chapter 3: differentiation 28

29 Point elasticity ( i) ε < p ( ii) ε > p 1 R is increasing i.e., R ( p) > 0 1 R is decreasing i.e., R ( p) < 0 Point elasticity Corollary Under the same hypotheses of Theorem 3.11 but with a negative change in p, then ( i) ε < p ( ii) ε > p 1 R is decreasing i.e., R ( p) < 0 1 R is increasing i.e., R ( p) > 0 Point elasticity Eercise. Show that given a demand function q = f(p), the own price elasticity is invariant under a change of measurement, i.e., when q is measured in a different scale, the own price elasticity of demand with respect to the new measurement is the same as that of the old measurement. chapter 3: differentiation 29

30 Rates of growth Rates of growth Definition. Let y = f( ), y, R ++ where f is differentiable. The instantaneous rate of growth of y is given by ρ = 1 dy y y d Rates of growth Theorem Let u = f() and v = g() be differentiable functions of. Let u,v > 0 for all > 0. Then ( i) ρ = ρ + ρ uv u v ( ) = u/ v u v ii ρ ρ ρ chapter 3: differentiation 30

31 Rates of growth Eample. Let 2 3(4) t t =, y = 5e Find the growth rate of the following epressions: ( i) v = y ( ii) w = y 1 ( iii) z = + y Rates of growth Eercise. Show that if u and v are differentiable functions of, then u v ρ = ρ + ρ u + v u + v u u + v v and u v ρ = ρ ρ u v u v u u v v Rates of growth Corollary Let Then rt A( t) = A(0) e, t R ρ = A( t) r ++ Remark. The above result is the reason we call this function an eponential growth function (given r > 0). chapter 3: differentiation 31

32 Rates of growth Eercise. Let w = 10, z = 20, ρ w = 0.02, ρ = 0.05 where ρ w is the growth rate of w and ρ z the growth rate of z. find the growth rates of the following: ( i) u = w z z ( ii) v = w z Partial derivative Partial derivative Definition. Let f be a function of several variables, say 1, 2,..., n. The partial derivative of f at j is given as y f( 1,..., j + h,..., n) f( 1,..., n) = lim h 0 h j if this limit eists. If all partial derivatives with respect to the n variables eist, then the n-tuple chapter 3: differentiation 32

33 Partial derivative y y y,,..., 1 2 n is called the gradient of f at the point ( 1, 2,..., n ). This is also denoted f and read as del f. Geometric interpretation Consider the function z = f(, y) Observe that f forms a surface 3 in R. Let l be a line on the y-plane Note that this line is a constant on the plane, say y = c. This line is actually an intersection of the plane parallel to the z-plane. Geometric interpretation Also, the said plane with the function f forms a curve BDC. Observe that at the point D, a tangent line can be defined. Its slope is precisely z/, assuming the value of y constant (in particular, y = c). chapter 3: differentiation 33

34 Geometric interpretation z f(,y) l y Geometric interpretation In particular, applying the definition of the partial derivative of f at (holding y constant), we have z f( + h, y) f(, y) f = = lim h 0 h For this reason, partial derivatives are equivalent to the ceteris paribus assumption in economics. Partial derivative Eample. Consider a Cobb- Douglas production function in three variables: Q α β γ = AK L N, A, α, β, γ > 0 α + β + γ = 1 where Q denotes output, K denotes capital, L denotes labor, and N denotes land. The partial derivatives are given by chapter 3: differentiation 34

35 Partial derivative Q α 1 β γ = αak L N K α α β γ α = AK L N = Q K K Similarly, we will get Q α β 1 γ β = β AK L N = Q L L Q AK α L β N γ 1 γ = γ = Q N N Partial derivative Definition. Let f be a function of several variables, say 1, 2,..., n. The partial elasticity of y with respect to j is defined as ε yj j = y y j y j = y j Partial derivative where y j = marginal function with respect to j and y = j average function with respect to j chapter 3: differentiation 35

