#### Transcript and x - Cengage

3 Differentiation Basic Rules of Differentiation The Product and Quotient Rules The Chain Rule Marginal Functions in Economics Higher Order Derivatives 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 We’ve learned that to find the rule for the derivative f ′of a function f, we first find the difference quotient lim h0 f ( x h) f ( x) h But this method is tedious and time consuming, even for relatively simple functions. This chapter we will develop rules that will simplify the process of finding the derivative of a function. Rule 1: Derivative of a Constant d We will use the notation f ( x ) dx To mean “the derivative of f with respect to x at x.” 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 We can see geometrically why the derivative of a constant must be zero. The graph of a constant function is a straight line parallel to the x axis. Such a line has a slope that is constant with a value of zero. Thus, the derivative of a constant must be zero as well. y f(x) = c x Rule 1: Derivative of a Constant We can use the definition of the derivative to demonstrate this: f ( x h) f ( x) f ( x ) lim h0 h cc lim h0 h lim 0 h0 0 Rule 2: The Power Rule Rule 2: The Power Rule If n is any real number, then d n n 1 x nx dx Rule 2: The Power Rule Lets verify this rule for the special case of n = 2. If f(x) = x2, then f ( x ) d 2 f ( x h) f ( x) x lim h0 dx h ( x h)2 x 2 x 2 2 xh h 2 x 2 lim lim h0 h0 h h 2 xh h 2 h(2 x h ) lim lim h0 h0 h h lim(2 x h ) 2 x h0 Rule 2: The Power Rule Practice Examples: If f(x) = x, then d f ( x ) x 1 x11 x 0 1 dx If f(x) = x8, then f ( x ) d 8 x 8 x 81 8 x 7 dx If f(x) = x5/2, then f ( x ) d 5/2 5 5/21 5 3/2 x x x dx 2 2 Example 2, page 159 Rule 2: The Power Rule Practice Examples: Find the derivative of f ( x) x d f ( x ) dx d 1/2 x x dx 1 1/21 x 2 Example 3, page 159 1 2 x 1 1/2 x 2 Rule 2: The Power Rule Practice Examples: 1 Find the derivative of f ( x ) 3 x f ( x ) d 1 d 1/3 x 3 dx x dx 1 x 1/31 3 1 4 / 3 1 x 4/3 3 3x Example 3, page 159 Rule 3: Derivative of a Constant Multiple Function Rule 3: Derivative of a Constant Multiple Function If c is any constant real number, then d d cf ( x ) c f ( x ) dx dx Rule 3: Derivative of a Constant Multiple Function Practice Examples: 3 Find the derivative of f ( x ) 5x d f ( x ) 5 x 3 dx d 3 5 x dx 5 3x 2 15 x 2 Example 4, page 160 Rule 3: Derivative of a Constant Multiple Function Practice Examples: 3 Find the derivative of f ( x ) x f ( x) d 3 x 1/ 2 dx 1 3/ 2 3 x 2 Example 4, page 160 3 2 x 3/ 2 Rule 4: The Sum 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 Practice Examples: Find the derivative of f ( x ) 4 x 5 3x 4 8 x 2 x 3 d 5 4 2 4 x 3 x 8 x x 3 dx d 5 d 4 d 2 d d 4 x 3 x 8 x x 3 dx dx dx dx dx f ( x ) 4 5x 4 3 4 x 3 8 2 x 1 0 20 x 4 12 x 3 16 x 1 Example 5, page 161 Rule 4: The Sum Rule Practice Examples: Find the derivative of t2 5 g (t ) 3 5 t d t2 5 d 1 2 g (t ) 3 t 5t 3 dt 5 t dt 5 1 d 2 d 3 t 5 t 5 dt dt 1 2t 5 3t 4 5 2t 15 2t 5 75 4 5 t 5t 4 Example 5, page 161 Applied Example: Conservation of a Species A group of marine biologists at the Neptune Institute of Oceanography recommended that a series of conservation measures be carried out over the next decade to save a certain species of whale from extinction. After implementing the conservation measure, the population of this species is expected to be N (t ) 3t 3 2t 2 10t 600 (0 t 10) where N(t) denotes the population at the end of year t. Find the rate of growth of the whale population when t = 2 and t = 6. How large will the whale population be 8 years after implementing the conservation measures? Applied Example 7, page 162 Applied Example: Conservation of a Species Solution The rate