Mathematics for Business (Finance) Instructor: Prof. Ken Tsang Room E409-R11 Email: kentsang@uic.edu.hk 1 Chapter 2: Differentiation(微分): Basic Concepts In this Chapter, we will encounter some important concepts. The Derivative (导数) Product (乘法准则)and Quotient Rules(除式准则), Higher-Order Derivatives(高次导数 ) The Chain Rule(链锁法则) Marginal Analysis(边际分析), Implicit Differentiation(隐函数求导). 2 Section 2.1 The Derivative (导数) Calculus is the mathematics of change, and the primary tool for studying change is a procedure called differentiation(微分). In this section, we will introduce this procedure and examine some of its uses, especially in computing rates of change. Rate of changes, for example velocity, acceleration, the rate of growth of a population, and many others, are described mathematically by derivatives. 3 Example 1 If air resistance is neglected, an object dropped from a great height will fall s(t ) 16t 2 feet in t seconds. What is the object’s instantaneous velocity after t=2 seconds? Solution: Average rate of change of s(t) over the time period [2,2+h] by the ratio V di s tan cetrave l e d s( 2 h) s( 2) ave e l aps e dti m e ( 2 h) 2 16( 2 h) 2 16( 2) 2 64 16h h Compute the instantaneous velocity by the limit Vins limVave lim(64 16 h) 64 h 0 h 0 That is, after 2 seconds, the object is traveling at the rate of 64 feet per second. 4 Rates of Change (变化率) How to determine instantaneous rate of change or rate of change of f(x) at x=c ? Find the average rate of change of f(x) as x varies from x=c to x=c+h rateave f ( x ) f (c h) f (c ) f (c h) f (c ) x (c h) c h Compute the instantaneous rate of change of f(x) at x=c by finding the limiting value of the average rate as h tends to 0 f (c h) f (c ) rateins limrateave lim h 0 h 0 h 5 Rates of Change (Linear function) A linear function y(x)=mx+b changes at the constant rate m with respect to the independent variable x. That is the rate of change of y(x) is given by the slope or steepness of its graph S l ope rateof ch an ge ch an gei n y ch an gei n x y y2 y1 x x2 x1 6 Rates of Change (non-Linear function) For the function that is nonlinear, the rate of change is not constant but varies with x. In particular, the rate of change at x=c is given by the steepness of the graph of f(x) at the point (c,f(c)), which can be measured by the slope of the tangent line to the graph at p. 7 The slope of secant line The secant line(割线): A line that intersects the curve at the point x and point x+h The average rate of change can be interpreted geometrically as the slope of the secant line from the point (x,f(x)) to the point (x+h,f(x+h)). 8 The Derivative (导数) A Difference Quotient for the function f(x): The express f ( x h) f ( x ) h The Derivative of a Function: The derivative of the function f(x) with respect to x is the function f’(x) given by f ( x h) f ( x ) f ( x) lim h 0 h The process of computing the derivative is called differentiation. f(x) is differentiable at x=c if f’(x) exists 9 Example 2 Find the derivative of the function f ( x) 2x2 16x 35 Solution: The difference quotient for f(x) is f ( x h) f ( x) 2( x h) 2 16( x h) 35 2 x 2 16x 35 h h 4 xh 2h 2 16h 4 x 2h 16 h Thus, the derivative of f(x) is the function f ( x h) f ( x ) f ( x) lim lim(4 x 2h 16) 4 x 16 h 0 h 0 h 10 Slope as a Derivative: The slope of the tangent line to the curve y=f(x) at the point (c,f(c)) is mtan f (c) Instantaneous Rate of Change as a Derivative: The rate of change of f(x) with respect to x when x=c is given by f’(c) Remarks: Since the slope of the tangent line at (a,f(a)) is f’(a), the equation of the tangent line is y f (a) f (a)(x a) The point-slope form 11 Example 3 What is the equation of the tangent line to the curve y x at the point where x=4 ? Solution: According to the definition of derivative, we have f ( x h) f ( x ) xh x f ( x) lim lim h 0 h 0 h h ( x h x )( x h x ) 1 1 lim lim h 0 h 0 h( x h x ) xh x 2 x Thus, the slope of the tangent line to the curve at the point where x=4 is f’(4)=1/4 . So the point-slope form is y 2 1 ( x 4) y 1 x 1 4 4 12 Example 4 Find the equation of the tangent line to the function f ( x) 4x 8 x at x=16. Solution: We know that the equation of a tangent line is given by So, we will need the derivative of the function Now we need to evaluate the function and the derivative. Thus, the point-slope form for a tangent