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Introductory maths analysis chapter 12 official

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Introductory maths analysis chapter 12 official

1. 1. INTRODUCTORY MATHEMATICALINTRODUCTORY MATHEMATICAL ANALYSISANALYSISFor Business, Economics, and the Life and Social Sciences ©2007 Pearson Education Asia Chapter 12Chapter 12 Additional Differentiation TopicsAdditional Differentiation Topics
2. 2. ©2007 Pearson Education Asia INTRODUCTORY MATHEMATICAL ANALYSIS 0. Review of Algebra 1. Applications and More Algebra 2. Functions and Graphs 3. Lines, Parabolas, and Systems 4. Exponential and Logarithmic Functions 5. Mathematics of Finance 6. Matrix Algebra 7. Linear Programming 8. Introduction to Probability and Statistics
3. 3. ©2007 Pearson Education Asia 9. Additional Topics in Probability 10. Limits and Continuity 11. Differentiation 12. Additional Differentiation Topics 13. Curve Sketching 14. Integration 15. Methods and Applications of Integration 16. Continuous Random Variables 17. Multivariable Calculus INTRODUCTORY MATHEMATICAL ANALYSIS
4. 4. ©2007 Pearson Education Asia • To develop a differentiation formula for y = ln u. • To develop a differentiation formula for y = eu . • To give a mathematical analysis of the economic concept of elasticity. • To discuss the notion of a function defined implicitly. • To show how to differentiate a function of the form uv . • To approximate real roots of an equation by using calculus. • To find higher-order derivatives both directly and implicitly. Chapter 12: Additional Differentiation Topics Chapter ObjectivesChapter Objectives
5. 5. ©2007 Pearson Education Asia Derivatives of Logarithmic Functions Derivatives of Exponential Functions Elasticity of Demand Implicit Differentiation Logarithmic Differentiation Newton’s Method Higher-Order Derivatives 12.1) 12.2) 12.3) Chapter 12: Additional Differentiation Topics Chapter OutlineChapter Outline 12.4) 12.5) 12.6) 12.7)
6. 6. ©2007 Pearson Education Asia Chapter 12: Additional Differentiation Topics 12.1 Derivatives of Logarithmic Functions12.1 Derivatives of Logarithmic Functions • The derivatives of log functions are: ( )               += → hx h x h x x dx d / 0 1limln 1 lna. ( ) 0where 1 lnb. ≠= x x x dx d ( ) 0for 1 lnc. ≠⋅= u dx du u u dx d
7. 7. ©2007 Pearson Education Asia Chapter 12: Additional Differentiation Topics 12.1 Derivatives of Logarithmic Functions Example 1 – Differentiating Functions Involving ln x b. Differentiate . Solution: 2 ln x x y = ( ) ( ) ( ) ( ) ( ) 0for ln21 2)(ln 1 lnln ' 3 4 2 22 22 > − = −      = − = x x x x xx x x x x dx d xx dx d x y a. Differentiate f(x) = 5 ln x. Solution: ( ) ( ) 0for 5 ln5' >== x x x dx d xf
8. 8. ©2007 Pearson Education Asia Chapter 12: Additional Differentiation Topics 12.1 Derivatives of Logarithmic Functions Example 3 – Rewriting Logarithmic Functions before Differentiating a. Find dy/dx if . Solution: b. Find f’(p) if . Solution: ( )3 52ln += xy ( ) 2/5for 52 6 2 52 1 3 −> + =      + = x xxdx dy ( ) ( ) ( ) ( ) 3 4 2 3 1 2 1 3 1 41 2 1 31 1 1 2' + + + + + =       + +      + +      + = ppp ppp pf ( ) ( ) ( ) ( )( )432 321ln +++= ppppf
9. 9. ©2007 Pearson Education Asia Chapter 12: Additional Differentiation Topics 12.1 Derivatives of Logarithmic Functions Example 5 – Differentiating a Logarithmic Function to the Base 2 Differentiate y = log2x. Solution: Procedure to Differentiate logbu • Convert logbu to and then differentiate. b u ln ln ( ) ( )x x dx d x dx dy 2ln 1 2ln ln log2 =      =