of growth of the whale population at any time t is given by N (t ) 9t 2 4t 10 In particular, for t = 2, we have N (2) 9 2 4 2 10 34 2 And for t = 6, we have N (6) 9 6 4 6 10 338 2 Thus, the whale population’s rate of growth will be 34 whales per year after 2 years and 338 per year after 6 years. Applied Example 7, page 162 Applied Example: Conservation of a Species Solution The whale population at the end of the eighth year will be N 8 3 8 2 8 10 8 600 3 2 2184 whales Applied Example 7, page 162 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 dx g ( x ) g ( x ) Rule 5: The Product Rule The derivative of the product of two differentiable functions is given by d f ( x ) g ( x ) f ( x ) g ( x ) g ( x ) f ( x ) dx Rule 5: The Product Rule Practice Examples: Find the derivative of f ( x ) 2 x 2 1 f ( x ) 2 x 2 1 x 3 3 d 3 d 3 x 3 x 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 Practice Examples: 3 Find the derivative of f ( x ) x x 1 d 1/2 d 3 1/2 f ( x ) x x 1 x 1 x dx dx 3 1 x 3 x 1/2 x1/2 1 3x 2 2 1 5/2 x 3x 5/2 3x 2 2 7 5/2 x 3x 2 2 Example 2, page 172 Rule 6: The Quotient Rule The derivative of the quotient of two differentiable functions is given by d f ( x) g ( x) f ( x) f ( x) g ( x) 2 dx g ( x) g ( x ) g x 0 Rule 6: The Quotient 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 Practice Examples: Find the derivative of x2 1 f ( x) 2 x 1 d 2 d 2 2 x 1 dx x 1 x 1 dx x 1 f ( x ) 2 2 x 1 2 Example 4, page 173 2 2 x 1 2 x x 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 The sales ( in millions of dollars) of DVDs of a hit movie 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. How fast are the sales changing at: ✦ The time the DVDs are released (t = 0)? ✦ And two years from the date of release (t = 2)? Applied Example 6, page 174 Applied Example: Rate of Change of DVD Sales Solution The rate of change at which the sales are changing at time t is given by d 5t S (t ) 2 dt t 1 t Applied Example 6, page 174 2 1 5 5t 2t t 2 1 5t 5 10t 2 t 2 1 2 2 2 5 1 t 2 t 2 1 2 Applied Example: Rate of Change of DVD Sales Solution The rate of change at which the sales are changing when the DVDs are released (t = 0) is 2 5 1 0 5 1 S (0) 5 2 2 2 1 0 1 That is, sales are increasing by $5 million per year. Applied Example 6, page 174 Applied Example: Rate of Change of DVD Sales Solution The rate of change two years after the DVDs are released (t = 2) is 2 5 1 2 5 1 4 15 3 S (2) 0.6 2 2 2 25 5 4 1 2 1 That is, sales are decreasing by $600,000 per year. Applied Example 6, page 174 3.3 The Chain Rule h( x ) d g f ( x ) g f ( x ) f ( x ) dx dy dy du dx du dx Deriving Composite Functions Consider the function h( x ) x x 1 2 2 To compute h′(x), we can first expand h(x) 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 h( x ) 4 x 3 6 x 2 6 x 2 But how should we derive a function like H(x)? H ( x ) x x 1 2 100 Deriving Composite Functions Note that H ( x ) x x 1 2 100 is a composite function: H(x) is composed of two simpler functions f ( x) x 2 x 1 g ( x ) x100 and So that H ( x ) g f ( x ) f ( x ) 100 x x 1 2 100 We can use this to find the derivative of H(x). Deriving Composite Functions To find the derivative of the composite function H(x): We let u = f(x) = x2 + x + 1 and y = g(u) = u100. Then we find the derivatives of each of these functions du f ( x ) 2 x 1 dx and dy g (u ) 100u 99 du The ratios of these derivatives suggest that dy dy du 100u 99 2 x 1 dx du dx Substituting x2 + x + 1 for u we get 99 dy 2 H ( x ) 100 x x 1 2 x 1 dx Rule 7: The Chain Rule If h(x) = g[f(x)], then h( x) d g f ( x) g f ( x) f ( x) dx Equivalently, if we write y = h(x) = g(u), where u = f(x), then dy dy du dx du dx The Chain Rule for Power Functions Many composite functions have