line is: y 32 3( x 16) y 3x 16 13 Example 5 A manufacturer determines that when x thousand units of a particular commodity are produced, the profit generated will be p( x) 400x 2 6800x 12000 dollars. At what rate is profit changing with respect to the level of production x when 9 thousand units are produced? Solution: We find that p ( x h) p ( x ) p( x) lim h 0 h 400( x h) 2 6800( x h) 12000 (400x 2 6800x 12000) lim h 0 h 400h 2 800hx 6800h lim 800x 6800 h 0 h 14 Thus, when the level of production is x=9, the profit is changing at the rate of p(9) 800(9) 6800 400dollars per thousand units. Which means that the tangent line to the profit curve y=p(x) is sloped downward at point Q where x=9. Therefore, the profit curve must be falling at Q and profit must be decreasing when 9 thousand units are being produced. 15 Significance of the Sign of the derivative f’(x): If the function f is differentiable at x=c, then f is increasing at x=c if f’(c)>0 and f is decreasing at x=c if f’(c)<0 16 Alternative Derivative Notation Given a function y=f(x) all of the following are equivalent and represent the derivative of f(x) with respect to x. If we want to evaluate the derivative at x=a all of the following are equivalent 17 Differentiability(可微) and Continuity (连续) Continuity of a Differentiable Function: If the function f(x) is differentiable at x=c, then it is also continuous at x=c. Notice that a continuous function may not be differentiable Graphs of four functions that are not differentiable at x=0. 18 Section 2.2 Techniques of Differentiation The Constant Rule: For any constant c, we have d c 0 dx That is, the derivative of a constant is zero Proof: Since f(x+h)=c for all x f ( x) lim d c 0 dx h 0 f ( x h) f ( x ) cc lim 0 h 0 h h 19 The Power Rule For any real number n d [ x n ] nx n 1 dx In words, to find the derivative of xn, reduce the exponent n of x by 1 and multiply your new power of x by original exponent. For Example d ( x 3 ) 3x 2 dx 1 1 d d 1 ( x) (x 2 ) x 2 dx dx 2 20 The constant Multiple Rule If c is a constant and f(x) is differentiable then so is cf(x) and d d cf ( x) c f ( x) dx dx That is, the derivative of a multiple is the multiple of the derivative For Example d d 4 (3 x ) 3 ( x 4 ) 3(4 x 3 ) 12 x 3 dx dx d 7 d 1 3 / 2 7 3 / 2 1/ 2 ( ) (7 x ) 7( x ) x dx x dx 2 2 21 The Sum Rule: If f(x) and g(x) are differentiable then so is the sum of s(x)=f(x)+g(x) and d d d [ f ( x) g ( x)] [ f ( x)] [ g ( x)] dx dx dx That is, the derivative of a sum is the sum of the separate derivative For Example d d d 2 2 ( x 7) (x ) (7) 2( x 3 ) 0 2 x 3 dx dx dx d d d (2 x 5 3x 7 ) 2 ( x 5 ) 3 ( x 7 ) 2(5 x 4 ) 3( 7 x 8 ) dx dx dx 10x 4 21x 8 22 Example 6 It is estimated that x months form now, the population of a certain community will be p( x) x 20x 8000 2 a. At what rate will the population be changing with respect to time 15 months from now? b. By how much will the population actually change during the 16th month? 23 Solution: a. The rate of change of the population with respect to time is the derivative of the population function. That is Rate of change p( x) 2 x 20 The rate of change of the population 15 months from now will be p(15) 2(15) 20 50 people per month b. The actual change in the population during the 16th month is p(16) p(15) 8576 8525 51 people Note: Since the rate varies during the month, the actual change in population during 16th month differs from the monthly rate of the change at the beginning of the month. 24 The Derivative as a Rate of Change (Review) 25 Relative and Percentage Rates of Change Relative and Percentage Rates of Change: The relative rate of change of a quantity Q(x) with respect to x is given by the ratio Relativerateof changeof Q(x) Q( x) dQ / dx Q( x) Q The corresponding percentage rate of change of Q(x) with respect to x is Percentagerate 100Q( x) of changeof Q( x) Q( x) 26 Example 7 The gross domestic product (GDP) of a certain country was N (t ) t 2 5t 106 billion dollars t years after 1995. a. At what rate was the GDP changing with respect to time in 2005? b. At what percentage rate was the GDP changing with respect to time in 2005? Solution: a. The rate of change of the GDP is the derivative N’(t)=2t+5. The rate of change in 2005 was N’(10)=2(10)+5=25 billion dollars per year. b. The percentage rate of change of the GDP in 2005 was 