Chapter 3

Chapter 3 - 3 Differentiation Differentiation Basic Rules...

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Unformatted text preview: 3 Differentiation Differentiation Basic Rules of Differentiation The Product and Quotient Rules The Chain Rule Marginal Functions in Economics Higher Order Derivatives 3.1 3.1 Basic Rules of Differentiation 1. 2. 3. 4. Derivative of a Constant The Power Rule Derivative of a Constant Multiple Function The Sum Rule Four Basic Rules Four We’ve learned that to find the rule for the derivative f ′of a We’ve derivative of function f, we first find the difference quotient we difference lim h→0 f ( x + h) − f ( x ) h But this method is tedious and time consuming, even for But and even relatively simple functions. relatively This chapter we will develop rules that will simplify the This process of finding the derivative of a function. process Rule 1: Derivative of a Constant Rule We will use the notation We d [ f ( x )] dx To mean “the derivative of f with respect to x at x.” To “the with Rule 1: Derivative of a constant d ( c) = 0 dx The derivative of a constant function is equal to zero. Rule 1: Derivative of a Constant Rule We can see geometrically why the derivative of a constant We geometrically must be zero. must The graph of a constant function is a straight line parallel The constant to the x axis. to axis Such a line has a slope that is constant with a value of zero. Such slope zero Thus, the derivative of a constant must be zero as well. y f( x ) = c x Rule 1: Derivative of a Constant Rule We can use the definition of the derivative to We definition demonstrate this: demonstrate f ′( x ) = lim f ( x + h) − f ( x) h→0 h c−c = lim h→0 h = lim 0 h→0 =0 Rule 2: The Power Rule Rule Rule 2: The Power Rule If n is any real number, then If dn x ) = nx n −1 ( dx Rule 2: The Power Rule Rule Lets verify this rule for the special case of n = 2. Lets If f(x) = x2, then If f ′( x ) = d2 f ( x + h) − f ( x ) x ) = lim ( h→0 dx h ( x + h )2 − x 2 x 2 + 2 xh + h 2 − x 2 = lim = lim h→0 h→0 h h 2 xh + h 2 h(2 x + h ) = lim = lim h→0 h→0 h h = lim(2 x + h ) = 2 x h→0 Rule 2: The Power Rule Rule Practice Examples: If f(x) = x, then If If f(x) = x8, then If If f(x) = x5/2, then If d f ′( x ) = ( x ) = 1 ⋅ x1−1 = x 0 = 1 dx f ′( x ) = f ′( x ) = d8 ( x ) = 8 ⋅ x8−1 = 8 x 7 dx d 5/2 5 5/2−1 5 3/2 ( x ) = 2⋅x = 2 x dx Example 2, page 159 Rule 2: The Power Rule Rule Practice Examples: Find the derivative of f ( x ) = x d f ′( x ) = dx () d 1/2 x = (x ) dx 1 −1/2 =x 2 1 1/2−1 =x 2 = 1 2x Example 3, page 159 Rule 2: The Power Rule Rule Practice Examples: 1 Find the derivative of f ( x ) = Find 3 x f ′( x ) = d 1 d −1/3 3 = dx ( x ) dx x 1 = − x −1/3−1 3 1 −4 / 3 1 =− x = − 4/3 3 3x Example 3, page 159 Rule 3: Derivative of a Constant Multiple Function Rule Rule 3: Derivative of a Constant Multiple Function If c is any constant real number, then If d d cf ( x )] = c [ f ( x )] [ dx dx Rule 3: Derivative of a Constant Multiple Function Rule Practice Examples: Find the derivative of f ( x ) = 5 x 3 d f ′( x ) = ( 5 x 3 ) dx d3 =5 (x ) dx = 5 ( 3x 2 ) = 15 x 2 Example 4, page 160 Rule 3: Derivative of a Constant Multiple Function Rule Practice Examples: Practice 3 Find the derivative of f ( x ) = x f ′( x) = d ( 3x −1/ 2 ) dx 1 −3/ 2 = 3 − x 2 3 = − 3/ 2 2x Example 4, page 160 Rule 4: The Sum Rule Rule Rule 4: The Sum Rule d d d [ f ( x ) ± g ( x )] = [ f ( x )] ± [ g ( x )] dx dx dx Rule 4: The Sum Rule Rule Practice Examples: Find the derivative of f ( x ) = 4 x 5 + 3x 4 − 8 x 2 + x + 3 d f ′( x ) = ( 4 x 5 + 3x 4 − 8 x 2 + x + 3) dx d5 d4 d2 d d = 4 ( x ) + 3 ( x ) − 8 ( x ) + ( x ) + ( 3) dx dx dx dx dx = 4 ( 5x 4 ) + 3 ( 4 x3 ) − 8 ( 2 x ) + 1 + 0 = 20 x 4 + 12 x 3 − 16 x + 1 Example 5, page 161 Rule 4: The Sum Rule Rule Practice Examples: Find the derivative of t2 5 g (t ) = + 3 5t d t2 5 d 1 2 g ′(t ) = + 3 = t + 5t −3 dt 5 t dt 5 1d 2 d −3 = ⋅ (t ) +5 (t ) 5 dt dt 1 ( 2t ) + 5 ( −3t −4 ) 5 2t 15 2t 5 − 75 = − 4= 5t 5t 4 = Example 5, page 161 Applied Example: Conservation of a Species Applied A group of marine biologists at the Neptune Institute of group Oceanography recommended that a series of conservation measures be carried out over the next decade to save a measures certain species of whale from extinction. certain After implementing the conservation measure, the After population of this species is expected to be population N (t ) = 3t 3 + 2t 2 − 10t + 600 (0 ≤ t ≤ 10) where N(t) denotes the population at the end of year t. where population Find the rate of growth of the whale population when Find rate t = 2 and t = 6. How large will the whale population be 8 years after How population implementing the conservation measures? implementing Applied Example 7, page 162 Applied Example: Conservation of a Species Applied Solution The rate of growth of the whale population at any time t is The rate whale given by given N ′(t ) = 9t 2 + 4t − 10 In particular, for t = 2, we have In N ′(2) = 9 ( 2 ) + 4 ( 2 ) − 10 = 34 2 And for t = 6, we have And N ′(6) = 9 ( 6 ) + 4 ( 6 ) − 10 = 338 2 Thus, the whale population’s rate of growth will be 34 Thus, rate 34 whales per year after 2 years and 338 per year after 6 years. 