the special form h(x) = g[f(x)] where g is defined by the rule g(x) = xn (n, a real number) so that h(x) = [f(x)]n In other words, the function h is given by the power of a function f. Examples: h( x ) x x 1 2 100 H ( x) 1 5 x 3 3 G( x) 2 x 2 3 The General Power Rule If the function f is differentiable and h(x) = [f(x)]n (n, a real number), then d n n 1 h( x ) f ( x ) n f ( x ) f ( x ) dx The General Power Rule Practice Examples: Find the derivative of G( x ) x 2 1 Solution 1/2 2 Rewrite as a power function: G( x ) x 1 Apply the general power rule: 1/2 d 1 2 2 G( x ) x 1 x 1 2 dx 1/2 1 2 x 1 2 x 2 x x2 1 Example 2, page 184 The General Power Rule Practice Examples: 5 2 Find the derivative of f ( x ) x 2 x 3 Solution Apply the product rule and the general power rule: f ( x ) x 2 d 5 5 d 2 x 3 2 x 3 x 2 dx dx x 5 2 x 3 2 2 x 3 2 x 4 2 10 x 2 2 x 3 2 x 2 x 3 4 2 x 2 x 3 5 x 2 x 3 4 2 x 2 x 3 7 x 3 4 Example 3, page 185 5 5 The General Power Rule Practice Examples: Find the derivative of f ( x ) 4x Solution 1 2 7 2 f ( x) 4 x 7 2 Rewrite as a power function: Apply the general power rule: f ( x ) 2 4 x 7 2 Example 5, page 186 16 x 4x 2 7 3 3 8 x 2 The General Power Rule Practice Examples: 3 2 x 1 Find the derivative of f ( x ) 3 x 2 Solution Apply the general power rule and the quotient rule: 2x 1 d 2x 1 f ( x) 3 3 x 2 dx 3 x 2 2 2x 1 3 3 x 2 2 3x 2 2 2 x 1 3 2 3x 2 2 2 x 1 6 x 4 6 x 3 3 2 x 1 3 2 4 3x 2 3x 2 3 x 2 2 Example 6, page 186 Applied Problem: Arteriosclerosis Arteriosclerosis begins during childhood when plaque forms in the arterial walls, blocking the flow of blood through the arteries and leading to heart attacks, stroke and gangrene. Applied Example 8, page 188 Applied Problem: Arteriosclerosis Suppose the idealized cross section of the aorta is circular with radius a cm and by year t the thickness of the plaque is h = g(t) cm then the area of the opening is given by A = p (a – h)2 cm2 Further suppose the radius of an individual’s artery is 1 cm (a = 1) and the thickness of the plaque in year t is given by h = g(t) = 1 – 0.01(10,000 – t2)1/2 cm Applied Example 8, page 188 Applied Problem: Arteriosclerosis Then we can use these functions for h and A h = g(t) = 1 – 0.01(10,000 – t2)1/2 A = f(h) = p (1 – h)2 to find a function that gives us the rate at which A is changing with respect to time by applying the chain rule: dA dA dh f (h ) g (t ) dt dh dt 1 2 1/2 2p (1 h )( 1) 0.01 10,000 t ( 2t ) 2 0.01 t 2p (1 h ) 1/2 10,000 t 2 0.02p (1 h )t 10,000 t 2 Applied Example 8, page 188 Applied Problem: Arteriosclerosis For example, at age 50 (t = 50), h g (50) 1 0.01(10,000 2500)1/2 0.134 So that dA 0.02p (1 0.134)50 0.03 dt 10,000 2500 That is, the area of the arterial opening is decreasing at the rate of 0.03 cm2 per year for a typical 50 year old. Applied Example 8, page 188 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 analysis is the study of the rate of change of economic quantities. These may have to do with the behavior of costs, revenues, profit, output, demand, etc. In this section we will discuss the marginal analysis of various functions related to: ✦ Cost ✦ Average Cost ✦ Revenue ✦ Profit ✦ Elasticity of Demand Applied Example: Rate of Change of Cost Functions Suppose the total cost in dollars incurred each week by Polaraire for manufacturing x refrigerators is given by the total cost function C(x) = 8000 + 200x – 0.2x2 (0 x 400) a. What is the actual cost incurred for manufacturing the 251st refrigerator? b. Find the rate of change of the total cost function with respect to x when x = 250. c. Compare the results obtained in parts (a) and (b). Applied Example 1, page 194 Applied Example: Rate of Change of Cost Functions Solution a. The cost