100 N (10) 25 100 9.77% per year N (10) 256 27 Example 8 3 2 Let p( x) 2x 3x 12x 5 , find all x where p’(x)>0, p’(x)=0 and p’(x)<0. Solution: Based on the techniques of differentiation, we have p( x) 2(3 x 2 ) 3(2 x) 12 0 6 x 2 6 x 12 6( x 1)(x 2) So the solution is p’(x)=0 at x=-1 and 2 p’(x)>0 at x<-1 and x>2 p’(x)<0 at -1<x<2 28 Rectilinear Motion: Motion of an object along a straightline in some way. If the position at time t of an object moving along a straight line is given by s(t), then the object has ds velocity and accelerati on v(t ) s(t ) a(t ) v(t ) dt dv dt The object is advancing when v(t)>0, retreating when v(t)<0, and stationary when v(t)=0. It is accelerating when a(t)>0 and decelerating when a(t)<0 29 Example 9 The position at time t of an object moving along a line is given by s(t ) t 6t 9t 5 3 2 a. Find the velocity of the object and discuss its motion between times t=0 and t=4. b. Find the total distance traveled by the object between times t=0 and t=4. c. Find the acceleration of the object and determine when the object is accelerating and decelerating between times t=0 and t=4. 30 Solution: ds a. The velocity is v(t ) 3t 2 12t 9 . The object will be dt stationary when v(t ) 3t 2 12t 9 3(t 1)(t 3) 0 that is, at times t=1 and t=3. Otherwise, the object is either advancing or retreating, as described in the following table. Interval 0<t<1 1<t<3 Sign of v(t) Description of Motion + Advancing from s(0)=5 to s(1)=9 - Retreating from s(1)=9 to s(3)=5 + Advancing from s(3)=5 to s(4)=9 The motion of an object: s(t ) t 3 6t 2 9t 5 3<t<4 31 b. The object travels from s(0)=5 to s(1)=9, then back to s(3)=5, and finally to s(4)=9. Thus the total distance traveled by the object is D | 9 5 | | 5 9 | | 9 5 | 12 0 t 1 1 t 3 3 t 4 c. The acceleration of the object is dv a( t ) 6t 12 6( t 2) dt The object will be accelerating [a(t)>0] when 2<t<4 and decelerating [a(t)<0] when 0<t<2. 32 The Motion of a Projectile: An important example of rectilinear motion . Suppose an object is projected (e.g. thrown, fired, or dropped) vertically in such a way that the only acceleration acting on the object is the constant downward acceleration g due to gravity (that means air resistance is negligible). Thus, the height of the object at time t is given by the formula 1 2 H (t ) gt V0t H 0 2 where H 0 andV0 are the initial height and velocity of the object, respectively. 33 Example 10 Suppose a person standing at the top of a building 112 feet high throws a ball vertically upward with an initial velocity of 96 ft/sec. a. Find the ball’s height and velocity at time t. b. When does the ball hit the ground and what is its impact velocity? c. When is the velocity 0? What is the significance of this time? d. How far does the ball travel during its flight? 34 Solution: a. Since g 32, V0 96 and H0 112, the height of the ball above the ground at time t is H (t ) 16t 2 96t 112 feet. The dH velocity at time t is v(t ) 32t 96 ft/sec dt b. The ball hits the ground when H=0. Solve the equation we find t=7 and t=-1(discard). The impact velocity is v(7)=-128 ft/sec (The negative sign means the direction of motion is different with the direction of initial velocity. ) c. Set v(t)=0, solve t=3. Thus, the ball is at its highest point when t=3 seconds. d. The ball starts at H(0)=112 feet and rises to a maximum height of H(3)=256. So The total distance travelled=(256-112)+256=400 feet 35 Section 2.3 Product and Quotient Rules; Higher-Order Derivative The derivative of a product of functions is not the product of separate derivative!! Similarly, the derivative of a quotient of functions is not the quotient of separate derivative. Suppose we have two function f(x)=x3 and g(x)=x6 36 The Product Rule(乘法准则) If the two functions f(x) and g(x) are differentiable at x, then we have the derivative of the product P(x)=f(x)g(x) is d f ( x) g ( x) f ( x) d [ g ( x)] g ( x) d [ f ( x)] dx dx dx or equivalently, ( fg ) fg gf Example 11 Using the product rule to find the derivative of the function y 3 x 2 (2 x x 2 ) Solution: 1 2 3 y( x) x (2 x x 2 ) 3 x 2 (2 2 x) 3 2 5 2 5 2 5 4 3 2 3 10 8 x x 2x 3 2x 3 x3 x3 3 3 3 3 37 Example 12 A manufacturer determines that t months after a new product is introduced to the market, x(t ) t 2 3t hundred units can be produced and then sold at a price of 3 2 p(t ) 2t 30 dollars per