10. 10. ©2007 Pearson Education Asia Chapter 12: Additional Differentiation Topics 12.2 Derivatives of Exponential Functions12.2 Derivatives of Exponential Functions • The derivatives of exponential functions are: ( ) dx du ee dx d uu =a. ( ) xx ee dx d =b. ( ) ( ) dx du bbb dx d uu lnc. = ( )( ) ( )( ) ( )( ) 0'for ' 1 d. 1 1 1 ≠= − − − xff xff xf dx d dy dx dx dy 1 e. =
11. 11. ©2007 Pearson Education Asia Chapter 12: Additional Differentiation Topics 12.2 Derivatives of Exponential Functions Example 1 – Differentiating Functions Involving ex a.Find . Solution: b. If y = , find . Solution: c. Find y’ when . Solution: x e x ( ) x xx e x e dx d xx dx d e dx dy − =+= −− 1 3ln2 ++= x eey xx eey =++= 00' ( )x e dx d 3 ( ) ( ) xxx ee dx d e dx d 333 == dx dy
12. 12. ©2007 Pearson Education Asia Chapter 12: Additional Differentiation Topics 12.2 Derivatives of Exponential Functions Example 3 – The Normal-Distribution Density Function Determine the rate of change of y with respect to x when x = μ + σ. ( ) ( ) ( )( )2 2 1 / 2 1 σµ σ −− == x e x xfy Solution: The rate of change is ( ) ( )( ) ( ) e e dx dy x x πσ σσ µσµ πσ σµ σµ 2 1 1 2 2 1 2 1 2 / 2 2 1 − =                   −+ −= −− +=
13. 13. ©2007 Pearson Education Asia Chapter 12: Additional Differentiation Topics 12.2 Derivatives of Exponential Functions Example 5 – Differentiating Different Forms Example 7 – Differentiating Power Functions Again Find . Solution: ( )xe xe dx d 22 ++ ( ) ( ) [ ]( ) x ex x eexxe dx d x e xexe 2 2ln2 2 1 2ln2 1 2ln12 +=       +=++ − − Prove d/dx(xa ) = axa−1 . Solution: ( ) ( ) 11ln −− === aaxaa axaxxe dx d x dx d
14. 14. ©2007 Pearson Education Asia Chapter 12: Additional Differentiation Topics 12.3 Elasticity of Demand12.3 Elasticity of Demand Example 1 – Finding Point Elasticity of Demand • Point elasticity of demand η is where p is price and q is quantity. ( ) dq dp q p q ==ηη Determine the point elasticity of the demand equation Solution: We have 0and0where >>= qk q k p 1 2 2 −=== − q k q k dq dp q p η
15. 15. ©2007 Pearson Education Asia Chapter 12: Additional Differentiation Topics 12.4 Implicit Differentiation12.4 Implicit Differentiation Implicit Differentiation Procedure 1. Differentiate both sides. 2. Collect all dy/dx terms on one side and other terms on the other side. 3. Factor dy/dx terms. 4. Solve for dy/dx.
16. 16. ©2007 Pearson Education Asia Chapter 12: Additional Differentiation Topics 12.4 Implicit Differentiation Example 1 – Implicit Differentiation Find dy/dx by implicit differentiation if . Solution: 73 =−+ xyy ( ) ( ) 2 2 3 31 1 013 7 ydx dy dx dy y dx dy dx d xyy dx d + = =−+ =−+
17. 17. ©2007 Pearson Education Asia Chapter 12: Additional Differentiation Topics 12.4 Implicit Differentiation Example 3 – Implicit Differentiation Find the slope of the curve at (1,2). Solution: ( )223 xyx −= ( ) ( )[ ] ( ) ( ) ( ) 2 7 2 443 223 2,1 2 32 22 223 = − −+ =       −−= −= dx dy xy xxyx dx dy x dx dy xy dx dy x xy dx d x dx d
18. 18. ©2007 Pearson Education Asia Chapter 12: Additional Differentiation Topics 12.5 Logarithmic Differentiation12.5 Logarithmic Differentiation Logarithmic Differentiation Procedure 1. Take the natural logarithm of both sides which gives . 2. Simplify In (f(x))by using properties of logarithms. 3. Differentiate both sides with respect to x. 4. Solve for dy/dx. 5. Express the answer in terms of x only. ( )( )xfy lnln =