36 Partial derivative Eample. We obtain the partial elasticities of the Cobb-Douglas function Q = α β γ AK L N Using our previous results on partial derivatives, we first calculate the partial elasticity of Q with respect to K: Partial derivative Remark. Suppose we linearize the Cobb-Douglas function by taking logarithms: Q = α β γ AK L N lnq = lna + αlnk + βlnl + γ lnn Partial derivative Definition. Consider a demand function q = f( p, r, y) where p is the own price of q, r is the price of a good related to q (say w), and y is the income of a particular consumer. We have the following terms that will describe q in relation to r and y: chapter 3: differentiation 36

37 Partial derivative ε ε ε ε qr qr qy qy r q = < 0 q r r q = > 0 q r y q = < 0 q y y q = > 0 q y q and w are complements q and w are substitutes q is an inferior good q is a normal good Partial derivative For these reasons, we call ε qp the own price elasticity of demand, ε qr the cross price elasticity of demand, and ε qy the income elasticity of demand. Note that in the last two, we are only after the signs of the elasticities. Partial derivative Eample. Consider a demand function q = ep r y where p, r, and y are the same variables defined previously. chapter 3: differentiation 37

38 Partial derivative Eercise. Given the following demand function of a good q with own price p and income y, q = 100 2p + 1 y 50 find the own price elasticity of demand and the income elasticity of demand, when p = 20 and y = 5000, respectively. Partial derivative Remark. Eistence of partial derivatives does not necessarily imply continuity (in the case of functions of several variables). Consider the mapping y, (, y ) (0,0) 2 2 f(, y) = + y 0, (, y) = (0,0) Partial derivative By definition, f(,0) f(0,0) f (0,0) = lim 0 0 (0) = lim = 0 chapter 3: differentiation 38

39 Partial derivative Also by definition, f(0, y) f(0,0) f y(0,0) = lim y 0 y 0 (0) y y = lim y 0 y 0 = 0 Partial derivative These two cases imply that f (0,0) f (0,0) eist. However note that at the line = y, we have 2 y lim = lim (, y) (0,0) 2 2 (, ) (0,0) y 1 1 = lim = (, ) (0,0) 2 2 y Partial derivative and at the line = y, we have 2 y lim = lim (, y) (0,0) 2 2 (, ) (0,0) y 1 1 = lim = (, ) (0,0) 2 2 This means, f is not continuous at (0,0) since lim y f (0,0) = 0 (, y) (0,0) y chapter 3: differentiation 39

40 Higher order partials Higher-order partials Definition. Let f be a function of several variables, say 1, 2,..., n. We call y = f, = 1,..., j j n j the first order partial derivatives of f (or simply, the first partials of f). Since the first partials are functions, we again define their partials: Higher-order partials j y = f, = 1,..., j j n j j called the second order own partial derivatives of f (or simply, the second own partials of f). Also, we can take the partials with respect to some other variable k and obtain chapter 3: differentiation 40

41 Higher-order partials k j y = f, j, k = 1,..., n jk j y = f,, = 1,..., k j k n j k called the second order cross partial derivatives of f (or simply, the second cross partials of f). Higher-order partials Theorem [W. H. Young] Let f be a function of several variables, say 1, 2,..., n. Suppose that f have continuous second order cross partial derivatives. Then y y = f = f = j k kj k j j k j, k = 1,..., n Higher-order partials Eample. Consider again the Cobb-Douglas production function Q α β γ = AK L N, A, α, β, γ > 0 α + β + γ = 1 where Q is output, K is capital, L is labor, and N is land. chapter 3: differentiation 41

42 Higher-order partials Eercise. Verify Young s theorem given the following functions: ( ) 2 2 ( ii) y z = e ln ( iii) + z = 4y + e ( iv) i z = + y y z = 4y 2y y The differential The differential y y = f() f( 0 + ) f ( 0 ) f( 0 ) chapter 3: differentiation 42