unit . a. Express the revenue R(t) for this product as a function of time . b. At what rate is revenue changing with respect to time after 4 months? Is revenue increasing or decreasing at this time? 38 Solution: a. The revenue is given by R(t ) x(t ) p(t ) (t 2 3t )(2t 3/ 2 30) hundred dollars. b. The rate of change of revenue R(t) with respect to time is given by the derivative R’(t), which we find using the product rule: d d 2t 3 / 2 30 (2t 3 / 2 30) [t 2 3t ] dt dt 3 (t 2 3t ) 2( t 1/ 2 ) (2t 3 / 2 30)[2t 3] 2 R(t ) (t 2 3t ) At time t=4, the revenue is changing at the rate R’(4)=-14 Thus, after 4 months, the revenue is changing at the rate of 14 hundred dollars per month. It is decreasing at that time since R’(4) is negative. 39 The Quotient Rule(除式准则) If the two functions f(x) and g(x) are differentiable at x, then the derivative of the quotient Q(x)=f(x)/g(x) is given by d d d f ( x) [ ] dx g ( x) or equivalently, g ( x) dx [ f ( x)] f ( x) dx [ g ( x)] 2 g ( x) if g ( x) 0 f ' gf ' fg ' ( ) g g2 Example 13 Using the quotient rule to find the derivative of the function Solution: W ( z ) 3(2 z ) (3z 9)(1) 15 2 (2 z ) (2 z ) 2 40 Example 14 A biologist models the effect of introducing a toxin to a bacterial colony by the function t 1 P (t ) t2 t 4 where P is the population of the colony (in millions) t hours after the toxin is introduced. a. At what rate is the population changing when the toxin is introduced? Is the population increasing or decreasing at this time? b. At what time does the population begin to decrease? By how much does the population increase before it begins to decline? 41 Solution: a. The rate of change of the population with respect to time is given d d by the derivative, that is (t 2 t 4) [t 1] (t 1) [t 2 t 4] P(t ) dt (t 2 t 4) 2 dt (t 2 t 4)(1) (t 1)(2t 1) t 2 2t 3 2 2 2 (t t 4) (t t 4) 2 That is the population is initially changing at the rate of P’(0)=0.1875 million bacteria per hour, and it is increasing since P’(0)>0. b. Since the numerator of P’(t) can be factored as –(t-1)(t+3), the denominator and the factor t+3 are both positive for all t≥0, which means that for 0≤t<1 P(t) is increasing, and for t>1, P(t) is decreasing. Therefore, before the population begins to decline, it increases by P(1)-P(0)=1/3-1/4=1/12 million. 42 A Word of Advice: The quotient rule is somewhat cumbersome, so don’t use it unnecessarily. Example 15 Differentiate the function y 2 x 4 x 1 3x 2 3 5 x . Solution: Don’t use the quotient rule! Instead, rewrite the function as y 2 2 1 4 x x 1 x 1 3 3 5 and then apply the power rule term by term to get dy 2 1 ( 2 x 3 ) 0 0 ( 1) x 2 dx 3 3 4 1 4 1 1 x 3 x 2 3 2 3 3 3x 3 x 43 The Second Derivative (二次导数) In applications, it may be necessary to compute the rate of change of a function that is itself a rate of change. For example, the acceleration of a car is the rate of change with respect to time of its velocity. The second derivative of a function is the derivative of its derivative. If y=f(x), the second derivative is denoted by 2 d y or f ( x) 2 dx The second derivative gives the rate of change of the rate of change of the original function. Note: The derivative f’(x) is called the first derivative. 44 Alternate Notation: There is some alternate notation for higher order derivatives as well. df d2 f f ( x) dx f ( x) dx2 Note: Before computing the second derivative of a function, always take time to simplify the first derivative as much as possible. Example 16 An efficiency study of the morning shift at a certain factory indicates that an average worker who arrives on the job at 8:00 A.M. will have produced Q(t ) t 3 6t 2 24t units t hours later. a. Compute the worker’s rate of production at 11:00 A.M. b. At what rate is the worker’s rate of production changing with respect to time at 11:00 A.M.? 45 Solution: a. The worker’s rate of production is the first derivative R(t ) Q(t ) 3t 2 12t 24 of the output Q(t). The rate of production at 11:00 A.M. (t=3) is R(3) Q(3) 33 units per hour b. The rate of change of the rate of production is the second derivative R(t ) Q(t ) 6t 12 of the output function. At 11:00 A.M., this rate is R(3) Q(3) 6 units per hour per hour The minus sign indicates that the worker’s rate of production is decreasing; that is, the worker is slowing down. The rate of this decrease in efficiency at 