PowerLecture_Chapter09

PowerLecture_Chapter - 9 The Derivative Limits One-sided Limits and Continuity The Derivative Basic Rules of Differentiation The Product and

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Unformatted text preview: 9 The Derivative Limits One-sided Limits and Continuity The Derivative Basic Rules of Differentiation The Product and Quotient Rules: The Higher-Order Derivatives The Chain Rule Differentiation of Exponential and Differentiation Logarithmic Functions Logarithmic Marginal Functions in Economics 9.1 9.1 Limits y 400 f ( x) 300 200 100 10 20 30 40 50 60 x Introduction to Calculus Introduction Historically, the development of calculus by Isaac Newton Historically, Isaac and Gottfried W. Leibniz resulted from the investigation Gottfried of the following problems: of 1. Finding the tangent line to a curve at a given point on the Finding tangent curve: curve: y T t Introduction to Calculus Introduction Historically, the development of calculus by Isaac Newton Historically, Isaac and Gottfried W. Leibniz resulted from the investigation Gottfried of the following problems: of 1. Finding the area of a planar region bounded by an Finding area arbitrary curve. arbitrary y R t Introduction to Calculus Introduction The study of the tangent-line problem led to the The tangent-line creation of differential calculus, which relies on the differential which concept of the derivative of a function. derivative The study of the area problem led to the creation of The area integral calculus, which relies on the concept of the integral which anti-derivative, or integral, of a function. anti-derivative or integral Example: A Speeding Maglev Example: From data obtained in a test run conducted on a From test prototype of maglev, which moves along a straight prototype which monorail track, engineers have determined that the position of the maglev (in feet) from the origin at time t position from time is given by is s = f(t) = 4t2 (0 ≤ t ≤ 30) 30) f is called the position function of the maglev. position The position of the maglev at time t = 0, 1, 2, 3, … , 10 is 0, The position f(0) = 0 (0) f(1) = 4 (1) f(2) = 16 (2) f(3) = 36 … f(10) = 400 (3) But what if we want to find the velocity of the maglev at But velocity any given point in time? point Example: A Speeding Maglev Example: Say we want to find the velocity of the maglev at t = 2. Say velocity We may compute the average velocity of the maglev We average over an interval of time, such as [2, 4] as follows: interval such [2, Distance covered f (4) − f (2) = Time elapsed 4−2 4(42 ) − 4(22 ) = 2 64 − 16 = 2 = 24 or 24 feet/second. or 24 This is not the velocity of the maglev at exactly t = 2, This not exactly but it is a useful approximation. approximation Example: A Speeding Maglev Example: We can find a better approximation by choosing a We better smaller interval to compute the speed, say [2, 3]. smaller [2, More generally, let t > 2. Then, the average velocity of let Then, average the maglev over the time interval [2, t] is given by time Distance covered f (t ) − f (2) = Time elapsed t−2 4(t 2 ) − 4(22 ) = t−2 4(t 2 − 4) = t−2 Example: A Speeding Maglev Example: 4(t 2 − 4) Average velocity = t−2 By choosing the values of t closer and closer to 2, we By closer we obtain average velocities of the maglev over smaller and average smaller time intervals. smaller The smaller the time interval, the closer the average The smaller time the closer velocity becomes to the instantaneous velocity of the train instantaneous at t = 2, as the table below demonstrates: t 2.5 2.1 2.01 2.001 2.0001 Average Velocity 18 16.4 16.04 16.004 16.0004 The closer t gets to 2, the closer the average velocity gets The closer the closer average to 16 feet/second. 16 Thus, the instantaneous velocity at t = 2 seems to be Thus, instantaneous 16 feet/second. 16 Intuitive Definition of a Limit Intuitive Consider the function g, which gives the average velocity Consider g, average of the maglev: of 4(t 2 − 4) g (t ) = t −2 Suppose we want to find the value that g(t) approaches Suppose approaches as t approaches 2. approaches ✦ We take values of t approaching 2 from the right We approaching from (as we did before), and we find that g(t) approaches approaches 16: 16 t 2.5 2.1 2.01 2.001 2.0001 g (t ) 18 16.4 16.04 16.004 16.0004 ✦ Similarly, we take values of t approaching 2 from the left, Similarly, approaching from and we find that g(t) also approaches 16: approaches 16 t 1.5 1.9 1.99 1.999 1.9999 g(t) 14 15.6 15.96 15.996 15.9996 Intuitive Definition of a Limit Intuitive We have found that as t approaches 2 from either side, We approaches from g(t) approaches 16. approaches 16 In this situation, we say that the limit of g(t) as t In the approaches 2 is 16. approaches 16 This is written as 4(t 2 − 4) lim g (t ) = lim = 16 t →2 t →2 t −2 Observe that t = 2 is not in the domain of g(t) . Observe not But this does not matter, since t = 2 does not play any But does since role in computing this limit. role Limit of a Function Limit The function f has a limit L as x approaches a, written The function limit approaches lim f ( x ) = L x→a if the value of f(x) can be made as close to the number if close L as we please by taking x values sufficiently close to values sufficiently (but not equal to) a. not Examples Examples Let f(x) = x3. Evaluate Let Evaluate lim f ( x ). x→2 Solution Solution You can see in the graph You that f(x) can be as close can close to 8 as we please by taking x sufficiently close to 