338 Applied Example 7, page 162 Applied Example: Conservation of a Species Applied Solution The whale population at the end of the eighth year will be The whale eighth N ( 8) = 3 ( 8) + 2 ( 8) − 10 ( 8) + 600 3 2 = 2184 whales Applied Example 7, page 162 3.2 3.2 The Product and Quotient Rules d [ f ( x ) g ( x )] = f ( x ) g ′( x ) + g ( x ) f ′( x ) dx d f ( x ) g ( x ) f ′( x ) − f ( x ) g ′( x ) 2 g ( x) = dx g ( x )] [ Rule 5: The Product Rule Rule The derivative of the product of two differentiable The functions is given by functions d [ f ( x ) g ( x )] = f ( x ) g ′( x ) + g ( x ) f ′( x ) dx Rule 5: The Product Rule Rule Practice Examples: Find the derivative of f ( x ) = ( 2 x 2 − 1) ( x 3 + 3) f ′( x ) = ( 2 x 2 − 1) d3 d x + 3) + ( x 3 + 3) ( 2 x 2 − 1) ( dx dx = ( 2 x 2 − 1) ( 3x 2 ) + ( x 3 + 3) ( 4 x ) = 6 x 4 − 3x 2 + 4 x 4 + 12 x = x ( 10 x 3 − 3x + 12 ) Example 1, page 172 Rule 5: The Product Rule Rule Practice Examples: Find the derivative of f ( x ) = x 3 3 ( x +1 ) d 1/2 d3 1/2 f ′( x ) = x ( x + 1) + ( x + 1) dx x dx 1 −1/2 = x x + ( x1/2 + 1) 3x 2 2 3 1 5/2 x + 3x 5/2 + 3x 2 2 7 = x 5/2 + 3x 2 2 = Example 2, page 172 Rule 6: The Quotient Rule Rule The derivative of the quotient of two differentiable The functions is given by functions d f ( x) g ( x) f ′( x) − f ( x) g ′( x) g ( x) = 2 dx g ( x) ] [ ( g ( x ) ≠ 0) Rule 6: The Quotient Rule Rule Practice Examples: Find the derivative of x f ( x) = 2x − 4 d d ( 2 x − 4) ( x) − x ( 2 x − 4) dx dx f ′( x ) = 2 2 x − 4) ( ( 2 x − 4 ) ( 1) − x ( 2 ) = 2 2 x − 4) ( = Example 3, page 173 2x − 4 − 2x ( 2 x − 4) 2 =− 4 ( 2 x − 4) 2 Rule 6: The Quotient Rule Rule Practice Examples: Find the derivative of 2 x2 + 1 f ( x) = 2 x −1 d2 d2 2 ( x − 1) dx ( x + 1) − ( x + 1) dx ( x − 1) f ′( x ) = 2 2 ( x − 1) (x = = Example 4, page 173 2 − 1) ( 2 x ) − ( x 2 + 1) ( 2 x ) (x 2 − 1) 2 2 x3 − 2 x − 2 x3 − 2 x (x 2 − 1) 2 =− (x 4x 2 − 1) 2 Applied Example: Rate of Change of DVD Sales Applied The sales ( in millions of dollars) of DVDs of a hit movie The sales t years from the date of release is given by 5t S (t ) = 2 t +1 Find the rate at which the sales are changing at time t. Find rate sales How fast are the sales changing at: ✦ The time the DVDs are released (t = 0)? The 0) ✦ And two years from the date of release (t = 2)? And 2) Applied Example 6, page 174 Applied Example: Rate of Change of DVD Sales Applied Solution The rate of change at which the sales are changing at The rate sales time t is given by d 5t S ′(t ) = 2 dt t + 1 (t = = 2 + 1) ( 5) − ( 5t ) ( 2t ) (t 2 + 1) 2 2 5t + 5 − 10t 2 (t 2 + 1) 2 = 5( 1 − t2 ) (t 2 + 1) 2 Applied Example 6, page 174 Applied Example: Rate of Change of DVD Sales Applied Solution The rate of change at which the sales are changing when The rate the DVDs are released (t = 0) is the 1 − ( 0 ) 2 5 1 5 = ( ) =5 ′(0) = S 2 2 2 ( 1) ( 0 ) + 1 That is, sales are increasing by $5 million per year. That increasing $5 Applied Example 6, page 174 Applied Example: Rate of Change of DVD Sales Applied Solution The rate of change two years after the DVDs are The rate released (t = 2) is released 1 − ( 2 ) 2 5 1 − 4 5 ) = − 15 = − 3 = −0.6 = ( S ′(2) = 2 2 2 25 5 4 + 1) ( ( 2 ) + 1 That is, sales are decreasing by $600,000 per year. That decreasing $600,000 Applied Example 6, page 174 3.3 3.3 The Chain Rule d h′( x ) = g ( f ( x ) ) = g ′ [ f ( x )] f ′( x ) dx dy dy du = ⋅ dx du dx Deriving Composite Functions Deriving Consider the function h ( x ) = x 2 + x + 1 ( ) 2 To compute h′(x), we can first expand h(x) To we expand h( x ) = ( x + x + 1) = ( x 2 + x + 1) ( x 2 + x + 1) 2 2 = x 4 + 2 x 3 + 3x 2 + 2 x + 1 and then derive the resulting polynomial derive h′( x ) = 4 x 3 + 6 x 2 + 6 x + 2 But how should we derive a function like H(x)? But H ( x ) = ( x + x + 1) 2 100 Deriving Composite Functions Deriving Note that H ( x ) = ( x + x + 1) Note 2 100 is a composite function: composite g ( x ) = x100 100 H(x) is composed of two simpler functions f ( x) = x2 + x + 1 So that and H ( x ) = g [ f ( x )] = [ f ( x )] = ( x + x + 1) 2 100 We can use this to find the derivative of H(x). Deriving Composite Functions Deriving To find the derivative of the composite function H(x): To find composite We let u = f(x) = x2 + x + 1 and y = g(u) = u100. We Then we find the derivatives of each of these functions Then find du = f ′( x ) = 2 x + 1 dx and dy = g ′(u ) = 100u 99 du The ratios of these derivatives suggest that The ratios dy dy du = ⋅ = 100u 99 ( 2 x + 1) dx du dx Substituting x2 + x + 1 for u we get we 99 dy 2 H ′( x ) = = 100 ( x + x + 1) ( 2 x + 1) dx Rule 7: The Chain Rule Rule If h(x) = g[f(x)], then If h′( x) = d g ( f ( x) ) = g ′ ( f ( x) ) f ′( x) dx Equivalently, if we write y = h(x) = g(u), Equivalently, where u = f(x), then dy