43 The differential Definition. Let y = f() be differentiable. The differential of y, denoted dy, is dy = f ( ) where is in the domain of f and is an arbitrary increment of. The differential of, denoted d, is d = The differential Eample. A manufacturer s total cost function is 3 2 q q C( q) = + 300q where q is the level of production. Obtain the actual and approimate change of cost if q increases from 6 to 6.1 units. The differential Eercise. Consider again the manufacturer s total cost function: 3 2 q q C( q) = + 300q Obtain the actual and approimate change of cost if q increases from 6 to 7 units. Compare this result with the one in the previous eample. chapter 3: differentiation 43

44 The differential Proposition [Properties of differentials] Let c be a constant, and let u and v be differentiable functions of some variable. Then ( i) dc = 0 ( ii) d( cu) = c du ( iii) d( u + v) = du + dv ( iv) d( uv) = udv + vdu The differential vdu udv v d( u/ v) =, v 0 2 v ( ) u u ( vi) d c = c c du c > ( ) ( ln ), 0 du (log c ),, 0 ulnc ( vii) d u = u c > The differential Eample. Recall that if y = f() is a differentiable function with > 0, then ε = y dy y d chapter 3: differentiation 44

45 The differential Remark. Consider a differentiable function z = f(,y). From the definition of the differential, The differential Hence, we have the following definition: Definition. Let f be a function of several variables, say 1, 2,..., n. Suppose that all the partial derivatives eist. The total differential dy is given by n n f dy = d = f d j j j j= 1 j j= 1 The differential Theorem [Chain rule for differentials] Let f be a function of several variables, say 1, 2,..., n. Suppose that all the partial derivatives are continuous, and every j = u j (t) is continuously differentiable. Then n dy f dj = = dt dt n j= 1 j j= 1 f u ( t) j j chapter 3: differentiation 45

46 The differential Eample. Let where 2 2 z = + y = t y = t + 3, 3 The differential Eample. Let where z = ln( + y) = w, y = w 2 The differential Eample. [Rates of growth] Consider the Cobb-Douglas function Q = AK( t) L( t) N( t) α β γ chapter 3: differentiation 46

47 The differential Eample. Recall our previous result on elasticities: dy d ε = = (lny) y y d d(ln ) If = (t), and y = y(t) The differential Eercise. Find the differential of the function y = f( ) = + If = g(w), find the derivative dy/dw. The differential Eercise. In general, we can etend the notion of the total derivative as follows: if y = f( 1, 2,..., n ) is differentiable with respect to each k and each k is a differentiable function given by g ( z, z,..., z ) k = k 1 2 m chapter 3: differentiation 47

48 The differential then we obtain the following partial derivatives of y with respect to z 1, given by y y 1 y 2 y = z z z z n 1 With respect to z 2, we obtain y y 1 y 2 y = z z z z n 2 n n The differential We do the above until z m and get y y 1 y 2 y = z z z z m 1 m 2 m n m In general, we have the m equations, given by n y y j =, k = 1,2,..., m z z k j= 1 j k n The differential Eample. [Equilibrium analysis] Consider a demand function for a certain good q, given by q = D( p, r, y), D, D < 0 D p y where p is the price of that good, r is the price of a relevant good (substitutes or complements), and y is consumer s income. Let the supply function be chapter 3: differentiation 48

49 The differential q = S( p, v, w), S > 0 where v and w are the prices of the inputs used in producing q. In equilibrium, q = q = q* D S S D( p*, r, y) = S( p*, v, w) where p* and q* denote equilibrium values of p and q, respectively. p The differential Taking total differentials, we have Taylor s theorem chapter 3: differentiation 49

50 Taylor s approimation Theorem [Taylor s theorem for differentiation (1715)] Let f() have continuous derivatives up to the (n+1) th order on an open interval (a,b). For every pair of points, 0 in (a,b), there is a p between, 0 such that f ( ) f( ) = f( ) + ( ) + R n ( j) 0 j 0 0 n+ 1 j= 1 j! Taylor s approimation where f ( p) R ( ) ( n + 1)! ( n+ 1) = n+ 1 n+ 1 0 Definition. In the Taylor approimation above, we call R n+1 the Lagrange form of the remainder of the approimation [1765]. Taylor s approimation We call the polynomial f ( ) j f( ) = f( ) + ( ) n ( j) j= 1 j! ( n+ 1) f ( p) ( 1 0 ) n+ + ( n + 1)! the Taylor polynomial around 0. chapter 3: differentiation 50