11:00 A.M. is 6 units per hour per hour. 46 The acceleration a(t) of an object moving along a straight line is the derivative of the velocity v(t). Thus, the acceleration may be d 2s thought of as the second derivative of position; that is a (t ) 2 dt Example 17 If the position of an object moving along a straight line is given 3 2 by s(t ) t 3t 4 at time t, find its velocity and acceleration. Solution: ds 3t 2 6t 4 The velocity of the object is v(t ) dt and its acceleration is dv d 2 s a(t ) 6t 6 2 dt dt 47 The nth Derivative: For any positive integer n, the nth derivative of a function is obtained from the function by differentiating successively n times. If the original function is y=f(x) the nth derivative is denoted by dny dxn or f (n) ( x) Example 18 Find the fifth derivative of the function y=1/x Solution: dy dx d3y 3 dx d5y 5 dx d 1 d2y d 2 1 2 2 3 (x ) x 2 ( x ) 2 x dx x dx2 dx x3 d 6 d4y d 24 3 4 4 5 (2 x ) 6 x 4 ( 6 x ) 24 x dx x dx4 dx x5 d 120 (24x 5 ) 120x 6 6 dx x 48 Section 2.4 The Chain Rule 链锁法则 Suppose the total manufacturing cost at a certain factory is a function of the number of units produced, which in turn is a function of the number of hours the factory has been operating. If C, q, t, denote the cost, units produced and time respectively, then dC rateof changeof cost dq with respect tooutput (dollars per unit) dq rat eof changeof output (unit sper hour) dt with respect totime The product of these two rates is the rate of change of cost with respect to time that is dC dC dq dt dq dt (dollars per hour) 49 The Chain Rule: If y=f(u) is a differentiable function of u and u=g(x) is in turn a differentiable function of x, then the composite function y=f(g(x)) is a differentiable function of x whose derivative is given by the product dy dy du dx du dx or, equivalently, by dy f ( g ( x )) g ( x ) dx Note: One way to remember the chain rule is to pretend the derivative dy/du and du/dx are quotients and to “cancel” du. 50 Example 19 dy Find if y ( x 2 2)3 3( x 2 2)2 1 dx Solution: We rewrite the function as y u 3 3u 2 1, where u x 2 2 . Thus, dy 3u 2 6u du and du 2x dx and according to the chain rule, dy dy du (3u 2 6u )(2 x) dx du dx [3( x 2 2) 2 6( x 2 2)](2 x) replace u with x 2 2 3( x 2 2)[x 2 ](2 x) 6 x 3 ( x 2 2) 51 Example 20 The cost of producing x units of a particular commodity is 1 2 C ( x ) x 4 x 53 3 dollars, and the production level t hours into a particular production run is x(t ) 0.2t 2 0.03t units At what rate is cost changing with respect to time after 4 hours? 52 Solution: We find that dC 2 x4 dx 3 and dx 0.4t 0.03 dt So according to the chain rule, we have dC dC dx 2 x 4 (0.4t 0.03) dt dx dt 3 When t=4, the level of production is x(4)=3.32 units, and by substituting t=4 and x=3.32 into the formula for dC dC dt , we get dt t 4 2 (3.32) 4[0.4(4) 0.03] 10.1277 3 Thus, after 4 hours, cost is increasing at the rate of approximately $10.13 per hour. 53 Let’s look at the functions then we can write the function as a composition. differentiate a composition function using the Chain Rule. The derivative is then 54 In general, we differentiate the outside function leaving the inside function alone and multiple all of this by the derivative of the inside function F ( x) ( x) f ( g ( x)) g derivativeof outside function leaveinside function alone derivativeof inside function Exercise a. Based on the chain rule, find the derivative of f ( x) ( x 2 x)3 with respect to x b. Find the derivative of f ( x) 1 1 x2 1 55 The General Power Rule: For any real number n and differentiable function h, we have d n n 1 d [h( x)] n[h( x)] [h( x)] dx dx Think of [h(x)]n as the composite function [h( x)]n g[h( x)] where g u n By the chain rule, we have d d n n 1 d [h( x)] g[h( x)] g [h( x)]h( x) n[h( x)] [h( x)] dx dx dx 56 Example 21 An environmental study of a certain suburban community suggests that the average daily level of carbon monoxide in the air will be c ( p ) 0.5 p 17 parts per million when the population is p thousand. It is estimated that t years from now, the population of the community will be p(t ) 3.1 0.1t 2 thousand. At what rate will the carbon monoxide level be changing with respect to time 3 years from now? 2 57 Solution: The goal is to find dc/dt when t=3. Since 1 1 dc 1 1 (0.5 p 2 17) 2 [0.5(2 p)] p(0.5 p 2 17) 2 dp 2 2 and dp 0.2t dt it follows