incurred in producing the 251st refrigerator is C(251) – C(250) = [8000 + 200(251) – 0.2(251)2] – [8000 + 200(250) – 0.2(250)2] = 45,599.8 – 45,500 = 99.80 or $99.80. Applied Example 1, page 194 Applied Example: Rate of Change of Cost Functions Solution b. The rate of change of the total cost function C(x) = 8000 + 200x – 0.2x2 with respect to x is given by C´(x) = 200 – 0.4x So, when production is 250 refrigerators, the rate of change of the total cost with respect to x is C´(x) = 200 – 0.4(250) = 100 or $100. Applied Example 1, page 194 Applied Example: Rate of Change of Cost Functions Solution c. Comparing the results from (a) and (b) we can see they are very similar: $99.80 versus $100. ✦ This is because (a) measures the average rate of change over the interval [250, 251], while (b) measures the instantaneous rate of change at exactly x = 250. ✦ The smaller the interval used, the closer the average rate of change becomes to the instantaneous rate of change. Applied Example 1, page 194 Applied Example: Rate of Change of Cost Functions Solution The actual cost incurred in producing an additional unit of a good is called the marginal cost. As we just saw, the marginal cost is approximated by the rate of change of the total cost function. For this reason, economists define the marginal cost function as the derivative of the total cost function. Applied Example 1, page 194 Applied Example: Marginal Cost Functions A subsidiary of Elektra Electronics manufactures a portable music player. Management determined that the daily total cost of producing these players (in dollars) is C(x) = 0.0001x3 – 0.08x2 + 40x + 5000 where x stands for the number of players produced. a. Find the marginal cost function. b. Find the marginal cost for x = 200, 300, 400, and 600. c. Interpret your results. Applied Example 2, page 195 Applied Example: Marginal Cost Functions Solution a. If the total cost function is: C(x) = 0.0001x3 – 0.08x2 + 40x + 5000 then, its derivative is the marginal cost function: C´(x) = 0.0003x2 – 0.16x + 40 Applied Example 2, page 195 Applied Example: Marginal Cost Functions Solution b. The marginal cost for x = 200, 300, 400, and 600 is: 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 Solution c. From part (b) we learn that at first the marginal cost is decreasing, but as output increases, the marginal cost increases as well. This is a common phenomenon that occurs because of several factors, such as excessive costs due to overtime and high maintenance costs for keeping the plant running at such a fast rate. Applied Example 2, page 195 Applied Example: Marginal Revenue Functions Suppose the relationship between the unit price p in dollars and the quantity demanded x of the Acrosonic model F loudspeaker system is given by the equation p = – 0.02x + 400 (0 x 20,000) a. Find the revenue function R. b. Find the marginal revenue function R′. c. Compute R′(2000) and interpret your result. Applied Example 5, page 199 Applied Example: Marginal Revenue Functions Solution a. The revenue function is given by R(x) = px = (– 0.02x + 400)x = – 0.02x2 + 400x Applied Example 5, page 199 (0 x 20,000) Applied Example: Marginal Revenue Functions Solution b. Given the revenue function R(x) = – 0.02x2 + 400x We find its derivative to obtain the marginal revenue function: R′(x) = – 0.04x + 400 Applied Example 5, page 199 Applied Example: Marginal Revenue Functions Solution c. When quantity demanded is 2000, the marginal revenue will be: R′(2000) = – 0.04(2000) + 400 = 320 Thus, the actual revenue realized from the sale of the 2001st loudspeaker system is approximately $320. Applied Example 5, page 199 Applied Example: Marginal Profit Function Continuing with the last example, suppose