Tutorial 2: Comparative Statics

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Tutorial 2: Comparative Statics ECO42F 20 Derivatives and Rules of Differentiation For each of the functions below: (a) Find the difference quotient. (b) Find the derivative dx. (c) Find f (4) and f (3).. y 4x 2 + 9 2. y 5x 2 4x 3. y 5x 2 Find the derivative of each of the following: 4. y f (x) x 3 5. y f (x) 7x 6 6. y f (u) 4u /2 8. y x 2 (4x + 6) 9. y (ax b) ( cx 2) 30. y 5x4 x 2 + 4x 20. y (2 3x) ( + x) (x + 2) 3. y x ( + x 2) 2 2 + x 2 7. y f (x) 63 8. y f (x) 3x 9. y f (x) x 4 0. y f (w) 9w 4. y f (u) au b 2. y f (x) cx 2 3. y f (w) 3 4 w4/3 4. y f (x) ax 2 +bx+c 5. y 7x 4 +2x 3 3x+37 6. y ( 9x 2 2 ) (3x + ) 7. y (3x + ) ( 6x 2 5x ) 2. y ( x 2 + 3 ) x 22. y 4x x + 5 23. y x + 7 x 24. y ax2 + b cx + d 25. y ( 3x 2 3 ) 3 26. y ( 8x 3 5 ) 9 27. y (ax + b) 4 28. y (6x + 3) 2 29. y x3 + 2x x + 32. y ( ) x 3 + x 2 2 + x 2 33. y f (t) ln at 34. y f (t) t 3 ln t 2 35. y e 2t+4 36. y e t2 + 37. y e ax2 +bx+c 38. y ln 8t 5 39. y ln (t + 9) 40. y ln x ln ( + x)

Find the derivative of each of the following by first taking the natural log of both sides: 4. y x 2 (x + 3) (2x + ) 42. y 3x (x + 2) (x + 4) 43. Given y u 3 + where u 5 x 2, find dx. 44. Given w ay 2 where y bx 2 + cx, find dw dx. 45. Are the following functions monotonic? (a) y x 6 + 5 (x > 0) (b) y 4x 5 + x 3 + 3x 46. Given the function f (x) ax + b, find the derivatives of (a) f (x) (b) xf (x) (c) f (x) (d) f (x) x 47. Given the total cost function C Q 3 5Q 2 + 4Q + 75, what is the variable cost function? Find the derivative of the variable cost function, and give the economic meaning of this derivative. 48. Given the average cost function AC Q 2 4Q+24, find the marginal cost function. 49. Given that average revenue is given by AR 60 3Q, find the marginal revenue curve. 50. If f (H, h) λ ( βh) h, where λ and β are positive constants, find f H and f h. 5. If f (T, t) T Qn + T Qnm (t T ) T 2 rqn, where Q, n, m and r are positive constants, find f T and f t. 2

Find y x and y x 2 for each of the following functions: 52. y 2x 3 x2 x 2 + 3x 2 2 54. y (2x + 3) (x 2 2) 53. y 7x + 5x x 2 2 9x3 2 55. y 4x + 3 x 2 2 Find f x and f y for each of the following functions: 56. f (x, y) x 2 + 5xy y 3 58. f (x, y) 2x 3y x + y 57. f (x, y) ( x 2 3y ) (x 2) 59. f (x, y) x2 xy 60. If the utility function of an individual takes the form U U (x, y) (x + 2) 2 (y + 3) 3, where U is total utility and x and y are the quantities of the two commodities consumed. (a) Find the marginal utility of each of the two commodities. (b) Find the value of the marginal utility of commodity x when 3 units of each commodity are consumed. 6. Given the national income model below Y C + I + G C a + b(y T ) a > 0, 0 < b < T d + ty d > 0, 0 < t < where Y is national income, C is (planned) consumption expenditure, I is investment expenditure, G is government expenditure and T is taxes. (a) Solve for Y, C and T using Cramer s Rule. (b) Find the government expenditure multiplier and the investment multiplier. (c) How is consumption affected by increased government spending. Do your findings accord with economic logic? 3