from the chain rule that 1 dc dc dp 1 0.1 pt 2 p (0.5 p 17 ) 2 (0.2t ) dt dp dt 2 0.5 p 2 17 When t=3, p(3)=4 and substituting t=3 and p=4 into the formula for dc/dt, we get dc 0.1(4)(3) 1.2 dt 0.5(4) 17 2 5 0.24 partsper millionper year 58 Section 2.5 Marginal Analysis and Approximations Using Increments In economics, the use of the derivative to approximate the change in a quantity that results from a 1-unit increase in production is called marginal analysis 边际分析 For instance, suppose C(x) is the total cost of producing x units of a particular commodity. If x0 units are currently being produced, then the derivative C ( x0 h) C ( x0 ) C ( x0 ) lim h 0 h is called the marginal cost of producing x0 units. The limiting value that defines this derivative is approximately equal to the difference quotient of C(x) when h=1; that is C ( x0 1) C ( x0 ) C ( x0 ) C ( x0 1) C ( x0 ) 1 59 Marginal Cost(边际成本): If C(x) is the total cost of producing x units of a commodity. Then the marginal cost of producing x0 units is the derivative C( x0 ) , which approximates the additional cost C( x0 1) C( x0 ) incurred when the level of production is increased by one unit, from x0 to x0 +1, assuming x0 >>1. 60 Marginal Revenue (边际收入) and Marginal Profit(边际利润): Suppose R(x) is the revenue function generated when x units of a particular commodity are produced, and P(x) is the corresponding profit function, when x=x0 units are being produced, then The marginal revenue is R( x0 ), it approximates R( x0 1) R( x0 ) , the additional revenue generated by producing one more unit. The marginal profit is P ( x0 ) , it approximates P( x0 1) P( x0 ) , the additional profit obtained by producing one more unit, assuming x0 >>1. 61 Example 22 A manufacturer estimates that when x units of a particular commodity are produced, the total cost will be C ( x) 1 x 2 3x 98 dollars, and 8 1 furthermore, that all x units will be sold when the price is p( x) (75 x) 3 dollars per unit. a. Find the marginal cost and the marginal revenue. b. Use marginal cost to estimate the cost of producing the ninth unit. c. What is the actual cost of producing the ninth unit? d. Use marginal revenue to estimate the revenue derived from the sale of the ninth unit. e. What is the actual revenue derived from the sale of the ninth unit? 62 Solution: a. The marginal cost is C’(x)=(1/4)x+3. The total revenue is R( x) xp( x) x((75 x) / 3) 25x x 2 / 3 , the marginal revenue is R’(x)=25-2x/3. b. The cost of producing the ninth unit is the change in cost as x increases from 8 to 9 and can be estimated by the marginal cost C’(8)=8/4+3=$5. c. The actual cost of producing the ninth unit is C(9)-C(8)=$5.13 which is reasonably well approximated by the marginal cost C’(8) d. The revenue obtained from the sale of the ninth unit is approximated by the marginal revenue R’(8)=25-2(8)/3=$19.67 e. The actual revenue obtained from the sale of the ninth unit is R(9)-R(8)=$19.33. 63 Marginal analysis is an important example of a general Incremental approximation procedure 64 Based on the fact that since f ( x0 ) lim h 0 f ( x0 h) f ( x0 ) h then for small h, the derivative f ( x0 ) is approximately equal to the difference quotient f ( x h) f ( x ) . We indicate h this approximation by writing 0 f ( x0 ) f ( x0 h) f ( x0 ) h 0 or, equivalent ly, f ( x0 h) f ( x0 ) f ( x0 )h Approximation by Increment If f(x) is differentiable at x=x0 and △x is a small change in x, then f ( x0 x) f ( x0 ) f ( x0 )x Or, equivalently, if f f ( x0 x ) f ( x0 ) , then f f ( x0 )x 65 Example 23 During a medical procedure, the size of a roughly tumor is estimated by measuring its diameter and using the formula V 4 R 3 to compute its volume. If the diameter 3 is measured as 2.5 cm with a maximum error of 2%, how accurate is the volume measurement? Solution: A sphere of radius R and diameter x=2R has volume 4 4 x 3 1 1 3 3 V R ( ) x ( 2.5) 3 8.181 cm 3 3 3 2 6 6 The error made in computing this volume using the diameter 2.5 when the actual diameter is 2.5+△x is V V (2.5 x) V (2.5) V (2.5)x To be continued 66 The corresponding maximum error in the calculation of volume is Maximumerrorin volume V [V (2.5)]x [V (2.5)](0.02(2.5)) Since V ( x) 1 1 (3x) 2 x 2 6 2 V (2.5) 1 (2.5) 2 9.817 2 it follows that Maximumerrorin volume (9.817)(0.05) 0.491 3 cm Thus, at worst, the calculation of the