the total cost (in dollars) of producing x units of the Acrosonic model F loudspeaker system is C(x) = 100x + 200,000 a. Find the profit function P. b. Find the marginal profit function P′. c. Compute P′ (2000) and interpret the result. Applied Example 6, page 199 Applied Example: Marginal Profit Function Solution a. From last example we know that the revenue function is R(x) = – 0.02x2 + 400x ✦ Profit is the difference between total revenue and total cost, so the profit function is 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 Solution b. Given the profit function P(x) = – 0.02x2 + 300x – 200,000 we find its derivative to obtain the marginal profit function: P′(x) = – 0.04x + 300 Applied Example 6, page 199 Applied Example: Marginal Profit Function Solution c. When producing x = 2000, the marginal profit is P′(2000) = – 0.04(2000) + 300 = 220 Thus, the profit to be made from producing the 2001st loudspeaker is $220. Applied Example 6, page 199 Elasticity of Demand Economists are frequently concerned with how strongly do changes in prices cause quantity demanded to change. The measure of the strength of this reaction is called the elasticity of demand, which is given by percentage change in quantity demanded E ( p) percentage change in price Note: Since the ratio is negative, economists use the negative of the ratio, to make the elasticity be a positive number. Elasticity of Demand Suppose the price of a good increases by h dollars from p to (p + h) dollars. The percentage change of the price is Percentage change in price = Change in price Price 100 h 100 p The percentage change in quantity demanded is 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 One good way to measure the effect that a percentage 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 E ( p) Percentage change in quantity demanded Percentage change in price f ( p h) f ( p ) f ( p) h p f ( p h) f ( p ) 100 f ( p) h p 100 f ( p h) f ( p ) h f ( p) p Elasticity of Demand We have f ( p h) f ( p ) h E ( p) f ( p) p If f is differentiable at p, then, when h is small, f ( p h) f ( p) f ( p ) h Therefore, if h is small, the ratio is approximately equal to f ( p) pf ( p) E ( p) f ( p) f ( p) p Economists call the negative of this quantity the elasticity of demand. Elasticity of Demand Elasticity of Demand If f is a differentiable demand function defined by x = f(p) , then the elasticity of demand at price p is given by pf ( p ) E ( p) f ( p) Note: Since the ratio is negative, economists use the negative of the ratio, to make the elasticity be a positive number. Applied Example: Elasticity of Demand Consider the demand equation for the Acrosonic model F loudspeaker system: p = – 0.02x + 400 (0 x 20,000) a. Find the elasticity of demand E(p). b. Compute E(100) and interpret your result. c. Compute E(300) and interpret your result. Applied Example 7, page 201 Applied Example: Elasticity of Demand Solution a. Solving the demand equation for x in terms of p, we get 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 Solution b. When p = 100 the elasticity of demand is 100 1 E (100) 400 100 3 ✦ This means that for every 1% increase in price we can expect to see a 1/3% decrease in quantity demanded. ✦ Because the response (change in quantity demanded) is less than the action (change in price), we say demand is inelastic. ✦ Demand is said to be inelastic whenever E(p) < 1. Applied Example 7, page 201 Applied Example: Elasticity of Demand Solution c. When p = 300 the elasticity of demand is 300 E (300) 3 400 300 ✦ This means that for every 1% increase in price we can expect to see a 3% decrease in quantity demanded. ✦ Because the response (change in quantity demanded) is greater than the action (change in price), we say demand is elastic. ✦ Demand is said to be elastic whenever E(p) > 1. ✦ Finally, demand is said to be unitary whenever E(p) = 1. Applied Example 7, page 201 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 The derivative f ′ of a function f is also a function. As such, f ′ may also be differentiated. Thus, the function f ′ has a derivative f ″ at a point x in the domain of f if the limit of the quotient f ( x h ) f ( x ) h exists as h approaches zero. The function f ″ obtained in this manner is called the second derivative of the function f, just as the derivative f ′ of f is often called the first derivative of f. By the same token, you may consider the third, fourth, fifth, etc. derivatives of a function f. Higher-Order Derivatives Practice Examples: Find the third derivative of the function f(x) = x2/3 and determine its domain. Solution 2 1 4/3 2 4/3 2 1/3 We have f ( x ) x and f ( x ) x x 3 3 3 9 So the required derivative is 2 4 7/3 8 7/3 8 f ( x) x x 9 3 27 27 x 7/3 The domain of the third derivative is the set of all real numbers except x = 0. Example 1, page 208 Higher-Order Derivatives Practice Examples: Find the second derivative of the function f(x) = (2x2 +3)3/2 Solution Using the general power rule we get the first derivative: 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 Practice Examples: Find the second derivative of the function f(x) = (2x2 +3)3/2 Solution Using the product rule we get the second derivative: 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 6 2 x 3 2 Example 2, page 209 1/2 6 4 x 2 3 2 x2 3 1/2 6 2 x 3 2 1/2 2 x 2 2 x 2 3 Applied Example: Acceleration of a Maglev The distance s (in feet) covered by a maglev moving along a straight track t seconds after starting from rest is given by the function s = 4t2 (0 t 10) What is the maglev’s acceleration after 30 seconds? Solution The velocity of the maglev t seconds from rest is given by v ds d 4t 2 8t dt dt The acceleration of the maglev t seconds from rest is given by the rate of change of the velocity of t, given by d d ds d 2 s d a v 2 8t 8 dt dt dt dt dt or 8 feet per second per second (ft/sec2). Applied Example 4, page 209 3.6 Implicit Differentiation and Related Rates Rocket x Spectator Launch Pad 4000 ft y Differentiating Implicitly Up to now we have dealt with functions in the form y = f(x) That is, the dependent variable y has been expressed explicitly in terms of the independent variable x. However, not all functions are expressed explicitly. For example, consider x2 y + y – x2 + 1 = 0 This equation expresses y implicitly as a function of x. Solving for y in terms of x we get ( x 2 1) y x 2 1 x2 1 y f ( x) 2 x 1 which expresses y explicitly. Differentiating Implicitly Now, consider the equation y4 – y3 – y + 2x3 – x = 8 With certain restrictions placed on y and x, this equation defines y as a function of x. But in this case it is difficult to solve for y in order to express the function explicitly. How do we compute dy/dx in this case? The chain rule gives us a way to do this. Differentiating Implicitly Consider the equation y2 = x. To find dy/dx, we differentiate both sides of the equation: d d 2 y x dx dx Since y is a function of x, we can rewrite y = f(x) and find: d d 2 2 y f ( x ) dx dx 2 f ( x ) f ( x ) dy 2y dx Example 1, page 216 Using the chain rule Differentiating Implicitly Therefore the above equation is equivalent to: dy 2y 1 dx Solving for dy/dx yields: dy 1 dx 2 y Example 1, page 216 Steps for Differentiating Implicitly To find dy/dx by implicit differentiation: 1. Differentiate both sides of the equation with respect to x. (Make sure that the derivative of any term involving y includes the factor dy/dx) 2. Solve the