51 Taylor s approimation If 0 = 0, then the approimation given by f (0) f( ) = f(0) +! n ( j) j= 1 j is called the Maclaurin polynomial [1742]. ( n+ 1) f ( p) + ( n + 1)! j n+ 1 Taylor s approimation Definition. A Taylor approimation of a function y = f() at a point 0 where n = 1 is called a linear approimation of f around the point 0. If n = 2, we call the approimation a quadratic approimation of f around the point 0. Taylor s approimation Eample. Find the linear and quadratic approimations of y = e around the point 0 = 2. chapter 3: differentiation 51

52 Taylor s approimation Eample. Find the linear, quadratic, and cubic approimations around the point = 1 of the function g( ) = ln Taylor s approimation Eercise. Let I be an open interval. Let f be twice continuously differentiable and real-valued, and suppose that f (a) eists at a point a in I. Show that f( + h) 2 f( ) + f( h) f ( ) = lim h 0 2 h Taylor s approimation Eercise. Find the cubic, quadratic and linear approimations of the function 1 y = around the point 0 = 1. Further, show that the quadratic and cubic approimations are identical. chapter 3: differentiation 52

53 Taylor s approimation Remark. Why is there a theorem on approimation of functions? Taylor s approimation Remark. In Taylor s approimation, when n = 0, we have f( ) = f( ) + f ( p)( ) 0 0 which is equivalent to Hence we obtain f( ) f( 0) = f ( p) 0 Mean value theorem chapter 3: differentiation 53

54 Mean value theorem Theorem [Mean value theorem for differentiation] Let f be a continuous function on [a,b], differentiable on (a,b), and let a < b. Then there eists a c in (a,b) such that f( b) f( a) f ( c) = b a Mean value theorem f(b) y f'(c) f(b) f(a) b a f(a) a c b Mean value theorem Eample. Consider the function f 2 ( ) =, [0,1] chapter 3: differentiation 54

55 Mean value theorem Theorem [Cauchy s mean value theorem for differentiation] Let f and g be continuous on [a,b] and differentiable on (a,b). Suppose that g ( ) 0, ( a, b) Then, there is a c (a,b) such that f ( c) f( b) f( a) = g ( c) g( b) g( a) Mean value theorem Eercise. Let f() = 1. Show that there is no number p in the interval [ 1,1] such that f(1) f( 1) = f ( p)[1 ( 1)] Does this contradict the mean value theorem applied to the interval [ 1,1]? Mean value theorem Eercise. i. Prove Bernoulli s inequality for every > 1: α (1 + ) α + 1 α 1 α (1 + ) α < α 1 ii. Show that e + 1 R chapter 3: differentiation 55

56 Homogeneity Homogeneity Definition. A function f is said to be homogeneous of degree k iff f( t, t,..., t ) 1 2 k = t f(,,..., ), t > n If k = 1, we then say that the function f is linearlyhomogeneous. (note: k can be any real number) n Homogeneity Eample. Consider the function 2 2 f(, y) = + y + y chapter 3: differentiation 56

57 Homogeneity Theorem [L. Euler] Let f( 1, 2,..., n ) be differentiable and homogeneous of degree k. Then n f j 1 2 j= 1 j kf(,,.. ) called Euler s identity for homogeneous functions of degree k. n Homogeneity Eample. Consider again the Cobb-Douglas production function in three variables: Q = F( K, L, N) = AK L N α β γ Homogeneity Eercise. If a differentiable function y = f(,,..., ) 1 2 n is homogeneous of degree k, then n m= 1 ε ym = k chapter 3: differentiation 57

58 Homogeneity Definition. Let f be a production function of several inputs, say 1, 2,..., n, and let f be homogeneous of degree k. If k < 1, we say that f ehibits decreasing returns to scale. If k = 1, we say that f ehibits constant returns to scale. If k > 1, we say that f ehibits increasing returns to scale. Homogeneity Remark. Not all linear functions are linearly homogeneous. Consider the general linear function y = f( ) = a + b Homogeneity Definition. Let c be a nonzero real number and let f be homogeneous of degree k. Then the function g(,... ) = f(,..., ) + c 1 n 1 is called an affine function. Theorem If f is homogeneous of degree k, then f is homogeneous of degree k 1. n chapter 3: differentiation 58