dy du = ⋅ dx du dx The Chain Rule for Power Functions The Many composite functions have the special form Many composite special h(x) = g[f(x)] where g is defined by the rule where g(x) = xn so that so h(x) = [f(x)]n function f. function Examples: (n, a real number) In other words, the function h is given by the power of a In the h( x ) = ( x + x + 1) 2 100 H ( x) = (5− x ) 1 33 G( x) = 2 x2 + 3 The General Power Rule The If the function f is differentiable and If h(x) = [f(x)]n then (n, a real number), real d n n −1 h′( x ) = [ f ( x )] = n [ f ( x )] f ′( x ) dx The General Power Rule The Practice Examples: Find the derivative of G ( x ) = x 2 + 1 Find Solution Solution 1/2 Rewrite as a power function: G ( x ) = ( x 2 + 1) Rewrite power Apply the general power rule: Apply general −1/2 d 12 G ′( x ) = ( x + 1) x 2 + 1) ( 2 dx −1/2 12 = ( x + 1) ( 2 x ) 2 x = x2 + 1 Example 2, page 184 The General Power Rule The Practice Examples: 5 Find the derivative of f ( x ) = x 2 ( 2 x + 3) Find Solution Solution Apply the product rule and the general power rule: Apply product general f ′( x ) = x 2 2 d 5 5d ( 2 x + 3) + ( 2 x + 3) x 2 dx dx 4 = x ( 5) ( 2 x + 3) 4 ( 2 ) + ( 2 x + 3) ( 2 ) x 5 5 = 10 x 2 ( 2 x + 3) + 2 x ( 2 x + 3) = 2 x ( 2 x + 3) ( 5x + 2 x + 3) 4 = 2 x ( 2 x + 3) ( 7 x + 3) 4 Example 3, page 185 The General Power Rule The Practice Examples: Find the derivative of f ( x ) = Find Solution Solution −2 Rewrite as a power function: f ( x ) = ( 4 x 2 − 7 ) Rewrite power Apply the general power rule: Apply general f ′( x ) = −2 ( 4 x − 7 ) 2 −3 ( 4x 1 2 − 7) 2 ( 8x ) =− ( 4x 16 x 2 − 7) 3 Example 5, page 186 The General Power Rule The Practice Examples: 3 2x + 1 Find the derivative of f ( x ) = Find 3x + 2 Solution Solution Apply the general power rule and the quotient rule: Apply general quotient 2x + 1 d 2x + 1 f ′( x ) = 3 3x + 2 dx 3x + 2 2x + 1 = 3 3x + 2 2 2 ( 3x + 2 ) ( 2 ) − ( 2 x + 1) ( 3) 2 ( 3x + 2 ) 6 x + 4 − 6 x − 3 3 ( 2 x + 1) 2 2x + 1 = 3 = 2 4 3x + 2 ( 3x + 2 ) 3x + 2 ) ( 2 Example 6, page 186 Applied Problem: Arteriosclerosis Applied Arteriosclerosis begins during childhood when plaque Arteriosclerosis plaque forms in the arterial walls, blocking the flow of blood blocking through the arteries and leading to heart attacks, stroke arteries and gangrene. and Applied Example 8, page 188 Applied Problem: Arteriosclerosis Applied Suppose the idealized cross section of the aorta is circular Suppose cross with radius a cm and by year t the thickness of the plaque is cm thickness h = g(t) cm cm then the area of the opening is given by then area A = π (a – h)2 cm2 Further suppose the radius of an individual’s artery is 1 cm Further radius cm (a = 1) and the thickness of the plaque in year t is given by and thickness is h = g(t) = 1 – 0.01(10,000 – t2)1/2 cm Applied Example 8, page 188 Applied Problem: Arteriosclerosis Applied Then we can use these functions for h and A Then h = g(t) = 1 – 0.01(10,000 – t2)1/2 A = f(h) = π (1 – h)2 to find a function that gives us the rate at which A is rate changing with respect to time by applying the chain rule: changing time chain dA dA dh = ⋅ = f ′( h ) ⋅ g ′(t ) dt dh dt 1 2 −1/2 = 2π (1 − h )( −1) −0.01 ( 10,000 − t ) ( −2t ) 2 0.01t = −2π (1 − h ) 1/2 ( 10,000 − t 2 ) 0.02π (1 − h )t =− 10,000 − t 2 Applied Example 8, page 188 Applied Problem: Arteriosclerosis Applied For example, at age 50 (t = 50), For 50 h = g (50) = 1 − 0.01(10,000 − 2500)1/2 ≈ 0.134 So that dA 0.02π (1 − 0.134)50 = ≈ −0.03 dt 10,000 − 2500 That is, the area of the arterial opening is decreasing at the That area rate of 0.03 cm2 per year for a typical 50 year old. rate 0.03 cm 50 Applied Example 8, page 188 3.4 3.4 Marginal Functions in Economics E ( p) = − Percentage change in quantity demanded Percentage change in price f ( p + h) − f ( p ) ( 100 ) f ( p) =− h p ( 100 ) Marginal Analysis Marginal Marginal analysis is the study of the rate of change of economic quantities. economic These may have to do with the behavior of costs, revenues, These profit, output, demand, etc. profit, In this section we will discuss the marginal analysis of In various functions related to: various ✦ Cost ✦ Average Cost ✦ Revenue ✦ Profit ✦ Elasticity of Demand Applied Example: Rate of Change of Cost Functions Applied Suppose the total cost in dollars incurred each week by Suppose total Polaraire for manufacturing x refrigerators is given by the total cost function total C(x) = 8000 + 200x – 0.2x2 (0 ≤ x ≤ 400) a. What is the actual cost incurred for manufacturing the What 251st refrigerator? 