resulting equation for dy/dx in terms of x and y. Differentiating Implicitly Examples Find dy/dx for the equation y 3 y 2 x 3 x 8 Solution Differentiating both sides and solving for dy/dx we get d d 3 3 y y 2 x x 8 dx dx d d d d d 3 3 y y 2 x x 8 dx dx dx dx dx dy dy 3y 6x2 1 0 dx dx dy 2 2 3 y 1 1 6 x dx dy 1 6 x 2 2 dx 3 y 1 2 Example 2, page 216 Differentiating Implicitly Examples Find dy/dx for the equation x 2 y 3 6 x 2 y 12 Then, find the value of dy/dx when y = 2 and x = 1. Solution d 2 3 d d d 2 x y 6 x y 12 dx dx dx dx x2 d dy 3 3 d 2 y y x 12 x dx dx dx 3x 2 y 2 dy dy 2 xy 3 12 x dx dx dy 3x y 1 dx 2 xy 3 12 x dy 2 xy 3 12 x dx 1 3x 2 y 2 2 Example 4, page 217 2 Differentiating Implicitly Examples Find dy/dx for the equation x 2 y 3 6 x 2 y 12 Then, find the value of dy/dx when y = 2 and x = 1. 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 Examples Find dy/dx for the equation Solution x2 y2 x2 5 d 2 d 2 d 2 1/2 x y x 5 dx dx dx 1 2 dy 2 1/2 x y 2 x 2 y 2x 0 2 dx x y 2 2 1/2 dy 2 x 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 2x x y dx y 2 Example 5, page 219 2 1/2 x Related Rates Implicit differentiation is a useful technique for solving a class of problems known as related-rate problems. Here are some guidelines to solve related-rate problems: 1. Assign a variable to each quantity. 2. Write the given values of the variables and their rate of change with respect to t. 3. Find an equation giving the relationship between the variables. 4. Differentiate both sides of the equation implicitly with respect to t. 5. Replace the variables and their derivatives by the numerical data found in step 2 and solve the equation for the required rate of change. Applied Example: Rate of Change of Housing Starts A study prepared for the National Association of Realtors estimates that the number of housing starts in the southwest, N(t) (in millions), over the next 5 years is related to the mortgage rate r(t) (percent per year) by the equation 9n2 + r = 36 What is the rate of change of the number of housing starts with respect to time when the mortgage rate is 11% per year and is increasing at the rate of 1.5% per year? 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 instant in time, and we are required to find dN/dt. 1. Substitute r = 11 into the given equation: 9 N 2 11 36 25 N2 9 5 N 3 Applied Example 6, page 220 (rejecting the negative root) Applied Example: Rate of Change of Housing Starts Solution We are given that r = 11% and dr/dt = 1.5 at a certain instant in time, and we are required to find dN/dt. 2. Differentiate the given equation implicitly on both sides with respect to t: d d d 2 9 N 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 We are given that r = 11% and dr/dt = 1.5 at a certain instant in time, and we are required to find dN/dt. 3. Substitute N = 5/3 and dr/dt = 1.5 into this equation and solve for 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 housing starts is decreasing at rate of 50,000 units per year. Applied Example 6, page 220 Applied Example: Watching a Rocket Launch At a distance of 4000 feet from the launch site, a spectator is observing a rocket being launched. If the rocket lifts off vertically and is rising at a speed of 600 feet per second when it is at an altitude of 3000 feet, how fast is the distance between the rocket and the spectator changing at that instant? Rocket x Spectator Launch Pad 4000 ft Applied Example 8, page 221 y Applied Example: Watching a Rocket Launch Solution 1. Let y = altitude of the rocket x = distance between the rocket and the spectator at any time t. 2. We are told that at a certain instant in time y 3000 and dy 600 dt and are asked to find dx/dt at that instant. Applied Example 