62. Let the national income model be Y C + I + G C a + b(y T ) a > 0, 0 < b < G gy 0 < g < where Y is national income, C is (planned) consumption expenditure, I is investment expenditure, G is government expenditure and T is taxes. (a) Solve for Y, C and G and state what restriction is required on the parameters for a solution to exist. (b) Give the economic meaning of the parameter g. (c) Find the tax multiplier and the investment multiplier, and give the economic intuition behind their signs. (d) Show that an increase in tax has a negative impact on consumption. 63. Consider the following simple Keynesian macroeconomic model Y C + I + G C 200 + 0.8Y I 000 2000R where Y is national income, C is (planned) consumption expenditure, I is investment expenditure, G is government expenditure and R is the interest rate. Use Cramer s Rule to solve for Y, and evaluate the effect of a R50 billion decrease in government spending on national income. 64. Use Jacobian determinants to test the existence of functional dependence between the paired functions below: (a) y 3x 2 + x 2 y 2 9x 4 + 6x2 (x 2 + 4) + x 2 (x 2 + 8) + 2 (b) y 3x 2 + 2x2 2 y 2 5x + 4

2 General Function Models (Total Differentiation). Find the total differential, given: (a) y x ( x 2 + 3 ) (b) y (x 8) (7x + 5) (c) y x x 2 + (d) y 0x 3 2x3 2 (e) y 3x 2 + 4x3 2 (f) y 3x 2 + xz 2z 3 (g) y 2x + 9x x 2 + x 2 2 2. Given the consumption function C a + by (where a > 0, 0 < b <, Y > 0), find the income elasticity of consumption ε CY. 3. Find the total derivative dz, given (a) z f (x, y) 2x + xy y 2 where x g (y) 3y 2 (b) z f (x, y) (x + y) (x 2y) where x g (y) 2 7y 4. Find the rate of change of output with respect to time, if the production function is Q A(t)K α L β, where A(t) is an increasing function of t, K K 0 e at and L L 0 +bt. 5. Given an isoquant which represents a production function y f (x, x 2 ), where y is some constant level of output. Use total differentials to derive an expression for the marginal rate of technical substitution between the factors of production, x and x 2. 5

3 Derivatives of Implicit Functions. Given F (y, x) x 3 2x 2 y + 3xy 2 22 0, is an implicit function y f (x) defined around the point (y 3, x )? If so, find by the implicit function rule, and dx evaluate it at this point. 2. Given F (y, x) 2x 2 + 4xy y 4 + 67 0, is an implicit function y f (x) defined around the point (y 3, x )? If so, find by the implicit function rule, and dx evaluate it at this point. 3. For each of the given equations F (y, x) 0, find by (i) solving for y, (ii) the dx implicit function rule, and (iii) taking the total differential: (a) F (y, x) x 2 + xy 4 0 (b) F (y, x) xy 2 0 (c) F (y, x) 3x 3 y 2x 2 + 6 0 4. Given F (y, x) x 2 3xy + y 3 7 0, check that an implicit function y f (x) is defined around the point (y 3, x 4). Hence find by the implicit function rule. dx 5. Given F (y, x) ye y x 0, verify the conditions of the implicit function theorem. 6. Consider the function corresponding to the upper semi-circle of the set of points in the xy-plane satisfying x 2 + y 2 4. Find dx by: (a) explicitly writing out the function y f (x) and finding its derivative. (b) the implicit function rule. 7. Given F (y, x) 0x y + x 2 x 2 + y 2 x 2 0, find the value x 2 for which the implicit function y f (x, x 2 ) is defined around the point (y, x 2, x 2?)? Hence y find the values of and y at this point. x x 2 8. Assuming that the equation F (U, x, y) 0 implicitly defines a utility function U f (x, y), find an expression for the marginal rate of substitution. 6

9. The market for a single commodity is described by the following set of equations Q d Q s Q d D(P, G) Q s S(P, N) where G is the price of substitutes and N is the price of inputs, and G and N are exogenously given. The following assumptions are imposed D P S P < 0, D G > 0 > 0, S N < 0 Use the implicit function theorem to show that the system implicitly defines the functions P (G, N) and Q (G, N). Hence use the implicit function rule to find and sign the derivatives P G, Q G, P N 0. Consider the system of equations and Q N. F (x, y; a) x 2 + axy + y 2 0 F 2 (x, y; a) x 2 + y 2 a 2 + 3 0 around the point (x 0, y, a 2). What is the impact of a change in a on x and y?. Let the national income model be written in the form Y C I 0 G 0 0 C α β (Y T ) 0 T γ δy 0 where the endogenous variables are Y (national income), C ((planned) consumption expenditure) and T (taxation), and the exogenous variables are I 0 (investment expenditure) and G 0 (government expenditure). (a) Find the government expenditure multiplier by the implicit function rule. (b) Find the nonincome tax multiplier by the implicit function rule. 7