251 refrigerator? b. Find the rate of change of the total cost function with Find rate respect to x when x = 250. c. Compare the results obtained in parts (a) and (b). Compare (a) (b) Applied Example 1, page 194 Applied Example: Rate of Change of Cost Functions Applied Solution a. The cost incurred in producing the 251st refrigerator is The 251 C(251) – C(250) = [8000 + 200(251) – 0.2(251)2] (251) – [8000 + 200(250) – 0.2(250)2] [8000 = 45,599.8 – 45,500 = 99.80 or $99.80. or $99.80 Applied Example 1, page 194 Applied Example: Rate of Change of Cost Functions Applied Solution a. The rate of change of the total cost function The rate C(x) = 8000 + 200x – 0.2x2 with respect to x is given by with C´(x) = 200 – 0.4x 200 So, when production is 250 refrigerators, the rate of So, 250 change of the total cost with respect to x is C´(x) = 200 – 0.4(250) 200 = 100 or $100. or $100 Applied Example 1, page 194 Applied Example: Rate of Change of Cost Functions Applied Solution a. Comparing the results from (a) and (b) we can see they are Comparing (a) (b) very similar: $99.80 versus $100. very $99.80 $100 ✦ This is because (a) measures the average rate of change This (a) average over the interval [250, 251], while (b) measures the [250, while (b) instantaneous rate of change at exactly x = 250. instantaneous ✦ The smaller the interval used, the closer the average rate smaller the interval the closer of change becomes to the instantaneous rate of change. of becomes instantaneous Applied Example 1, page 194 Applied Example: Rate of Change of Cost Functions Applied Solution The actual cost incurred in producing an additional unit The additional of a good is called the marginal cost. marginal As we just saw, the marginal cost is approximated by the As marginal approximated rate of change of the total cost function. rate total For this reason, economists define the marginal cost For function as the derivative of the total cost function. derivative total Applied Example 1, page 194 Applied Example: Marginal Cost Functions Applied A subsidiary of Elektra Electronics manufactures a subsidiary portable music player. portable Management determined that the daily total cost of Management daily producing these players (in dollars) is C(x) = 0.0001x3 – 0.08x2 + 40x + 5000 where x stands for the number of players produced. where number a. Find the marginal cost function. b. Find the marginal cost for x = 200, 300, 400, and 600. Find c. Interpret your results. Applied Example 2, page 195 Applied Example: Marginal Cost Functions Applied Solution a. If the total cost function is: If total C(x) = 0.0001x3 – 0.08x2 + 40x + 5000 then, its derivative is the marginal cost function: then, its derivative marginal C´(x) = 0.0003x2 – 0.16x + 40 Applied Example 2, page 195 Applied Example: Marginal Cost Functions Applied Solution a. The marginal cost for x = 200, 300, 400, and 600 is: The marginal C´(200) = 0.0003(200)2 – 0.16(200) + 40 = 20 C´(300) = 0.0003(300)2 – 0.16(300) + 40 = 19 C´(400) = 0.0003(400)2 – 0.16(400) + 40 = 24 C´(600) = 0.0003(600)2 – 0.16(600) + 40 = 52 or $20/unit, $19/unit, $24/unit, and $52/unit, respectively. Applied Example 2, page 195 Applied Example: Marginal Cost Functions Applied Solution a. From part (b) we learn that at first the marginal cost is From (b) at marginal decreasing, but as output increases, the marginal cost decreasing but as the marginal increases as well. increases This is a common phenomenon that occurs because of This common several factors, such as excessive costs due to overtime and several such overtime high maintenance costs for keeping the plant running at maintenance such a fast rate. such Applied Example 2, page 195 Applied Example: Marginal Revenue Functions Applied Suppose the relationship between the unit price p in Suppose price dollars and the quantity demanded x of the Acrosonic quantity model F loudspeaker system is given by the equation model p = – 0.02x + 400 (0 ≤ x ≤ 20,000) a. Find the revenue function R. Find revenue b. Find the marginal revenue function R′. Find marginal c. Compute R′(2000) and interpret your result. Compute Applied Example 5, page 199 Applied Example: Marginal Revenue Functions Applied Solution a. The revenue function is given by The revenue R(x) = px = (– 0.02x + 400)x (– = – 0.02x2 + 400x 0.02 (0 ≤ x ≤ 20,000) Applied Example 5, page 199 Applied Example: Marginal Revenue Functions Applied Solution a. Given the revenue function Given revenue R(x) = – 0.02x2 + 400x 0.02 We find its derivative to obtain the marginal revenue We function: function R′(x) = – 0.04x + 400 Applied Example 5, page 199 Applied Example: Marginal Revenue Functions Applied Solution a. When quantity demanded is 2000, the marginal revenue When quantity the marginal will be: will R′(2000) = – 0.04(2000) + 400 = 320 Thus, the actual revenue realized from the sale of the Thus, actual 2001st loudspeaker system is approximately $320. loudspeaker Applied Example 5, page 199 Applied Example: Marginal Profit Function Applied Continuing with the last example, suppose the total cost Continuing total cost (in dollars) of producing x units of the Acrosonic model F loudspeaker system is loudspeaker C(x) = 100x + 200,000 100 a. b. c. Find the profit function P. Find profit Find the marginal profit function P′. Find marginal Compute P′ (2000) and interpret the result. Compute (2000) Applied Example 6, page 199 Applied Example: Marginal Profit Function