8, page 221 Applied Example: Watching a Rocket Launch Solution 3. Apply the Pythagorean theorem to the right triangle we find that x 2 y 2 40002 Therefore, when y = 3000, x 30002 40002 5000 Rocket x Spectator Launch Pad 4000 ft Applied Example 8, page 221 y Applied Example: Watching a Rocket Launch Solution 4. Differentiate x 2 y 2 40002 with respect to t, obtaining dx dy 2x 2y dt dt 5. Substitute x = 5000, y = 3000, and dy/dt = 600, to find dx 2 5000 2 3000 600 dt dx 360 dt Therefore, the distance between the rocket and the spectator is changing at a rate of 360 feet per second. Applied Example 8, page 221 3.7 Differentials y T f(x + x) f(x) y dy P x + x x x x Increments Let x denote a variable quantity and suppose x changes from x1 to x2. This change in x is called the increment in x and is denoted by the symbol x (read “delta x”). Thus, x = x2 – x1 Examples: Find the increment in x as x changes from 3 to 3.2. Solution Here, x1 = 3 and x2 = 3.2, so x = x2 – x1 = 3.2 – 3 = 0.2 Example 1, page 227 Increments Let x denote a variable quantity and suppose x changes from x1 to x2. This change in x is called the increment in x and is denoted by the symbol x (read “delta x”). Thus, x = x2 – x1 Examples: Find the increment in x as x changes from 3 to 2.7. Solution Here, x1 = 3 and x2 = 2.7, so x = x2 – x1 = 2.7 – 3 = – 0.3 Example 1, page 227 Increments Now, suppose two quantities, x and y, are related by an equation y = f(x), where f is a function. If x changes from x to x + x, then the corresponding change in y is called the increment in y. It is denoted y and is defined by y = f(x + x) – f(x) y f(x + x) y f(x) x + x x x Example 1, page 227 x Example Let y = x3. Find x and y when x changes a. from 2 to 2.01, and b. from 2 to 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 b. 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 We can obtain a relatively quick and simple way of approximating y, the change in y due to small change x. Observe below that near the point of tangency P, the tangent line T is close to the graph of f. Thus, if x is small, then dy is a good approximation of y. y T f(x + x) f(x) y dy P x + x x x x Differentials Notice that the slope of T is given by dy/x (rise over run). But the slope of T is given by f ′(x), so we have dy/x = f ′(x) or dy = f ′(x) x Thus, we have the approximation y ≈ dy = f ′(x)x The quantity dy is called the differential of y. y T f(x + x) f(x) y dy P x + x x x x The Differential Let y = f(x) define a differentiable function x. Then 1. The differential dx of the independent variable x is dx = x 2. The differential dy of the dependent variable y is dy = f ′(x)x = f ′(x)dx Example Approximate the value of 26.5 using differentials. Solution Let’s consider the function y = f(x) = x . Since 25 is the number nearest 26.5 whose square root is readily recognized, let’s take x = 25. We want to know the change in y, y, as x changes from x = 25 to x = 26.5, an increase of x = 1.5. So we find 1 y dy f ( x)x 2 x Therefore, 1 1.5 1.5 0.15 10 x 25 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 The total cost incurred in operating a certain type of truck on a 500-mile trip, traveling at an average speed of v mph, is estimated to be 4500 C (v ) 125 v v dollars. Find the approximate change in the total operating cost when the average speed is increased from 55 to 58 mph. Applied Example 5, page 230 Applied Example: Effect of Speed on Vehicular Operating Solution Total operating cost is given by 4500 C (v ) 125 v v With v = 55 an v = dv = 3, we find 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. This might explain why so many independent truckers often exceed the 55 mph speed limit. Applied Example 5, page 230 End of Chapter