19. 19. ©2007 Pearson Education Asia Chapter 12: Additional Differentiation Topics 12.5 Logarithmic Differentiation Example 1 – Logarithmic Differentiation Find y’ if . Solution: ( ) 4 22 3 1 52 + − = xx x y ( ) ( ) ( ) ( )x x xx xxxy xx x y 2 1 1 4 1 ln252ln3 1ln52lnln 1 52 lnln 2 4 223 4 22 3       + −−−= +−−−= + − =
20. 20. ©2007 Pearson Education Asia Chapter 12: Additional Differentiation Topics 12.5 Logarithmic Differentiation Example 1 – Logarithmic Differentiation       + −− −+ − = + −− − = + −− − = )1( 2 52 6 1 )52( ' )1(2 2 52 6 )2)( 1 1 ( 4 1 ) 1 (2)2)( 52 1 (3 ' 24 22 3 2 2 xx x xxxx x y x x xx x xxxy y Solution (continued):
21. 21. ©2007 Pearson Education Asia Chapter 12: Additional Differentiation Topics 12.5 Logarithmic Differentiation Example 3 – Relative Rate of Change of a Product Show that the relative rate of change of a product is the sum of the relative rates of change of its factors. Use this result to express the percentage rate of change in revenue in terms of the percentage rate of change in price. Solution: Rate of change of a function r is ( ) %100 ' 1%100 ' %100 ' %100 ' %100 ' ''' p p r r q q p p r r q q p p r r η+= += +=
22. 22. ©2007 Pearson Education Asia Chapter 12: Additional Differentiation Topics 12.6 Newton’s Method12.6 Newton’s Method Example 1 – Approximating a Root by Newton’s Method Newton’s method: ( ) ( ) ,...3,2,1 ' 1 =−=+ n xf xf xx n n nn Approximate the root of x4 − 4x + 1 = 0 that lies between 0 and 1. Continue the approximation procedure until two successive approximations differ by less than 0.0001.
23. 23. ©2007 Pearson Education Asia Chapter 12: Additional Differentiation Topics 12.6 Newton’s Method Example 1 – Approximating a Root by Newton’s Method Solution: Letting , we have Since f (0) is closer to 0, we choose 0 to be our first x1. Thus, ( ) ( ) 44 13 ' 3 4 1 − − =−=+ n n n n nn x x xf xf xx 25099.0,3When 25099.0,2When 25.0,1When 0,0When 4 3 2 1 ≈= ≈= == == xn xn xn xn ( ) 144 +−= xxxf ( ) ( ) 21411 11000 −=+−= =+−= f f ( ) ( ) 44' 14 3 4 −= +−= nn nnn xxf xxxf
24. 24. ©2007 Pearson Education Asia Chapter 12: Additional Differentiation Topics 12.7 Higher-Order Derivatives12.7 Higher-Order Derivatives For higher-order derivatives:
25. 25. ©2007 Pearson Education Asia Chapter 12: Additional Differentiation Topics 12.7 Higher-Order Derivatives Example 1 – Finding Higher-Order Derivatives a. If , find all higher-order derivatives. Solution: b. If f(x) = 7, find f(x). Solution: ( ) 26126 23 −+−= xxxxf ( ) ( ) ( ) ( ) ( ) 0 36''' 2436'' 62418' 4 2 = = −= +−= xf xf xxf xxxf ( ) ( ) 0'' 0' = = xf xf
26. 26. ©2007 Pearson Education Asia Chapter 12: Additional Differentiation Topics 12.7 Higher-Order Derivatives Example 3 – Evaluating a Second-Order Derivative Example 5 – Higher-Order Implicit Differentiation Solution: ( ) .4whenfind, 4 16 If 2 2 = + = x dx yd x xf ( ) ( ) 3 2 2 2 432 416 − − += +−= x dx yd x dx dy 16 1 4 2 2 = =x dx yd Solution: y x dx dy dx dy yx 4 082 − = =+ .44ifFind 22 2 2 =+ yx dx yd
27. 27. ©2007 Pearson Education Asia Chapter 12: Additional Differentiation Topics 12.7 Higher-Order Derivatives Example 5 – Higher-Order Implicit Differentiation Solution (continued): 32 2 3 22 2 2 4 1 16 4 getto 4 ateDifferenti ydx yd y xy dx yd y x dx dy −= + −= − =