2. sufficiently Therefore, Therefore, f( x ) = x y 8 3 6 4 2 lim x 3 = 8 x→2 –2 –1 1 –2 2 3 x Examples Examples x + 2 1 Let g ( x ) = Let if x ≠ 1 if x = 1 Evaluate lim g ( x ). x→1 Solution Solution You can see in the graph that You g(x) can be as close can close to 3 as we please by taking x sufficiently close to 1. sufficiently Therefore, Therefore, y g(x) 5 3 1 lim g ( x ) = 3 x→1 –2 –1 1 2 3 x Examples Examples Let f ( x ) = Let 1 x2 Evaluate lim f ( x ). Solution Solution The graph shows us that as x The approaches 0 from either approaches side, f(x) increases without side bound and thus does not bound not approach any specific real number. number. Thus, the limit of f(x) does Thus, not exist as x approaches 0. not approaches x→0 y 5 f ( x) = –2 –1 1 2 1 x2 x Theorem 1 Theorem Properties of Limits lim f ( x ) = L x→a Suppose Then, 1. lim [ x→a and lim g ( x ) = M x→a r f ( x )] = lim f ( x ) = Lr x →a r r, a real number 2. lim cf ( x ) = c lim f ( x ) = cL 2. x→a x →a 3. lim x →a [ f ( x ) ± g ( x )] = lim x→a 4. lim 4. x→a [ f ( x ) g ( x )] = lim x→a 5. 5. f ( x ) ± lim g ( x ) = L ± M x→a f ( x ) lim g ( x ) = LM x→a f ( x ) lim f ( x ) L lim = x→a = x →a g ( x ) lim g ( x ) M x→a c, a real number Provided that M ≠ 0 Provided Examples Examples Use theorem 1 to evaluate the following limits: Use theorem 3 lim x = lim x = 23 = 8 x→2 3 x→2 3/2 lim 5x = 5 lim x = 5(4)3/2 = 40 x→4 3/2 x→4 4 lim (5x − 2) = 5 lim x − lim 2 = 5(1)4 − 2 = 3 x→1 4 x→1 x→1 Examples Examples Use theorem 1 to evaluate the following limits: Use theorem lim 2x x →3 x + 7 = 2 lim x x→3 3 2 3 lim x 2 + 7 = 2(3)3 (3)2 + 7 = 216 x→3 lim (2 x 2 + 1) 2(2)2 + 1 9 2x + 1 = x→2 = = =3 lim x→2 lim ( x + 1) 2 +1 3 x +1 x→2 2 Indeterminate Forms Indeterminate 4( x 2 − 4) Let’s consider lim x →2 x − 2 which we evaluated earlier for the maglev example by which maglev looking at values for x near x = 2. near If we attempt to evaluate this expression by applying If Property 5 of limits, we get Property lim 4( x 2 − 4) 0 4( x − 4) x→2 lim = = x →2 x − 2 lim x − 2 0 x →2 2 In this case we say that the limit of the quotient f(x)/g(x) as In quotient x approaches 2 has the indeterminate form 0/0. approaches 0/0 This expression does not provide us with a solution to our This not solution problem. problem. Strategy for Evaluating Indeterminate Forms Strategy 1. Replace the given function with an appropriate one that takes on the same values as the same original function everywhere except at x = a. everywhere except 2. Evaluate the limit of this function as x approaches a. approaches Examples Examples 4( x 2 − 4) Evaluate lim x →2 x − 2 Solution As we’ve seen, here we have an indeterminate form 0/0. As indeterminate 0/0 We can rewrite 4( x − 2)( x + 2) We 4( x 2 − 4) = = 4( x + 2) x−2 x−2 x≠2 Thus, we can x 2 − 4) 4( say that lim x →2 x−2 = lim 4( x + 2) = 16 x →2 Note that 16 is the same value we obtained for the maglev Note 16 example through approximation. example approximation Examples Examples 4( x 2 − 4) Evaluate lim x →2 x − 2 Solution Notice in the graphs below that the two functions yield the Notice same graphs, except for the value x = 2: same except 4( x 2 − 4) f ( x) = x−2 y 20 20 16 8 4 –1 12 8 –2 16 12 –3 g ( x ) = 4( x + 2) y 4 1 2 3 x –3 –2 –1 1 2 3 x Examples Examples Evaluate 1+ h −1 lim h→0 h Solution As we’ve seen, here we have an indeterminate form 0/0. As indeterminate 0/0 We can rewrite (with the constraint that h ≠ 0): We constraint 1+ h −1 1+ h −1 1+ h +1 h 1 = ⋅ = = h h 1 + h + 1 h( 1 + h + 1) 1+ h +1 Thus, we can say that 1+ h −1 1 1 1 lim = lim = = h→0 h→0 1 + h + 1 h 1 +1 2 Limits at Infinity Limits There are occasions when we want to know whether f(x) There approaches a unique number as x increases without approaches unique bound. bound In the graph below, as x increases without bound, f(x) In without approaches the number 400. 400 We call the line y = 400 We y a horizontal asymptote. horizontal 400 In this case, we can say In f ( x) that that 300 lim f ( x ) = 400 x→∞ and we call this a limit and limit of a function at infinity. 200 100 10 20 30 40 50 60 x Example Example 2x2 Consider the function f ( x ) = 1 + x2 Determine what happens to f(x) as x gets larger and larger. Determine larger Solution We can pick a sequence of values of x and substitute them We sequence in the function to obtain the following values: in x 1 2 5 10 100 1000 f(x) 1 1.6 1.92 1.98 1.9998 1.999998 As x gets larger and larger, f(x) gets closer and closer to 2. As larger closer Thus, we can say that 2x2 lim x→∞ 1 + x 2 =2 Limit of a Function at Infinity Limit The function f has the limit L as x increases without The limit bound (as x approaches infinity), written bound approaches lim f ( x ) = L x→∞ if f(x) can be made arbitrarily close to L by taking x if close large enough. large Similarly, the function f has the limit M as x decreases Similarly, limit without bound (as x approaches negative infinity), without approaches ), written written lim f ( x ) = M x→−∞ if f(x) can be made arbitrarily close to M by taking x if close large enough in absolute value. large Examples Examples −1 Let f ( x ) = Let 1 Evaluate if x < 0 if x ≥ 0 lim f ( x ) lim f ( x ) and x→∞ x→−∞ Solution Graphing f(x) reveals that Graphing y lim f ( x ) = 1 x→∞ lim f ( x ) = −1 x→−∞ f ( x) 1 –3 3 –1 x Examples Examples Let g ( x ) = Let Evaluate 1 x2 lim g ( x ) and lim g ( x ) x→∞ x→−∞ Solution Graphing g(x) reveals that Graphing y 1 g ( x) = 2 x lim g ( x ) = 0 x→∞ lim g ( x ) = 0 x→−∞ –3 –2 –1 1 2 3 x Theorem 2 