Applied Solution a. From last example we know that the revenue function is From revenue R(x) = – 0.02x2 + 400x ✦ Profit is the difference between total revenue and total difference total cost, so the profit function is cost P(x) = R(x) – C(x) = (– 0.02x2 + 400x) – (100x + 200,000) = – 0.02x2 + 300x – 200,000 Applied Example 6, page 199 Applied Example: Marginal Profit Function Applied Solution a. Given the profit function Given profit P(x) = – 0.02x2 + 300x – 200,000 we find its derivative to obtain the marginal profit we derivative function: function P′(x) = – 0.04x + 300 0.04 Applied Example 6, page 199 Applied Example: Marginal Profit Function Applied Solution a. When producing x = 2000, the marginal profit is When the marginal P′(2000) = – 0.04(2000) + 300 0.04(2000) = 220 220 Thus, the profit to be made from producing the 2001st Thus, profit producing the 2001 loudspeaker is $220. loudspeaker $220 Applied Example 6, page 199 Elasticity of Demand Elasticity Economists are frequently concerned with how strongly do Economists how changes in prices cause quantity demanded to change. prices cause quantity to The measure of the strength of this reaction is called the The elasticity of demand, which is given by elasticity percentage change in quantity demanded E ( p) = − percentage change in price Note: Since the ratio is negative, economists use the negative Since negative economists of the ratio, to make the elasticity be a positive number. of to positive Elasticity of Demand Elasticity Suppose the price of a good increases by h dollars from p to Suppose price increases (p + h) dollars. The percentage change of the price is The percentage of Percentage change in price = Change in price Price ( 100 ) = h ( 100 ) p The percentage change in quantity demanded is The percentage in Percentage change in quantity demanded = Change in quantity demanded Quantity demanded at price p ( 100) f ( p + h) − f ( p) = ( 100 ) f ( p) Elasticity of Demand Elasticity One good way to measure the effect that a percentage One change in price has on the percentage change in the quantity demanded is to look at the ratio of the latter to the former. We find former. Percentage change in quantity demanded Percentage change in price E ( p) = f ( p + h) − f ( p ) ( 100 ) f ( p) = h p ( 100 ) f ( p + h) − f ( p ) h = f ( p) p f ( p + h) − f ( p ) f ( p) = h p Elasticity of Demand Elasticity We have f ( p + h) − f ( p ) h E ( p) = f ( p) p f ( p + h) − f ( p) ≈ f ′( p ) h If f is differentiable at p, then, when h is small, If differentiable then, Therefore, if h is small, the ratio is approximately equal to Therefore, f ′( p ) pf ′( p ) E ( p) = = f ( p) f ( p) p Economists call the negative of this quantity the elasticity of demand. Economists negative elasticity Elasticity of Demand Elasticity Elasticity of Demand If f is a differentiable demand function defined by differentiable x = f(p) , then the elasticity of demand at price p is elasticity given by given pf ′( p ) E ( p) = − f ( p) Note: Since the ratio is negative, economists use the negative of Since negative economists the ratio, to make the elasticity be a positive number. the to positive Applied Example: Elasticity of Demand Applied Consider the demand equation for the Acrosonic model F Consider demand loudspeaker system: loudspeaker p = – 0.02x + 400 a. b. c. (0 ≤ x ≤ 20,000) (0 Find the elasticity of demand E(p). Find elasticity Compute E(100) and interpret your result. Compute interpret Compute E(300) and interpret your result. Compute interpret Applied Example 7, page 201 Applied Example: Elasticity of Demand Applied Solution a. Solving the demand equation for x in terms of p, we get Solving demand x = f(p) = – 50p + 20,000 From which we see that f ′(p) = – 50 Therefore, pf ′( p ) p( −50) E ( p) = − =− f ( p) −50 p + 20,000 p = 400 − p Applied Example 7, page 201 Applied Example: Elasticity of Demand Applied Solution a. When p = 100 the elasticity of demand is When elasticity ( 100) E (100) = 400 − ( 100 ) 1 = 3 ✦ This means that for every 1% increase in price we can This 1% increase price expect to see a 1/3% decrease in quantity demanded. 1/3% decrease quantity ✦ Because the response (change in quantity demanded) is Because response quantity is less than the action (change in price), we say demand is less action price), inelastic. inelastic ✦ Demand is said to be inelastic whenever E(p) < 1. Demand inelastic Applied Example 7, page 201 Applied Example: Elasticity of Demand Applied Solution a. When p = 300 the elasticity of demand is When elasticity ( 300) E (300) = 400 − ( 300 ) =3 ✦ This means that for every 1% increase in price we can This 1% increase price expect to see a 3% decrease in quantity demanded. 