volume as 8.181 3 cm is off by 0.491 , so the actual volume V must satisfy 7.690 V 8.672 67 The percentage change of a quantity expresses the change in that quantity as a percentage of its size prior to the change. In particular, changein quant it y Percent ageof change 100 size of quant it y 68 Example 24 The GDP of a certain country was N (t ) t 2 5t 200 billion dollars t years after 1997. Use calculus to estimate the percentage change in the GDP during the first quarter of 2005. Solution: Use the formula N (t )t Percentagechangein N 100 N (t ) with t=8, △t=0.25 and N’(t)=2t+5, to get N (8)0.25 N (8) [2(8) 5](0.25) 100 2 1.73% (8) 5(8) 200 P ercent agechangein N 100 69 Differentials(微分): The differential of x is dx=△x, and if y=f(x) is a differentiable function of x, then dy=f’(x)dx is the differential of y. 70 Example 25 In each case, find the differential of y=f(x). a. f ( x) x3 7 x 2 2 2 2 f ( x ) ( x 5 )( 3 x 2 x ) b. Solution: a. dy f ( x)dx [3x2 7(2x)]dx (3x2 14x)dx b. By the product rule, dy f ( x)dx [(x 2 5)(1 4 x) (2 x)(3 x 2 x 2 )]dx (8 x 3 3x 2 14x 5)dx 71 Section 2.6 Implicit Differentiation(隐 函数求导) and Related Rates So far the functions have all been given by equations of the form y=f(x). A function in this form is said to be in explicit form. For example, the functions 3 x 1 y x 2 3x 1 y 2x 3 and y 1 x 2 are all functions in explicit form(显式表达式). Sometimes practical problems will lead to equations in which the function y is not written explicitly in terms of the independent variable x. For example, the equations such as x 2 y3 6 5 y3 xy and x2 y 2 y3 3x 2 y are said to be in implicit form(隐式表达式). 72 Example 26 Find dy/dx if x y y x 2 2 3 Solution: We are going to differentiate both sides of the given equation with respect to x. Firstly, we temporarily replace y by f(x) and rewrite the x 2 f ( x) ( f ( x))2. Secondly, x3 equation as we differentiate both sides of this equation term by term with respect to x: d 2 [ x f ( x) ( f ( x))2 ] dx d 2 df 2 df x dx f ( x) dx ( x ) 2 f ( x) dx d 2 [ x f ( x )] dx d [( f ( x )) 2 ] dx d 3 [x ] dx 3 x2 d 3 (x ) dx To be continued 73 Thus, we have df df df 2 x f ( x)(2 x) 2 f ( x) 3x gat her all t erms dx dx dx df 2 df x 2 f ( x) 3 x 2 2 xf ( x) on one side of t heequat ion dx dx df [ x 2 2 f ( x)] 3 x 2 2 xf ( x) combinet erms dx df 3 x 2 2 xf ( x) df 2 solve for dx x 2 f ( x) dx 2 Finally, replace f(x) by y to get dy 3 x 2 2 xy dx x2 2 y 74 Implicit Differentiation: Suppose an equation defines y implicitly as a differentiable function of x. To find df/dx 1. Differentiate both sides of equation with respect to x. remember that y is really a function of x and use the chain rule when differentiating terms containing y. 2. Solve the differentiated equation algebraically for dy/dx. Exercise Find dy/dx by implicit differentiation where x3 y 3 xy 75 Computing the slope of a tangent line by implicit differentiation Example 27 Find the slope of the tangent line to the circle x 2 y 2 25 at the point (3,4). What is the slope at the point (3,-4)? Solution: Differentiating both sides of the equation with respect to x, we have 2x 2 y dy dy x 0 dx dx y Thus, the slope at (3,4) is The slope at (3,-4) is dy 3 dx (3, 4) 4 dy 3 3 dx (3, 4) 4 4 76 Application to Economics Example 28 Suppose the output at a certain factory is Q 2x3 x2 y y3 units, where x is the number of hours of skilled labor used and y is the number of hours of unskilled labor. The current labor force consists of 30 hours of skill labor and 20 hours of unskilled labor. Question: Use calculus to estimate the change in unskilled labor y that should be made to offset a 1-hour increase in skilled labor x so that output will be maintained at its current level. 77 Solution: If output is to be maintained at the current level, which is the value of Q when x=30 and y=20, the relationship between skilled labor x and unskilled labor y is given by the equation 80,000 Q(30,20) 2x3 x 2 y y 3 The goal is to estimate the change in y that corresponds to a 1-unit increase in x when x and y are related by above equation. As we know, the change in y caused by a 1-unit increase in x can be approximated by the derivative dy/dx. Using implicit differentiation, dy dy we have 0 6x2 x2 2 xy 3 y 2 dx (x2 3y2 ) dx dy 6 x 2 2 xy dx dy 6 x 2 2 xy 2 dx x 3y2 Now evaluate this derivative when x=30 and y=20 to conclude that dy Change in y dx x 30 y 20 6(30) 2 2(30)(20) 3.14 hours (30) 2 3(20) 2 78 In certain practical problems, x and y are related by an equation and can be regarded as a function of a third variable t, which often represents time. Then implicit differentiation can be used to relate dx/dt to dy/dt. This kind of problem is said to involve related rates. A procedure for solving related rates problems 1. Find a formula relating the variables. 