Theorem Properties of Limits All properties of limits listed in Theorem 1 are valid when a is replaced by ∞ or –∞ . In addition, we have the following properties for limits to infinity: For all n > 0, For 1 lim =0 x→∞ x n and 1 provided that n is defined. x 1 lim =0 x→−∞ x n Examples Examples x2 − x + 3 Evaluate lim Evaluate x→∞ 2 x3 + 1 Solution Solution The limits of both the numerator and denominator do not The limits numerator denominator exist as x approaches infinity, so property 5 is not exist property not applicable. applicable. We can find the solution instead by dividing numerator We dividing and denominator by x3: and by 11 3 − 2+ 3 ( x 2 − x + 3) / x 3 xx x = 0−0+0 = 0 = 0 lim = lim x→∞ (2 x 3 + 1) / x 3 x→∞ 1 2+0 2 2+ 3 x Examples Examples 3x 2 + 8 x − 4 Evaluate lim Evaluate x→∞ 2 x2 + 4 x − 5 Solution Solution Again, we see that property 5 does not apply. Again, property not So we divide numerator and denominator by x2: So divide and by 84 −2 (3x + 8 x − 4) / x x x = 3+ 0−0 = 3 lim 2 = lim x→∞ (2 x + 4 x − 5) / x 2 x→∞ 45 2+ − 2 2+0−0 2 xx 2 2 3+ Examples Examples 2 x 3 − 3x 2 + 1 Evaluate lim Evaluate x→∞ 2 x + 2x + 4 Solution Solution Again, we see that property 5 does not apply. Again, property not But dividing numerator and denominator by x2 does not But dividing and by not 1 help in this case: help 2x − 3 + 2 3 2 2 lim x→∞ (2 x − 3x + 1) / x x = lim x→∞ 24 ( x 2 + 2 x + 4) / x 2 1+ + 2 xx In other words, the limit does not exist. In the We indicate this by writing 2 3 lim x→∞ 2 x − 3x + 1 =∞ 2 x + 2x + 4 9.2 9.2 One-Sided Limits and Continuity y a b c d x One-Sided Limits One-Sided x −1 x +1 Consider the function f ( x ) = if x < 0 if x ≥ 0 Its graph shows that Its f does not have a does not limit as x approaches limit zero, because zero because approaching from each side results in different values. values y f ( x) 1 –1 1 –1 x One-Sided Limits One-Sided x −1 x +1 Consider the function f ( x ) = if x < 0 if x ≥ 0 If we restrict x to be greater If restrict greater than zero (to the right of zero), than we see that f(x) approaches 1 as approaches close as we please as x approaches 0. approaches In this case we say that the In right-hand limit of f as x right-hand approaches 0 is 1, written approaches lim f ( x ) = 1 x→0+ y f ( x) 1 –1 1 –1 x One-Sided Limits One-Sided x −1 x +1 Consider the function f ( x ) = if x < 0 if x ≥ 0 Similarly, if we restrict x to be Similarly, restrict less than zero (to the left of zero), less we see that f(x) approaches –1 approaches –1 as close as we please as x approaches 0. approaches In this case we say that the In left-hand limit of f as x left-hand approaches 0 is – 1, written approaches lim f ( x ) = −1 x→0− y f ( x) 1 –1 1 –1 x One-Sided Limits One-Sided The function f has the right-hand limit L as x The right-hand approaches from the right, written approaches lim f ( x ) = L x→a + if the values of f(x) can be made as close to L as we if please by taking x sufficiently close to (but not equal to) a and to the right of a. to Similarly, the function f has the left-hand limit L as x Similarly, left-hand approaches from the left, written approaches lim f ( x ) = L x→a − if the values of f(x) can be made as close to L as we if please by taking x sufficiently close to (but not equal to) a and to the left of a. to Theorem 3 Theorem Properties of Limits The connection between one-sided limit and the two-sided limit defined earlier is given by the following theorem. Let f be a function that is defined for all values of x Let close to x = a with the possible exception of a itself. Then Then lim f ( x ) = L x →a if and only if lim f ( x ) = lim f ( x ) = L x→a − x→a + Examples Show that lim f ( x ) exists by studying the one-sided Show one-sided x →0 limits of f as x approaches 0: limits approaches x f ( x) = −x Solution Solution For x > 0, we find For if x > 0 if x ≤ 0 y lim f ( x ) = 0 x→0+ And for x ≤ 0, we find And lim f ( x ) = 0 1 x→0− Thus, Thus, lim f ( x ) = 0 x →0 f ( x) 2 –2 –1 1 2 x Examples Examples Show that lim g ( x ) does not exist. Show x →0 −1 g ( x) = 1 if x < 0 if x ≥ 0 Solution For x < 0, we find For y lim g ( x) = −1 x →0− And for x ≥ 0, we find And lim g ( x) = 1 g ( x) 1 x →0+ Thus, Thus, x lim g ( x) does x →0 not exist. not –1 Continuous Functions Continuous Loosely speaking, a function is continuous at a given point if Loosely function its graph at that point has no holes, gaps, jumps, or breaks. no holes gaps jumps or breaks Consider, for example, the graph of f Consider, y a x This function is discontinuous at the following points: This discontinuous ✦ At x = a, f is not defined (x = a is not in the domain of f ). At not not ). Continuous Functions Continuous Loosely speaking, a function is continuous at a given point if Loosely function its graph at that point has no holes, gaps, jumps, or breaks. no holes gaps jumps or breaks Consider, for example, the graph of f Consider, y a b x This function is discontinuous at the following points: This discontinuous ✦ At x = b, f(b) is not equal to the limit of f(x) as x approaches b. At not of as approaches Continuous Functions Continuous Loosely speaking, a function is continuous at a given point if Loosely function its graph at that point has no holes, gaps, jumps, or breaks. no holes gaps jumps or breaks Consider, for example, the graph of f Consider, y b c x This function is discontinuous at the following points: This discontinuous ✦ At x = c, the function does not have a limit, since the left-hand At does since left-hand and right-hand limits are not equal. right-hand not Continuous Functions Continuous Loosely speaking, a function is continuous at a given point if Loosely function its graph at that point has no holes, gaps, jumps, or breaks. no holes gaps jumps or breaks Consider, for example, the graph of f Consider, y c d x This function is discontinuous at the following points: This discontinuous ✦ At x = d, the limit of the function does not exist, resulting in a At limit resulting break in the graph. break Continuity of a Function at a Number Continuity A function f is continuous at a number x = a function continuous if the following conditions are satisfied: conditions 1. f(a) is defined. 