59 Homogeneity Eample. Consider the function y = f( ) = 3 Homogeneity Eercise. Determine the degree of homogeneity of the following functions. Verify the results using Euler s theorem (Theorem 3.22): ( i) z = f(, y) = ( ii) z = f(, y) = y y y + y 3 3 Indeterminate forms chapter 3: differentiation 59

60 Indeterminate forms Remark. Consider a ratio a/b. Note that in all of the theorems involving ratios, we always put the condition that b 0, for otherwise, we say that the epression is undefined. Loosely speaking, a = if b = 0 a 0 b Indeterminate forms Given a ratio a/b, and both a = b = 0, we say that a/b is indeterminate. Also, if a = b =, we say that a/b is indeterminate, since we noted in the remark after Theorem 2.23 that 1 Indeterminate forms Definition. An indeterminate form of a function h() is where f(a) = g(a) = 0 with f( ) f( a) 0 h( ) = h( a) g( ) = g( a) = 0 Remark. The function h() also takes an indeterminate form when f(b) = g(b) = for some b. chapter 3: differentiation 60

61 Indeterminate forms Remark. Other indeterminate forms include () i ( ii) 0 ( iii) 1 ( iv) 0 0 ( v) ( vi) 0 ( vii) 0 Indeterminate forms Theorem [G.F.A. L Hospital (1696)] Let f() and g() have continuous n th -order derivatives on some open interval (a,b), and let c be in that interval. If f(c) = g(c) = 0 (or if f(c) = g(c) = ), then f ( ) f = ( lim L lim ) = L c g ( ) c g( ) Indeterminate forms Remark. L Hospital s original problem is given by lim a a a aa a where aa = a 2. a 4 3 chapter 3: differentiation 61

62 Indeterminate forms Eample. Find the limit if it eists: lim h 0 h y h What happens if y = 1? h Indeterminate forms Eercise. Find the following limits, if they eist: ( i) lim 2 0 ( ) 0 e t t 1 a b ii lim, a, b > 0 a b e e ( iii) lim a + b c + d Indeterminate forms Eample. Evaluate lim e e chapter 3: differentiation 62

63 Indeterminate forms Remark. A generalization of the preceding eample is the following theorem. Theorem a a > 1 lim = 0 a Proof. Danao [2001], pp Indeterminate forms Remark. Successive applications of the L Hospital s rule does not always lead to a limit that is finite or infinite. For eample, consider the function f( ) = Indeterminate forms Eample. Consider the constant elasticity of substitution function p f(, y) = A a + (1 a) y where p A > 0, a (0,1), p R* 1/ p chapter 3: differentiation 63

64 Indeterminate forms Remark. Recall the definition of the derivative: if f is a differentiable function, then f( + h) f( ) f ( ) = lim h 0 h Indeterminate forms Eercise. i. Show that if c > 0, then c c 1 lnc lim = c c c 1 + lnc ii. Provide an alternative proof for the eistence of Euler s number: 1 e = lim + 1 n n n Implicit function theorems chapter 3: differentiation 64

65 Implicit function theorems Theorem [Implicit function theorem] Let F(,y) be defined on an open disk D containing the point ( 0,y 0 ) and suppose that ( a) F(, y ) = ( ) continuous on b F F D ( c) Fy 0 y0 y (, ) 0 Implicit function theorems Then, there eists a function f defined on an open interval I containing 0 such that ( i) y = f( ) on I ( ii) F(, f( )) = 0 on I ( iii) fhas a continuous derivative on I Implicit function theorems Remark. The implicit function theorem guarantees the eistence of f, only on some open interval I. Further, it does not say anything about the form of the function f; it merely guarantees its eistence, given the conditions above. chapter 3: differentiation 65