3% decrease quantity ✦ Because the response (change in quantity demanded) is Because response quantity is greater than the action (change in price), we say demand greater action price), is elastic. elastic ✦ Demand is said to be elastic whenever E(p) > 1. Demand elastic ✦ Finally, demand is said to be unitary whenever E(p) = 1. Finally, unitary Applied Example 7, page 201 3.5 3.5 Higher Order Derivatives 2 4 8 −7/3 8 f ′′′( x ) = − − x −7/3 = x= 9 3 27 27 x 2 3 x dv d ds d 2 s d a= = = 2 = ( 8t ) = 8 dt dt dt dt dt Higher-Order Derivatives Higher-Order The derivative f ′ of a function f is also a function. The is As such, f ′ may also be differentiated. As may Thus, the function f ′ has a derivative f ″ at a point x in the Thus, domain of f if the limit of the quotient domain f ′( x + h ) − f ′( x ) h exists as h approaches zero. exists approaches The function f ″ obtained in this manner is called the The second derivative of the function f, just as the derivative f ′ second just of f is often called the first derivative of f. first By the same token, you may consider the third, fourth, By third fourth fifth, etc. derivatives of a function f. fifth etc. Higher-Order Derivatives Higher-Order Practice Examples: Find the third derivative of the function f(x) = x2/3 and Find third determine its domain. domain Solution 2 2 1 2 We have f ′( x ) = x −1/3 and f ′′( x ) = − x −4/3 = − x −4/3 We 3 3 3 9 So the required derivative is 24 8 −7/3 8 ′′′( x ) = − − x −7/3 = f x= 9 3 27 27 x 7/3 The domain of the third derivative is the set of all real The domain third numbers except x = 0. numbers Example 1, page 208 Higher-Order Derivatives Higher-Order Practice Examples: Find the second derivative of the function f(x) = (2x2 +3)3/2 Find second (2 Solution Using the general power rule we get the first derivative: general first 1/2 1/2 3 2 2 f ′( x ) = ( 2 x + 3) ( 4 x ) = 6 x ( 2 x + 3) 2 Example 2, page 209 Higher-Order Derivatives Higher-Order Practice Examples: Find the second derivative of the function f(x) = (2x2 +3)3/2 Find second (2 Solution Using the product rule we get the second derivative: product second 1/2 1/2 d d 2 2 f ′′( x ) = 6 x ⋅ ( 2 x + 3) + ( 2 x + 3) ⋅ ( 6 x ) dx dx −1/2 1/2 1 2 2 = 6 x ⋅ ( 2 x + 3) ( 4 x ) + ( 2 x + 3) ⋅ 6 2 = 12 x ( 2 x + 3) 2 2 −1/2 + 6 ( 2 x + 3) 2 1/2 = 6 ( 2 x + 3) 2 −1/2 = Example 2, page 209 6 ( 4 x 2 + 3) 2x2 + 3 2 x 2 + ( 2 x 2 + 3) Applied Example: Acceleration of a Maglev Applied The distance s (in feet) covered by a maglev moving along The distance feet covered a straight track t seconds after starting from rest is given seconds by the function s = 4t2 (0 ≤ t ≤ 10) (0 What is the maglev’s acceleration after 30 seconds? What acceleration 30 Solution The velocity of the maglev t seconds from rest is given by The velocity ds d = ( 4t 2 ) = 8t dt dt The acceleration of the maglev t seconds from rest is given The acceleration by the rate of change of the velocity of t, given by rate velocity v= d d ds d 2 s d a = v = = 2 = ( 8t ) = 8 dt dt dt dt dt Applied Example 4, page 209 or 8 feet per second per second (ft/sec2). or 3.6 3.6 Implicit Differentiation and Related Rates Rocket x Spectator Launch Pad 4000 ft y Differentiating Implicitly Differentiating Up to now we have dealt with functions in the form Up y = f(x) That is, the dependent variable y has been expressed That dependent explicitly in terms of the independent variable x. explicitly independent However, not all functions are expressed explicitly. However, are expressed For example, consider x 2y + y – x 2 + 1 = 0 This equation expresses y implicitly as a function of x. This implicitly Solving for y in terms of x we get Solving ( x 2 + 1) y = x 2 − 1 x2 − 1 y = f ( x) = 2 x +1 which expresses y explicitly. which explicitly Differentiating Implicitly Differentiating Now, consider the equation y4 – y3 – y + 2x3 – x = 8 With certain restrictions placed on y and x, this equation With this defines y as a function of x. defines as But in this case it is difficult to solve for y in order to But difficult solve express the function explicitly. express explicitly How do we compute dy/dx in this case? How in The chain rule gives us a way to do this. The chain Differentiating Implicitly Differentiating Consider the equation y2 = x. Consider To find dy/dx, we differentiate both sides of the equation: To dy/dx d d 2 ( y ) = dx ( x ) dx Since y is a function of x, we can rewrite y = f(x) and find: Since we d d 2 2 ( y ) = dx [ f ( x )] dx = 2 f ( x ) f ′( x ) dy = 2y dx Example 1, page 216 Using the chain rule Differentiating Implicitly Differentiating Therefore the above equation is equivalent to: dy 2y =1 dx Solving for dy/dx yields: Solving dy/dx dy 1 = dx 2 y Example 1, page 216 Steps for Differentiating Implicitly Steps To find dy/dx by implicit differentiation: To by 1. Differentiate both sides of the equation with respect to x. respect (Make sure that the derivative of any term (Make involving y includes the factor dy/dx) involving dy/dx 1. Solve the resulting equation for dy/dx in for dy/dx terms of x and y. Differentiating Implicitly Differentiating Examples Find dy/dx for the equation y 3 − y + 2 x 3 − x = 8 Find dy/dx Solution Differentiating both sides and solving for dy/dx we get solving we d3 d y − y + 2 x 