2. lim f ( x ) exists. x→a 3. lim f ( x ) = f (a ) 3. x→a If f is not continuous at x = a, then f is said to If then be discontinuous at x = a. discontinuous Also, f is continuous on an interval if f is Also, continuous continuous at every number in the interval. continuous Examples Examples Find the values of x for which the function is continuous: Find continuous f ( x) = x + 2 Solution The function f is continuous everywhere because the three The continuous conditions for continuity are satisfied for all values of x. y 5 f ( x) = x + 2 4 3 2 1 – 2 –1 1 2 x Examples Examples Find the values of x for which the function is continuous: Find continuous x2 − 4 g ( x) = x−2 Solution The function g is discontinuous at x = 2 because g is not The defined at that number. It is continuous everywhere else. defined y x2 − 4 g ( x) = x−2 5 4 3 2 1 – 2 –1 1 2 x Examples Examples Find the values of x for which the function is continuous: Find continuous x + 2 h( x ) = 1 if x ≠ 2 if x = 2 Solution The function h is continuous everywhere except at x = 2 The continuous except where it is discontinuous because because h (2) = 1 ≠ lim h ( x ) = 4 x→2 5 y y = h( x ) 4 3 2 1 – 2 –1 1 2 x Examples Examples Find the values of x for which the function is continuous: Find continuous −1 if x < 0 F ( x) = 1 if x ≥ 0 Solution The function F is discontinuous at x = 0 because the limit The at of F fails to exist as x approaches 0. It is continuous fails approaches It everywhere else. y y = F ( x) 1 x –1 Examples Examples Find the values of x for which the function is continuous: Find continuous 1 if x > 0 G( x) = x −1 i f x ≤ 0 Solution The function G is discontinuous at x = 0 because the limit The at of G fails to exist as x approaches 0. It is continuous fails approaches It everywhere else. y y = G ( x) x –1 Properties of Continuous Functions Properties 1. The constant function f(x) = c is continuous everywhere. The constant continuous 2. The identity function f(x) = x is continuous everywhere. The identity continuous If f and g are continuous at x = a, then and are at 1. [f(x)]n, where n is a real number, is continuous at where continuous x = a whenever it is defined at that number. 2. f ± g is continuous at x = a. continuous 3. fg is continuous at x = a. fg continuous 4. f /g is continuous if g(a) ≠ 0. /g continuous Properties of Continuous Functions Properties Using these properties, we can obtain the following Using additional properties. additional 1. A polynomial function y = P(x) is continuous at every polynomial value of x. 2. A rational function R(x) = p(x)/q(x) is continuous at rational every value of x where q(x) ≠ 0. Examples Examples Find the values of x for which the function is continuous. Find values continuous f ( x ) = 3x 3 + 2 x 2 − x + 10 Solution The function f is a polynomial function of degree 3, so f(x) The polynomial so is continuous for all values of x. continuous Examples Examples Find the values of x for which the function is continuous. Find values continuous 8 x10 − 4 x 2 + 1 g ( x) = x2 + 1 Solution The function g is a rational function. The rational Observe that the denominator of g is never equal to zero. Observe denominator never Therefore, we conclude that g(x) is continuous for all Therefore, values of x. values Examples Examples Find the values of x for which the function is continuous. Find values continuous 4 x 3 − 3x 2 + 1 h( x ) = 2 x − 3x + 2 Solution The function h is a rational function. The rational In this case, however, the denominator of h is equal to zero In denominator equal at x = 1 and x = 2, which we can see by factoring. Therefore, we conclude that h(x) is continuous everywhere Therefore, except at x = 1 and x = 2. Intermediate Value Theorem Intermediate Let’s look again at the maglev example. Let’s maglev The train cannot vanish at any instant of time and cannot The cannot skip portions of track and reappear elsewhere. skip Intermediate Value Theorem Intermediate Mathematically, recall that the position of the maglev is a recall function of time given by f(t) = 4t2 for 0 ≤ t ≤ 30: for y y = 4t 2 s2 s3 s1 t1 t3 t2 t Suppose the position of the maglev is s1 at some time t1 and its Suppose position time position is s2 at some time t2. position time Then, if s3 is any number between s1 and s2, there must be at Then, is least one t3 between t1 and t2 giving the time at which the least maglev is at s3 (f(t3) = s3). Theorem 4 Theorem Intermediate Value Theorem Intermediate The Maglev example carries the gist of the The intermediate value theorem: intermediate If f is a continuous function on a closed interval [a, b] If continuous closed and M is any number between f(a) and f(b), then there is at least one number c in [a, b] such that f(c) = M. is y y f(b) f(b) y = f ( x) y = f ( x) M f(a) M f(a) a c b x a c1 c2 c3 b x Theorem 5 Existence