Tutorial 2: Comparative Statics SELECTED SOLUTIONS ECO42F 20 Derivatives and Rules of Di erentiation. (a) y x 8x 0 + 4x 2. (a) y x 0x 0 + 5x 3. (a) y x 5 (b) dx 8x (c) f 0 (4) 32; f 0 (3) 24: 4 (b) dx 0x 4 (c) f 0 (4) 36; f 0 (3) 26: (b) dx 5 (c) f 0 (4) 5; f 0 (3) 5: 4. f 0 (x) 3x 2 5. f 0 (x) 42x 5 6. f 0 (u) 2u 2 2 p u 7. f 0 (x) 0 8. f 0 (x) 3x 2 3 x 2 9. f 0 (x) 4x 5 4 x 5 0. f 0 (w) 36w 3. f 0 (u) abu b 2. f 0 (x) 2cx 3. f 0 (w) w 3 4. f 0 (x) 2ax + b 5. 6. 8. 20. dx 28x3 + 6x 2 3 dx 3 27x2 + 6x 2x (x + ) dx x (9x + 4) dx 33. f 0 (t) t 34. f 0 (t) 2t 2 ( + 3 ln t) 2 35. 36. 37. 38. 39. 40. dt 2e2t+4 dt 2tet2 + dx (2ax + b) eax2 +bx+c dt 5 t dt t + 9 dx x ( + x) 4. dx x (7x + 6) (x + 3) 2 (2x + ) 2 42. dx 3 8 x 2 (x + 2) 2 (x + 4) 2

47. V C Q 3 5Q 2 + 4Q (Recall that to nd variable costs, you drop the xed costs) dv C dq 3Q2 0Q + 4 This is the same as the marginal cost function. (Check this by di erentiating the total cost function - you will get the same result) 64. (a) jjj 0 The functions are dependent. (b) jjj 20x 2 The functions are independent. 2 General Function Models (Total Di erentiation). (a) 3 x 2 + dx (b) (4x 5) dx x 2 (c) (x 2 + ) 2 dx (d) 30x 2 dx 6x 2 2 dx 2 (e) 6x dx + 2x 2 2 dx 2 (f) (6x + z) dx + x 6z 2 dz (g) (2 + 9x 2 ) dx + (9x + 2x 2 ) dx 2 2. dc dy b and C Y a + by Y ) " CY dcdy CY b (a + by ) Y by (a + by ) 3. (a) dz (b) dz 0y + 9y2 02y 30 2

4. dq @Q @Q @Q da + dk + @A @K @L dl dq @Q da dt @A dt + @Q dk @K dt + @Q dl h @L dt K L i A 0 (t) h + A(t)K L i ak0 e at + h K L i A 0 (t) + a A K K 0e at + b A L h A(t)K L i [b] 5. We know that y will be constant along a given isoquant. The total di erential is given by Because y is constant, 0. Thus @f @x dx + @f @x 2 dx 2 @f @x dx + @f @x 2 dx 2 0 ) @f @x 2 dx 2 dx 2 dx 3 Derivatives of Implicit Functions @f dx @x @f@x MRT S @f@x 2. (a) Does the function F have continuous partial derivatives? Yes. F y 2x 2 + 6xy F x 3x 2 4xy + 3y 2 (b) Is F y 6 0 at a point satisfying F (y; x) 0? F (y 3; x ) () 3 2 () 2 (3) + 3 () (3) 2 22 6 + 27 22 0 and F y (y 3; x ) 2 () 2 + 6 () (3) 6 6 0 Therefore, an implicit function y f (x) is de ned around the point (y 3; x ) and we can use the implict function rule to nd dx F x 3x 2 4xy + 3y 2 F y 2x 2 + 6xy 3

At the point (y 3; x )! dx 3 () 2 4 () (3) + 3 (3) 2 2 () 2 + 6 () (3) 9 8 2. An implicit function y f (x) is de ned around the point (y 3; x ) and we can use the implict function rule to nd dx 4x + 4y 2 4x 4y 3 3 at the point (y 3; x ) : 4. (a) Does the function F have continuous partial derivatives? Yes. F y 3x + 3y 2 F x 2x 3y (b) Is F y 6 0 at a point satisfying F (y; x) 0? F (y 3; x 4) (4) 2 3 (4) (3) + (3) 3 7 0 and F y (y 3; x 4) 3 (4) + 3 (3) 2 5 6 0 Therefore, an implicit function y f (x) is de ned around the point (y 3; x 4) and we can use the implict function rule to nd dx F x 2x 3y F y 3x + 3y 2 5. (a) Does the function F have continuous partial derivatives? Yes. F y ( + y) e y F x (b) Is F y 6 0 at a point satisfying F (y; x) 0? From (a) we can see that F y 0, y 4