3 − x ) = ( 8) ( dx dx d d d d d 3 3 ( y ) − dx ( y ) + dx ( 2 x ) − dx ( x ) = dx ( 8) dx dy dy 3y − + 6x2 − 1 = 0 dx dx dy 3 y 2 − 1) = 1 − 6 x 2 ( dx dy 1 − 6 x 2 =2 dx 3 y − 1 2 Example 2, page 216 Differentiating Implicitly Differentiating Examples Find dy/dx for the equation x 2 y 3 + 6 x 2 = y + 12 Find dy/dx Then, find the value of dy/dx when y = 2 and x = 1. Then, Solution d 23 d d d x y ) + ( 6 x 2 ) = ( y ) + ( 12 ) ( dx dx dx dx x2 ⋅ d d dy y 3 ) + y 3 ⋅ ( x 2 ) + 12 x = ( dx dx dx dy dy 3 3x y + 2 xy + 12 x = dx dx 2 2 Example 4, page 217 dy ( 3x y − 1) dx = −2 xy 3 − 12 x dy 2 xy 3 + 12 x = dx 1 − 3x 2 y 2 2 2 Differentiating Implicitly Differentiating Examples Find dy/dx for the equation x 2 y 3 + 6 x 2 = y + 12 Find dy/dx Then, find the value of dy/dx when y = 2 and x = 1. Then, Solution Substituting y = 2 and x = 1 we find: dy 2 xy 3 + 12 x = dx 1 − 3x 2 y 2 2(1)(2)3 + 12(1) = 1 − 3(1)2 (2) 2 16 + 12 1 − 12 28 =− 11 = Example 4, page 217 Differentiating Implicitly Differentiating Examples Find dy/dx for the equation Find dy/dx Solution x2 + y2 − x2 = 5 d2 d d 2 1/2 x + y ) − ( x 2 ) = ( 5) ( dx dx dx 12 dy 2 −1/2 x + y ) 2x + 2 y − 2x = 0 ( 2 dx (x +y 2 2 −1/2 ) dy 2x + 2 y = 4x dx dy 2 2 1/2 x+ y = 2x ( x + y ) dx dy 2 2 1/2 y = 2x ( x + y ) − x dx dy 2 x ( x + y = dx y 2 2 1/2 ) −x Example 5, page 219 Related Rates Related Implicit differentiation is a useful technique for solving a Implicit class of problems known as related-rate problems. related-rate Here are some guidelines to solve related-rate problems: Here guidelines 1. Assign a variable to each quantity. Assign variable 2. Write the given values of the variables and their rate Write values of change with respect to t. of 3. Find an equation giving the relationship between the Find equation relationship variables. variables. 4. Differentiate both sides of the equation implicitly with implicitly respect to t. 5. Replace the variables and their derivatives by the variables derivatives numerical data found in step 2 and solve the equation step for the required rate of change. for Applied Example: Rate of Change of Housing Starts Applied A study prepared for the National Association of Realtors study estimates that the number of housing starts in the housing southwest, N(t) (in millions), over the next 5 years is related to the mortgage rate r(t) (percent per year) by the related mortgage equation equation 9n2 + r = 36 What is the rate of change of the number of housing starts What change housing with respect to time when the mortgage rate is 11% per mortgage 11% year and is increasing at the rate of 1.5% per year? increasing 1.5% Applied Example 6, page 220 Applied Example: Rate of Change of Housing Starts Applied Solution We are given that r = 11% and dr/dt = 1.5 at a certain We instant in time, and we are required to find dN/dt. dN/dt 1. Substitute r = 11 into the given equation: 1. 9 N 2 + ( 11) = 36 25 N2 = 9 5 N= 3 (rejecting the (rejecting negative root) negative Applied Example 6, page 220 Applied Example: Rate of Change of Housing Starts Solution We are given that r = 11% and dr/dt = 1.5 at a certain We instant in time, and we are required to find dN/dt. dN/dt 1. Differentiate the given equation implicitly on both 1. implicitly sides with respect to t: with d d d 9 N 2 ) + ( r ) = ( 36 ) ( dt dt dt dN dr 18 N + =0 dt dt Applied Example 6, page 220 Applied Example: Rate of Change of Housing Starts Solution Solution We are given that r = 11% and dr/dt = 1.5 at a certain We instant in time, and we are required to find dN/dt. dN/dt 1. Substitute N = 5/3 and dr/dt = 1.5 into this equation and 5/3 and dr/dt 1.5 solve for dN/dt: solve dN/dt 5 dN 18 + 1.5 = 0 3 dt dN 30 = −1.5 dt dN 1.5 =− dt 30 dN = −0.05 dt Thus, at the time under consideration, the number of Thus, Applied Example 6, page 220 housing starts is decreasing at rate of 50,000 units per year. decreasing units Applied Example: Watching a Rocket Launch Applied At a distance of 4000 feet from the launch site, a spectator At 4000 is observing a rocket being launched. is If the rocket lifts off vertically and is rising at a speed of If vertically speed 600 feet per second when it is at an altitude of 3000 feet, 600 altitude 3000 how fast is the distance between the rocket and the fast distance spectator changing at that instant? spectator Rocket x Spectator Launch Pad 4000 ft Applied Example 8, page 221 y Applied Example: Watching a Rocket Launch Applied Solution 1. Let y = altitude of the rocket altitude x = distance between the rocket and the spectator distance at any time t. at 1. We are told that at a certain instant in time dy y = 3000 and = 600 dt and are asked to find dx/dt at that instant. and dx/dt Applied Example 8, page 221 Applied Example: Watching a Rocket Launch Applied Solution 1. Apply the Pythagorean theorem to the right triangle we Apply find that x 2 = y 2 + 40002 find x = 30002 + 40002 = 5000 Therefore, when y = 3000, Therefore, Rocket x Spectator Launch Pad 4000 ft Applied