of Zeros of a Continuous Function Existence A special case of this theorem is when a continuous special function crosses the x axis. function axis If f is a continuous function on a closed interval [a, b], If continuous closed and if f(a) and f(b) have opposite signs, then there is at least one solution of the equation f(x) = 0 in the least interval (a, b). y y f(b) f(b) y = f ( x) y = f ( x) a f(a) c b x a c1 f(a) c2 c3 b x Example Example Let f(x) = x3 + x + 1. Let a. Show that f is continuous for all values of x. Show continuous b. Compute f(–1) and f(1) and use the results to deduce that Compute there must be at least one number x = c, where c lies in the where interval (–1, 1) and f(c) = 0. (–1, Solution a. The function f is a polynomial function of degree 3 and is The polynomial therefore continuous everywhere. continuous b. f (–1) = (–1)3 + (–1) + 1 = –1 and f (1) = (1)3 + (1) + 1 = 3 (–1) and (1) Since f (–1) and f (1) have opposite signs, Theorem 5 and opposite Theorem tells us that there must be at least one number x = c with –1 < c < 1 such that f(c) = 0. –1 9.3 9.3 The Derivative y f(x + h) f(x + h) – f(x) f ( x) A h x x+h x An Intuitive Example An Consider the maglev example from Section 2.4. Consider maglev The position of the maglev is a function of time given by The position time (0 ≤ t ≤ 30) (0 s = f(t) = 4t2 where s is measured in feet and t in seconds. where feet seconds Its graph is: s (ft) (ft) s = f (t ) 60 40 20 1 2 3 4 t (sec) (sec) An Intuitive Example An The graph rises slowly at first but more rapidly over time. The slowly at more over This suggests the steepness of f(t) is related to the speed of This steepness speed the maglev, which also increases over time. over If so, we might be able to find the speed of the maglev at If find speed any given time by finding the steepness of f at that time. steepness at But how do we find the steepness of a point at a curve? s (ft) (ft) s = f (t ) 60 40 20 1 2 3 4 t (sec) (sec) Slopes of Lines and of Curves The slope at a point of a curve is given by the slope of the The slope curve slope tangent to the curve at that point: tangent y A Suppose we want to find the Suppose slope at point A. slope The tangent line has the same The tangent slope as the curve does at slope curve point A. x Slopes of Lines and of Curves The slope at a point of a curve is given by the slope of the The slope curve slope tangent to the curve at that point: tangent y Slope = 1.8 ∆ y = 1.8 1.8 A ∆x = 1 The slope of the tangent in this The slope case is 1.8: 1.8 ∆y 1.8 Slope = = = 1.8 ∆x 1 x Slopes of Lines and of Curves The slope at a point of a curve is given by the slope of the The slope curve slope tangent to the curve at that point: tangent y Slope = 1.8 A x Slopes of Lines and of Curves To calculate accurately the slope of a tangent to a curve, we To accurately tangent curve we must make the change in x as small as possible: change small as y Slope = 1.8 Slope = ∆y 3 = = 0.75 ∆x 4 ∆y = 3 A As we let ∆ x get smaller, the As smaller the slope of the secant becomes slope of secant closer to the slope of the closer of tangent to the curve at that tangent point. point. ∆x = 4 x Slopes of Lines and of Curves To calculate accurately the slope of a tangent to a curve, we To accurately tangent curve we must make the change in x as small as possible: change small as y Slope = 1.8 Slope = ∆y 2.4 = = 0.8 ∆x 3 ∆ y = 2.4 2.4 A As we let ∆ x get smaller, the As smaller the slope of the secant becomes of secant closer to the slope of the closer of tangent to the curve at that tangent point. point. ∆x = 3 x Slopes of Lines and of Curves To calculate accurately the slope of a tangent to a curve, we To accurately tangent curve we must make the change in x as small as possible: change small as y Slope = 1.8 Slope = ∆ y = 2.2 2.2 A ∆y 2.2 = = 1.1 ∆x 2 As we let ∆ x get smaller, the As smaller the slope of the secant becomes of secant closer to the slope of the closer of tangent to the curve at that tangent point. point. ∆x = 2 x Slopes of Lines and of Curves To calculate accurately the slope of a tangent to a curve, we To accurately tangent curve we must make the change in x as small as possible: change small as y Slope = 1.8 A Slope = ∆y 1.5 = = 1.5 ∆x 1 As we let ∆ x get smaller, the As smaller the slope of the secant becomes of secant closer to the slope of the closer of tangent to the curve at that tangent point. point. ∆ y = 1.5 1.5 ∆x = 1 x Slopes of Lines and of Curves To calculate accurately the slope of a tangent to a curve, we To accurately tangent curve we must make the change in x as small as possible: change small as y Slope = 1.8 Slope = ∆y 0.00179 = ≈ 1.8 ∆x 0.001 As we let ∆ x get smaller, the As smaller the slope of the secant becomes of secant closer to the slope of the closer of tangent to the curve at that tangent point. point. A ∆ y = 0.00179 0.00179 ∆ x = 0.001 0.001 x Slopes of Lines and of Curves In general, we can express the slope of the secant as follows: In of Slope = ∆y f ( x + h ) − f ( x ) f ( x + h ) − f ( x ) = = ∆x ( x + h) − x h y f(x + h) f(x + h) – f(x) f (x ) A h x x+h x Slopes of Lines and of Curves Thus, as h approaches zero, the slope of the secant approaches Thus, approaches the slope of secant approaches the slope of the tangent to the curve at that point: slope of tangent y f(x + h) f(x + h) – f(x) f (x ) A h x x+h x Slopes of Lines and of Curves Thus, as h approaches zero, the slope of the secant approaches Thus, approaches the slope of secant approaches the slope of the