Thus we can nd combinations of (y 0 ; x 0 ) ; y 0 6 that satisfy F (y 0 ; x 0 ) 0 and F y (y 0 ; x 0 ) 6 0. For example, the point (y ; x e) F (y ; x e) () e e 0 and F y (y ; x e) ( + ) e 2e 6 0 Therefore, an implicit function y f (x) is de ned around the point (y ; x e) and we can use the implict function rule to nd dx F x F y ( + y) e y ( + y) e y 6. (a) (b) dx x 4 x2 2 dx x y 7. x 2 4 9. Rewrite the system as two equations by letting Q Q d Q s : Q D(P; G) Q S(P; N) or equivalently: F (P; Q; G; N) D(P; G) Q 0 F 2 (P; Q; G; N) S(P; N) Q 0 5

Check the conditions for the implicit function theorem: (a) (b) F P F Q F G @G F N 0 F 2 P F 2 Q F 2 G 0 F 2 N @N Therefore, continuous partial derivatives wrt all endogenous and exogenous variables exist. Therefore, jjj 6 0 jjj > 0 FP FQ FP 2 FQ 2 + Conditions for implicit function satis ed, and so system implicitly de nes the functions P (G; N) and Q (G; N) : Use the implicit function rule: 6

First, take the total di erential of each equation: df F P dp + F QdQ + F GdG + F NdN 0 ) dp dq + @G dg + 0 0 dp dq dg () @G df 2 F 2 P dp + F 2 QdQ + F 2 GdG + F 2 NdN 0 ) dp Putting equations () and (2) in matrix form dp dq dq + 0 + @N dn 0 dp dq dn (2) @N @G 0 dg + dn (3) 0 @N Note that the coe cient matrix is the Jacobian matrix J. To nd @Q and we partially di erentiate with respect with G, holding N @G @G constant which implies that dn 0. Setting dn 0 and dividing through by dg in (3) gives: 2 3 6 4 @G 7 @Q 5 @G 0 @G (Note the partial derivative signs - we are di erentiating with respect to G, holding N constant). Use Cramer s rule to solve for @G @G and @Q @G : > 0 @G 0 jjj @G 7

@Q @G > 0 @G 0 jjj @G To nd @Q and we partially di erentiate with respect with N, holding G @N @N constant which implies that dg 0. Setting dg 0 and dividing through by dn in (3) gives: 2 3 6 4 @N 7 0 @Q 5 @N @N (Note the partial derivative signs - we are di erentiating with respect to N, holding G constant). Use Cramer s rule to solve for @N @N and @Q @N : > 0 0 @N jjj @N @Q @N < 0 0 @N jjj @N 0. At the point (x 0; y ; a 2), the equations F 0 and F 2 0 are satis ed. 8

The F i functions possess continuous partial derivatives: Fx 2x + ay Fy ax + 2y Fa xy F 2 x 2x F 2 y 2y F 2 a 2a Thus, if the Jacobian jjj 6 0 at point (x 0; y ; a 2) we can use the implicit function theorem. First take the total di erential of the system (2x + ay) dx + (ax + 2y) + xyda 0 2xdx + 2y 2ada 0 Moving the exogenous di erential da to the RHS and writing in matrix form we get 2x + ay ax + 2y dx xy da 2x 2y 2a where the coe cient matrix of the LHS is the Jacobian jjj F x Fy 2x + ay ax + 2y 2x 2y At the point (x 0; y ; a 2), jjj 2 2 0 2 4 6 0 Therefore the implicit function rule applies and " # @x 2x + ay ax + 2y @a 2x 2y @y F 2 x F 2 y @a xy 2a 9

Use Cramer s rule to nd an expression for @x at the point (x 0; y ; a 2) : @a xy ax + 2y @x 2a 2y @a jjj 0 2 4 2 4 8 4 2 Use Cramer s rule to nd an expression for @y at the point (x 0; y ; a 2) : @a 2x + ay @x xy 2x 2a @a jjj 2 0 0 4 4 8 4 2. See Chiang, example 6, pages 203-204. (a) (b) @Y @G + @Y @ + 0