Example 8, page 221 y Applied Example: Watching a Rocket Launch Applied Solution 2 2 2 1. Differentiate x = y + 4000 with respect to t, obtaining Differentiate dx dy 2x = 2 y dt dt 1. Substitute x = 5000, y = 3000, and dy/dt = 600, to find Substitute 5000 and dy/dt dx 2 ( 5000 ) = 2 ( 3000 ) ( 600 ) dt dx = 360 dt Therefore, the distance between the rocket and the Therefore, distance spectator is changing at a rate of 360 feet per second. changing at feet Applied Example 8, page 221 3.7 3.7 Differentials y f(x + ∆x) f(x) P dy T ∆y x ∆x x + ∆x x Increments Increments Let x denote a variable quantity and suppose x changes Let changes from x1 to x2. This change in x is called the increment in x and is denoted This increment by the symbol ∆ x (read “delta x”). Thus, Thus, ∆ x = x2 – x1 Examples: Find the increment in x as x changes from 3 to 3.2. Find increment changes 3.2 Solution Here, x1 = 3 and x2 = 3.2, so Here, ∆ x = x2 – x1 = 3.2 – 3 = 0.2 Example 1, page 227 Increments Increments Let x denote a variable quantity and suppose x changes Let changes from x1 to x2. This change in x is called the increment in x and is denoted This increment by the symbol ∆ x (read “delta x”). Thus, Thus, ∆ x = x2 – x1 Examples: Find the increment in x as x changes from 3 to 2.7. Find increment changes 2.7 Solution Here, x1 = 3 and x2 = 2.7, so Here, ∆ x = x2 – x1 = 2.7 – 3 = – 0.3 Example 1, page 227 Increments Increments Now, suppose two quantities, x and y, are related by an Now, two are equation y = f(x), where f is a function. where If x changes from x to x + ∆ x, then the corresponding If changes then change in y is called the increment in y. change increment It is denoted ∆ y and is defined by It and ∆ y = f( x + ∆ x ) – f( x ) y f (x + ∆ x ) ∆y f( x ) x Example 1, page 227 ∆x x + ∆x x Example Example Let y = x3. Let Find ∆ x and ∆ y when x changes Find a. from 2 to 2.01, and a. from 2.01 and b. from 2 to 1.98. b. from 1.98 Solution a. Here, ∆ x = 2.01 – 2 = 0.01 Next, ∆y = f ( x + ∆x ) − f ( x ) = f (2.01) − f (2) = (2.01)3 − 23 = 8.120601 − 8 = 0.120601 a. Here, ∆ x = 1.98 – 2 = – 0.02 Next, ∆y = f ( x + ∆x ) − f ( x ) = f (1.98) − f (2) = (1.98)3 − 23 = 7.762392 − 8 = −0.237608 Example 2, page 228 Differentials Differentials We can obtain a relatively quick and simple way of We quick approximating ∆ y, the change in y due to small change ∆ x. approximating the Observe below that near the point of tangency P, the Observe near point the tangent line T is close to the graph of f. tangent close Thus, if ∆ x is small, then dy is a good approximation of ∆ y. Thus, small then dy good y f (x + ∆ x ) f (x ) P dy T ∆y x ∆x x + ∆x x Differentials Differentials Notice that the slope of T is given by dy/∆ x (rise over run). Notice dy/ (rise But the slope of T is given by f ′(x), so we have But so dy/∆ x = f ′(x) or dy = f ′(x) ∆ x dy/ Thus, we have the approximation Thus, approximation ∆ y ≈ dy = f ′(x)∆ x The quantity dy is called the differential of y. The dy differential y f (x + ∆ x ) f (x ) P dy T ∆y x ∆x x + ∆x x The Differential The Let y = f(x) define a differentiable function x. Then Let 1. The differential dx of the independent variable x is The differential dx independent dx = ∆ x 1. The differential dy of the dependent variable y is The differential dy dependent dy = f ′(x)∆ x = f ′(x)dx dx Example Example Approximate the value of Approximate 26.5 using differentials. Solution Let’s consider the function y = f(x) = x . Let’s Since 25 is the number nearest 26.5 whose square root is Since 25 readily recognized, let’s take x = 25. We want to know the change in y, ∆ y, as x changes from We as x = 25 to x = 26.5, an increase of ∆ x = 1.5. 25 26.5 an 1.5 So we find 1 1 ′( x)∆x = ∆y ≈ dy = f ⋅ ( 1.5 ) = ( 1.5 ) = 0.15 10 2 x x = 25 Therefore, 26.5 − 25 = ∆y ≈ 0.15 26.5 ≈ 25 + 0.15 = 5.15 Example 4, page 229 Applied Example: Effect of Speed on Vehicular Operating Applied The total cost incurred in operating a certain type of truck The on a 500-mile trip, traveling at an average speed of v mph, 500-mile average is estimated to be 4500 C (v ) = 125 + v + v dollars. Find the approximate change in the total operating cost Find change when the average speed is increased from 55 to 58 mph. average increased 55 mph. Applied Example 5, page 230 Applied Example: Effect of Speed on Vehicular Operating Applied Solution Total operating cost is given by 4500 C ( v ) = 125 + v + v With v = 55 an ∆ v = dv = 3, we find With dv 4500 ∆C ≈ dC = C ′( v )dv = 1 − 2 ( 3) v v =55 4500 = 1 − ( 3) ≈ −1.46 3025 so the total operating cost is found to decrease by $1.46. so total decrease $1.46 This might explain why so many independent truckers This often exceed the 55 mph speed limit. mph Applied Example 5, page 230 End of Chapter ...
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This note was uploaded on 07/05/2010 for the course MATH 2240 taught by Professor Crespo during the Spring '09 term at SPSU.

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