tangent to the curve at that point: slope of tangent y f(x + h) f(x + h) – f(x) f( x ) A h x x+h x Slopes of Lines and of Curves Thus, as h approaches zero, the slope of the secant approaches Thus, approaches the slope of secant approaches the slope of the tangent to the curve at that point: slope of tangent y f(x + h) f(x + h) – f(x) f( x ) A h x x+h x Slopes of Lines and of Curves Thus, as h approaches zero, the slope of the secant approaches Thus, approaches the slope of secant approaches the slope of the tangent to the curve at that point: slope of tangent y f(x + h) f( x ) f(x + h) – f(x) A h x x+h x Slopes of Lines and of Curves Thus, as h approaches zero, the slope of the secant approaches Thus, approaches the slope of secant approaches the slope of the tangent to the curve at that point: slope of tangent y f(x + h) f( x ) A h f(x + h) – f(x) xx + h x Slopes of Lines and of Curves Thus, as h approaches zero, the slope of the secant approaches Thus, approaches the slope of secant approaches the slope of the tangent to the curve at that point. slope of tangent Expressed in limit notation: Expressed limit The slope of the tangent line to the graph The of of f at the point P(x, f(x)) is given by lim h→0 if it exists. f ( x + h) − f ( x) h Average Rates of Change Average We can see that measuring the slope of the tangent line to slope of tangent a graph is mathematically equivalent to finding the mathematically rate of change of f at x. rate The number f(x + h) – f(x) measures the change in y that The change corresponds to a change h in x. change Then the difference quotient Then difference f ( x + h) − f ( x ) h measures the average rate of change of y with respect to x average over the interval [x, x + h]. In the maglev example, if y measures the position of the In maglev if position train at time x, then the quotient gives the average velocity then quotient average of the train over the time interval [x, x + h]. Average Rates of Change Average The average rate of change of f over the interval The average interval [x, x + h] or slope of the secant line to the graph of slope secant f through the points (x, f(x)) and (x + h, f(x + h)) is f ( x + h) − f ( x ) h Instantaneous Rates of Change Instantaneous By taking the limit of the difference quotient as h goes to of zero, evaluating zero, f ( x + h) − f ( x ) lim h→0 h we obtain the rate of change of f at x. we rate This is known as the instantaneous rate of change of f at x This instantaneous (as opposed to the average rate of change). (as average rate In the maglev example, if y measures the position of a In position train at time x, then the limit gives the velocity of the train velocity at time x. Instantaneous Rates of Change Instantaneous The instantaneous rate of change of f at x or slope The instantaneous slope of the tangent line to the graph of f at (x, f(x)) is of f ( x + h) − f ( x) lim h→0 h This limit is called the derivative of f at x . This derivative The Derivative of a Function The The derivative of a function f with respect to x is the The function f′′ (read “f prime”). (read f ( x + h) − f ( x) f ′( x ) = lim h→0 h The domain of f ′ is the set of all x where the limit exists. The domain all limit Thus, the derivative of function f is a function f ′ that Thus, derivative function gives the slope of the tangent line to the graph of f at any slope of point (x, f(x)) and also the rate of change of f at x. rate The Derivative of a Function The Four Step Process for Finding f ′(x) Four 1. Compute f(x + h). Compute 2. Form the difference f(x + h) – f(x). Form f ( x + h) − f ( x) . 3. Form the quotient 3. h f ( x + h) − f ( x) ′( x ) = lim . 4. Compute f Compute h→0 h Examples Examples Find the slope of the tangent line to the graph f(x) = 3x + 5 Find slope at any point (x, f(x)). Solution The required slope is given by the derivative of f at x. The derivative To find the derivative, we use the four-step process: To four-step Step 1. f(x + h) = 3(x + h) + 5 = 3x + 3h + 5. Step 2. f(x + h) – f(x) = 3x + 3h + 5 – (3x + 5) = 3h. Step 3. f ( x + h ) − f ( x ) 3h = = 3. h h Step 4. f ′( x ) = lim f ( x + h ) − f ( x ) = lim 3 = 3. Step h→0 h→0 h Examples Examples Find the slope of the tangent line to the graph f(x) = x2 at Find slope any point (x, f(x)). Solution The required slope is given by the derivative of f at x. The derivative To find the derivative, we use the four-step process: To four-step Step 1. f(x + h) = (x + h)2 = x2 + 2xh + h2. Step 2. f(x + h) – f(x) = x2 + 2xh + h2 – x2 = h(2x + h). Step 3. f ( x + h ) − f ( x ) h(2 x + h ) = = 2 x + h. h h Step 4. f ′( x ) = lim f ( x + h ) − f ( x ) = lim(2 x + h ) = 2 x . Step h→0 h→0 h Examples Examples Find the slope of the tangent line to the graph f(x) = x2 at Find slope any point (x, f(x)). The slope of the tangent line is given by f ′(x) = 2x. Now, find and interpret f ′(2). Now, find y Solution f ′(2) = 2(2) = 4. 5 This means that, at the point (2, This at (2, 4) 4 4)… 4) 3 … the slope of the tangent line the to the graph is 4. 2 4 1 1 –2 –1 0 1 2 x Applied Example: Demand for Tires Applied The management of Titan Tire Company has determined The that the weekly demand function of their Super Titan tires weekly is given by is p = f ( x ) = 144 − x 2 where p is measured in dollars and x is measured in where dollars thousands of tires. thousands Find the average rate of change in the unit price of a tire if Find average unit the quantity demanded is between 5000 and 6000 tires; 5000 6000 between 5000 and 5100 tires; and between 5000 and 5010 5000 5100 5000 5010 tires. tires. What is the instantaneous rate of change of the unit price What instantaneous when the quantity demanded is 5000 tires? 