2. Use implicit differentiation to find how the rates are related. 3. Substitute any given numerical information into the equation in step 2 to find the desired rate of change. 79 Example 29 The manager of a company determines that when q hundred units of a particular commodity are produced, the cost of production is C thousand dollars, where C 2 3q3 4275. When 1500 units are being produced, the level of production is increasing at the rate of 20 units per week. What is the total cost at this time and at what rate is it changing? 80 Solution: We want to find dC/dt when q=15 and dq/dt=0.2. Differentiating the equation C 2 3q 3 4275implicitly with respect to time, we get dC 2 dq 2C 33q 0 dt dt so that dC 9q 2 dq dt 2C dt When q=15, the cost C satisfies C 2 3(15)3 4275 C 2 4275 3(15)3 14400 C 120 and by substituting q=15, C=120 and dq/dt=0.2 into the formula for dC/dt, we obtain dC 9(15) 2 (0.2) 1.6875 thousand dollars per week. dt 2(120) 81 Example 30 A storm at sea has damaged an oil rig. Oil spills from the rupture at the constant rate of 60 ft 3 / min , forming a slick that is roughly circular in shape and 3 inches thick. a. How fast is the radius of the slick increasing when the radius is 70 feet? b.Suppose the rupture is repaired in such a way that flow is shut off instantaneously. If the radius of the slick is increasing at the rate of 0.2 ft/min when the flow stops. What is the total volume of oil that spilled onto the sea? 82 Solution: We can think of the slick as a cylinder of oil of radius r feet and thickness h=3/12=0.25 feet. Such a cylinder will have volume V r 2 h 0.25r 2 ft3 Differentiating implicitly in this equation with respect to time t, and since dV/dt=60 at all times, we get dV dr dr dr 0.25 2r 60 0.5r 0.5r dt dt dt dt a. We want to find dr/dt when r=70, so that dr 60 0.55 ft /min dt (0.5) (70) 60 191 feet b. Since dr/dt=0.2 at that instant, we have r 0.5 (0.2) Therefore, the total amount of oil spilled is V 0.25 (191) 2 28652ft3 83 Exercise A lake is polluted by waste from a plant located on its shore. Ecologists determine that when the level of pollute is x parts per million (ppm), there will be F fish of a certain species in the lake, where 32000 F 3 x When there are 4000 fish left in the lake, the pollution is increasing at the rate of 1.4 ppm/year. At what rate is the fish population changing at this time? 84 Summary Definition of the Derivative f ( x h) f ( x ) f ( x) lim h 0 h Interpretation of the Derivative Slope as a Derivative: The slope of the tangent line to the curve y=f(x) at the point (c,f(c)) is mtan f (c) Instantaneous Rate of Change as a Derivative: The rate of change of f(x) with respect to x when x=c is given by f’(c) 85 Sign of The Derivative If the function f is differentiable at x=c, then f is increasing at x=c if f (c ) >0 f is decreasing at x=c if f (c ) <0 Techniques of Differentiation d d d d [ x n ] nx n 1 c 0 cf ( x) c f ( x) dx dx dx dx d d d [ f ( x) g ( x)] [ f ( x)] [ g ( x)] dx dx dx d f ( x) g ( x) f ( x) d [ g ( x)] g ( x) d [ f ( x)] The Product Rule dx dx dx d d g ( x) [ f ( x)] f ( x) [ g ( x)] d f ( x) dx dx [ ] if g ( x) 0 The Quotient Rule 2 dx g ( x) g ( x) 86 The Chain Rule dy dy du dx du dx dy f ( g ( x )) g ( x ) dx d n n 1 d [h( x)] n[h( x)] [h( x)] The General Power Rule dx dx The Higher -order Derivative d2y dx2 or f ( x) The Second Derivative dny (n) or f ( x) The nth Derivative n dx Application of Derivative Tangent line, Rectilinear Motion, Projectile Motion 87 Marginal Analysis and Approximation by increments The marginal cost is C ( x0 ) , it approximates C( x0 1) C( x0 ) , the additional cost generated by producing one more unit. C ( x0 1) C ( x0 ) C ( x0 h) C ( x0 ) C ( x0 1) C ( x0 ) C ( x0 ) lim h 0 1 h f ( x0 x) f ( x0 ) f ( x0 )x Approximation by Increment Marginal Revenue Marginal Profit Implicit Differentiation and Related Rate 88