66 Implicit function theorems Theorem If f is a differentiable function of such that y = f(), and f is defined implicitly by the function F(,y) = 0 having continuous first partials and that F y (,y) is nonzero, then dy F (, y) = d F (, y) y Implicit function theorems Definition. Let z = f(,y). A level curve of f is defined by z 0 = f(,y), where z 0 is a constant. Remark. From z = f(,y), we can define an implicit function F(,y) = 0 given by F(, y) = f(, y) z = 0 0 Implicit function theorems Remark. Consider a level curve in the Cartesian plane, given by z = f(, y) Let f be homogeneous of degree k, i.e., Let f( t, ty) = t k f(, y), t > 0 z = f(, y ) chapter 3: differentiation 66

67 Implicit function theorems If in addition, f is also differentiable, then by a previous eercise, f is homogeneous of degree k 1. Observe that the slope of the level curve at the point ( 0,y 0 ) using Theorem 3.30 is given by dy f ( 0, y0) =, f y ( 0, y0 ) 0 d f (, y ) y 0 0 Implicit function theorems y f ( 0, y0) f (, y ) y 0 0 y 0 ( 0,y 0 ) (0,0) 0 z = f(, y ) Implicit function theorems Definition. A level curve of a utility function U = U( 1, 2 ), is called an indifference curve. The slope of an indifference curve is given by d U (, ) = d U (, ) The numerical value of this derivative is called the marginal rate of substitution. chapter 3: differentiation 67

68 Implicit function theorems Eercise. Consider a loglinear utility function U(, y) = 2ln + 4lny Obtain the marginal rate of substitution (MRS) if (,y) = (3,5) and interpret. Implicit function theorems Eample. Consider a utility function given by 2 U(, y) = + y Implicit function theorems Eercise. Verify that the utility functions given by 1 2 ( i) U(, y) = y ( ii) U(, y) = 3y + ln are quasilinear utility functions. Provide a sketch of the graph of each utility function. 2 chapter 3: differentiation 68

69 Implicit function theorems Definition. A level curve of a production function Q = F(K,L), is called an isoquant. The slope of an isoquant is given by dk dl F ( K, L) L = F ( K, L) The numerical value of this derivative is called the marginal rate of technical substitution. K Implicit function theorems Eercise. Consider a production function Q = F( K, L) = lnk + lnl Obtain the marginal rate of technical substitution (MRTS) if (K,L) = (4,9) and interpret. Implicit function theorems Definition. A level curve of a cost function C = C(w 1,w 2 ) is called a production possibilities frontier, and its slope is given by dw C ( w, w ) 2 w1 1 2 = dw C ( w, w ) 1 w2 1 2 The numerical value of this derivative is called the marginal rate of transformation. chapter 3: differentiation 69

70 Implicit function theorems Eercise. Consider a production possibilities frontier defined by Q = F( w, z) = w + 4z 2 2 Obtain the marginal rate of transformation (MRT) if (w,z) = (3,5) and interpret. Implicit function theorems Theorem Let F(,y,z) be defined on an open ball B containing the point ( 0,y 0,z 0 ) and suppose that ( a) F(, y, z ) = ( ) b F, F, F continuous on B y z ( c) F y z (,, ) 0 z Implicit function theorems Then, there eists a function f defined on an open disk D containing ( 0,y 0 ) such that ( i) z = f(, y) on D ( ii) F(, y, f(, y)) = 0 on D ( iii) fhas continuous partial derivatives on D chapter 3: differentiation 70

71 Implicit function theorems Theorem If f is a differentiable function of,y such that = f(,y), and f is defined implicitly by the function G(,y,z) = 0 having continuous first partials and that F z (,y,z) is nonzero, then dz G (, y, z) (,, ) dz G y z y = = d G (, y, z) dy G (, y, z) z z Implicit function theorems Eample. Consider again the Cobb-Douglas production function in three variables: Q = F( K, L, N) = AK L N α β γ To end... A man is like a fraction whose numerator is what he is and whose denominator is what he thinks of himself. The larger the denominator, the smaller the fraction. Leo Tolstoy chapter 3: differentiation 71