5000 Applied Example: Demand for Tires Applied Solution The average rate of change of the unit price of a tire if the The average quantity demanded is between x and x + h is f ( x + h ) − f ( x ) [144 − ( x + h )2 ] − (144 − x 2 ) = h h 144 − x 2 − 2 xh − h 2 − 144 + x 2 = h −2 xh − h 2 h( −2 x − h ) = = h h = −2 x − h Applied Example: Demand for Tires Applied Solution The average rate of change is given by –2x – h. The average –2 To find the average rate of change of the unit price of a To average tire when the quantity demanded is between 5000 and quantity 5000 6000 tires [5, 6], we take x = 5 and h = 1, obtaining 6000 [5, we −2(5) − 1 = −11 or –$11 per 1000 tires. or –$11 1000 Similarly, with x = 5, and h = 0.1, we obtain Similarly, and we −2(5) − 0.1 = −10.1 or –$10.10 per 1000 tires. –$10.10 1000 Finally, with x = 5, and h = 0.01, we get Finally, and we −2(5) − 0.01 = −10.01 or –$10.01 per 1000 tires. –$10.01 1000 Applied Example: Demand for Tires Applied Solution The instantaneous rate of change of the unit price of a tire The instantaneous when the quantity demanded is x tires is given by f ′( x ) = lim h→0 f ( x + h) − f ( x) = lim( −2 x − h ) = −2 x h→0 h In particular, the instantaneous rate of change of the price the per tire when quantity demanded is 5000 is given by –2(5), when 5000 –2(5) or –$10 per tire. –$10 Differentiability and Continuity Differentiability Sometimes, one encounters continuous functions that fail Sometimes, encounters to be differentiable at certain values in the domain of the to domain function f. For example, consider the continuous function f below: consider ✦ It fails to be differentiable at x = a , because the graph It fails makes an abrupt change (a corner) at that point. abrupt (It is not clear what the slope is at that point) slope is y a x Differentiability and Continuity Differentiability Sometimes, one encounters continuous functions that fail Sometimes, encounters to be differentiable at certain values in the domain of the to domain function f. For example, consider the continuous function f below: consider ✦ It also fails to be differentiable at x = b because the It fails slope is not defined at that point. slope y a b x Applied Example: Wages Applied Mary works at the B&O department store, where, on a Mary weekday, she is paid $8 an hour for the first 8 hours and paid $8 first $12 an hour of overtime. $12 overtime The function if 0 ≤ x ≤ 8 8 x f ( x) = 12 x − 32 if x > 8 gives Mary’s earnings on a weekday in which she worked gives earnings x hours. hours Sketch the graph of the function f and explain why it is not differentiable at x = 8. differentiable Applied Example: Wages Applied Solution 130 110 y if 0 ≤ x ≤ 8 8 x f ( x) = 12 x − 32 if x > 8 90 70 (8, 64) 50 30 10 2 4 6 8 10 12 x The graph of f has a corner at x = 8 and so is not The corner differentiable at that point. differentiable 9.4 9.4 Basic Rules of Differentiation d ( 4 x 5 + 3 x 4 − 8 x 2 + x + 3) dx d d d d d = 4 ( x 5 ) + 3 ( x 4 ) − 8 ( x 2 ) + ( x ) + ( 3) dx dx dx dx dx f ′( x ) = = 4 ( 5x 4 ) + 3 ( 4 x3 ) − 8 ( 2 x ) + 1 + 0 = 20 x 4 + 12 x 3 − 16 x + 1 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.” “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 + h) − f ( x) h→0 h c−c = lim h→0 h = lim 0 f ′( x ) = lim 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 ) = d8 ( x ) = 8 ⋅ x8−1 = 8 x 7 dx f ′( x ) = d 5/2 5 5/2−1 5 3/2 ( x ) = 2⋅x = 2 x dx 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 1/2−1 =x 2 = 1 2x 1 −1/2 =x 2 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 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 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 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 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 = 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: 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: 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 9.5 9.5 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 ) Rule 5: The Product Rule Rule Practice Examples: Find the derivative of f ( x ) = x 3 ( ) x +1 d 1/2 d3 1/2 f ′( x ) = x ( x + 1) + ( x + 1) dx x dx 3 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 = 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) ( = 2x − 4 − 2x ( 2 x − 4) 2 =− 4 ( 2 x − 4) 2 Rule 6: The Quotient Rule Rule Practice Examples: Find the derivative of 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) 2 (x = = 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: 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) 5t + 5 − 10t 2 (t 2 + 1) 2 2 2 = 5( 1 − t2 ) (t 2 + 1) 2 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: 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 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 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 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) −1/2 = 6 ( 2 x + 3) 2 x 2 + ( 2 x 2 + 3) 2 2 2 = 6 ( 4 x 2 + 3) 2x2 + 3 −1/2 + 6 ( 2 x + 3) 2 1/2 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 or 8 feet per second per second (ft/sec2). or 9.6 9.6 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 H(x) is composed of two simpler functions f ( x) = x2 + 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 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 (n, a real number) In other words, the function h is given by the power of a In the function f. function Examples: h( x ) = ( x + x + 1) 2 100 H ( x) = 1 (5− x ) 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 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 d 5 5d ( 2 x + 3) + ( 2 x + 3) x 2 dx dx = x ( 5) ( 2 x + 3) 2 4 ( 2 ) + ( 2 x + 3) ( 2 ) x = 10 x 2 ( 2 x + 3) + 2 x ( 2 x + 3) 4 = 2 x ( 2 x + 3) ( 5x + 2 x + 3) 4 = 2 x ( 2 x + 3) ( 7 x + 3) 4 5 5 The General Power Rule The Practice Examples: Find the derivative of f ( x ) = Find ( 4x 1 2 − 7) 2 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 =− 16 x ( 4x 2 − 7) 3 −3 ( 8x ) 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 2 2x + 1 d 2x + 1 f ′( x ) = 3 3x + 2 dx 3x + 2 2x + 1 = 3 3x + 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 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 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 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 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 9.7 9.7 Differentiation of the Exponential Differentiation and Logarithmic Functions and y (− 1 2 , e −1/2 ) 1 ( 1 2 , e −1/2 ) 2 2 f ( x) = e− x x –1 1 Rule 8 Rule Derivative of the Exponential Function The derivative of the exponential function with The base e is equal to the function itself: dx ( e ) = ex dx Examples Examples Find the derivative of the function f ( x ) = x 2 e x Find derivative Solution Using the product rule gives Using product d 2x d d x e ) = x2 ( ex ) + ex ( x2 ) ( dx dx dx = x 2e x + e x (2 x ) f ′( x ) = = xe x ( x + 2) Examples Examples ( Find the derivative of the function g (t ) = e + 2 Find derivative Solution Using the general power rule gives Using general 1/2 d 3t g ′(t ) = ( e + 2 ) ( et + 2 ) 2 dt 1/2 t 3t = ( e + 2) e 2 1/2 3t t = e ( e + 2) 2 t ) 3/2 Rule 9 Rule Chain Rule for Exponential Functions If f(x) is a differentiable function, then If d f ( x) e ) = e f ( x ) f ′( x ) ( dx Examples Examples Find the derivative of the function f ( x ) = e 2 x Find derivative Solution d f ′( x ) = e ( 2x) dx = e 2 x (2) 2x = 2e 2 x Examples Examples Find the derivative of the function y = e −3 x Find derivative Solution dy −3 x d =e ( 3 x ) dx dx = e −3 x ( −3) = −3e −3 x Examples Examples Find the derivative of the function g (t ) = e 2 t Find derivative Solution g ′(t ) = e 2 t 2 +t d ⋅ ( 2t 2 + t ) dt = (4t + 1)e 2 t 2 +t 2 +t Examples Examples Find the derivative of the function Find derivative y = xe −2 x Solution dy d −2 x −2 x d = x (e ) +e ( x) dx dx dx d = x e −2 x ( −2 x ) + e −2 x (1) dx = xe −2 x ( −2) + e −2 x = −2 xe −2 x + e −2 x = e −2 x (1 − 2 x ) Examples Examples et Find the derivative of the function g (t ) = Find derivative et + e − t Solution t −t d t td e +e ) (e ) −e et + e − t ) ( ( dt dt g ′(t ) = t −t 2 (e +e ) (e = = = t + e − t ) et − et ( et − e − t ) (e t +e e2t + 1 − e2t + 1 (e (e t +e 2 t +e ) −t 2 ) −t 2 ) −t 2 Rule 10 Rule Derivative of the Natural Logarithm The derivative of ln x is The ln d 1 ln x = dx x ( x ≠ 0) Examples Examples Find the derivative of the function f ( x ) = x ln x Find derivative Solution d d f ′( x ) = x ⋅ (ln x ) + ln x ⋅ ( x ) dx dx 1 = x ⋅ + ln x ⋅ (1) x = 1 + ln x Examples Examples Find the derivative of the function g ( x ) = Find derivative Solution d d (ln x ) − ln x ⋅ ( x ) dx g ′( x ) = dx x2 1 x ⋅ − ln x ⋅ (1) x = x2 1 − ln x = x2 x⋅ ln x x Rule 11 Rule Chain Rule for Logarithmic Functions If f(x) is a differentiable function, then If d f ′( x ) [ ln f ( x )] = dx f ( x) [ f ( x ) > 0] Examples Examples Find the derivative of the function f ( x ) = ln( x 2 + 1) Find derivative Solution d2 ( x + 1) f ′( x ) = dx 2 x +1 2x =2 x +1 Examples Examples Find the derivative of the function y = ln[( x 2 + 1)( x 3 + 2)6 ] Find derivative Solution y = ln[( x 2 + 1)( x 3 + 2)6 ] = ln( x 2 + 1) + ln( x 3 + 2)6 = ln( x 2 + 1) + 6ln( x 3 + 2) d2 d3 ( x + 1) ( x + 2) dy dx = + 6 dx 3 dx x2 + 1 x +2 2x 3x 2 =2 +6 3 x +1 x +2 2x 18 x 2 =2 +3 x +1 x + 2 9.8 9.8 Marginal Functions in Economics MC(251) = C(251) – C(250) (251) (251) = [8000 + 200(251) – 0.2(251)2] – [8000 + 200(250) – 0.2(250)2] [8000 = 45,599.8 – 45,500 = 99.80 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 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: 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: 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 C´ 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: 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: 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: 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: 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: 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´ 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: 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: 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: 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: 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: 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: 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: 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: 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: 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 End of Chapter ...
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This note was uploaded on 07/08/2011 for the course BUS 205 taught by Professor Michael during the Spring '11 term at University of Hawaii, Manoa.

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