SHORT-CUTS TO DIFFERENTIATION

Chapter Three SHORT-CUTS TO DIFFERENTIATION In Chapter, we efine the erivative function f () = lim h 0 f( + h) f() h an saw how the erivative represents a slope an a rate of change. We learne how to approimate the erivative of a function given graphically (by estimating the slope of the tangent at each point) an numerically (by fining the average rate of change of the function between ata values). We calculate the erivatives of an 3 eactly using the efinition. In this chapter we make a systematic stuy of the erivatives of functions given by formulas, such as power, polynomial, eponential, logarithmic, trigonometric, an hyperbolic functions. The chapter also contains general rules, such as the prouct, quotient, an chain rules, which allow us to ifferentiate combinations of functions. The chapter conclues with theorems about ifferentiable functions, incluing the Mean Value Theorem an its consequences.

0 Chapter Three SHORT-CUTS TO DIFFERENTIATION 3. POWERS AND POLYNOMIALS Derivative of a Constant Times a Function y Figure 3. shows the graph of y = f() an of three multiples: y = 3f(), y = f(), an y = f(). What is the relationship between the erivatives of these functions? In other wors, for a particular -value, how are the slopes of these graphs relate? y y y f() 3f() f()/ f() Slope = m Slope = 3m Slope = m/ Slope= m Figure 3.: A function an its multiples: Derivative of multiple is multiple of erivative Multiplying the value of a function by a constant stretches or shrinks the original graph (an reflects it in the -ais if the constant is negative). This changes the slope of the curve at each point. If the graph has been stretche, the rises have all been increase by the same factor, whereas the runs remain the same. Thus, the slopes are all steeper by the same factor. If the graph has been shrunk, the slopes are all smaller by the same factor. If the graph has been reflecte in the -ais, the slopes will all have their signs reverse. In other wors, if a function is multiplie by a constant, c, so is its erivative: Theorem 3.: Derivative of a Constant Multiple If f is ifferentiable an c is a constant, then [cf()] = cf (). Proof Although the graphical argument makes the theorem plausible, to prove it we must use the efinition of the erivative: cf( + h) cf() f( + h) f() [cf()] = lim = lim c h 0 h h 0 h f( + h) f() = c lim = cf (). h 0 h We can take c across the limit sign by the properties of limits (part of Theorem. on page 5). Derivatives of Sums an Differences Suppose we have two functions, f() an g(), with the values liste in Table 3.. Values of the sum f() + g() are given in the same table. Table 3. Sum of Functions f() g() f() + g() 0 00 0 00 0 0. 0. 30 0.4 30.4 3 60 0.6 60.6 4 00 0.8 00.8

3. POWERS AND POLYNOMIALS We see that aing the increments of f() an the increments of g() gives the increments of f() + g(). For eample, as increases from 0 to, f() increases by 0 an g() increases by 0., while f() + g() increases by 0. 00 = 0.. Similarly, as increases from 3 to 4, f() increases by 40 an g() by 0., while f() + g() increases by 00.8 60.6 = 40.. From this eample, we see that the rate at which f()+g() is increasing is the sum of the rates at which f() an g() are increasing. Similar reasoning applies to the ifference, f() g(). In terms of erivatives: Theorem 3.: Derivative of Sum an Difference If f an g are ifferentiable, then [f() + g()] = f () + g () an [f() g()] = f () g (). Proof Using the efinition of the erivative: [f( + h) + g( + h)] [f() + g()] [f() + g()] = lim h 0 h = lim h 0 = f () + g (). f( + h) f() h } {{ } Limit of this is f () g( + h) g() + }{{ h } Limit of this is g () We have use the fact that the limit of a sum is the sum of the limits, part of Theorem. on page 5. The proof for f() g() is similar. Powers of In Chapter we showe that f () = ( ) = an g () = (3 ) = 3. The graphs of f() = an g() = 3 an their erivatives are shown in Figures 3. an 3.3. Notice f () = has the behavior we epect. It is negative for < 0 (when f is ecreasing), zero for = 0, an positive for > 0 (when f is increasing). Similarly, g () = 3 is zero when = 0, but positive everywhere else, as g is increasing everywhere else. y f() = y 0 0 5 5 0 f () = Figure 3.: Graphs of f() = an its erivative f () = g () = 3 0 0 3 3 0 0 g() = 3 Figure 3.3: Graphs of g() = 3 an its erivative g () = 3

Chapter Three SHORT-CUTS TO DIFFERENTIATION Eample These eamples are special cases of the power rule which we justify for any positive integer n on page 3: The Power Rule For any constant real number n, (n ) = n n Problem 7 asks you to show that this rule hols for negative integral powers; such powers can also be ifferentiate using the quotient rule (Section 3.3). In Section 3.6 we inicate how to justify the power rule for powers of the form /n. Use the power rule to ifferentiate (a) (a) For n = 3: (b) For n = /: (c) For n = /3: 3, (b) /, (c) 3. ( ) 3 = ( 3 ) = 3 3 = 3 4 = 3 4. ( /) = (/) = / =. ( ) 3 = ( /3) = 3 ( /3) = 3 4/3 = 3 4/3. Eample Use the efinition of the erivative to justify the power rule for n = : Show ( ) = 3. Provie 0, we have ( ) = ( ) = lim h 0 ( ) (+h) = lim h h 0 h = lim h 0 h [ ( + h) ( + h) [ ( + h + h ] ) h h = lim h 0 h( + h) h = lim h 0 ( + h) ( + h) = 4 (Letting h 0) = 3. ] (Combining fractions over a common enominator) (Simplifying numerator) (Diviing numerator an enominator by h) (Multiplying out) The graphs of an its erivative, 3, are shown in Figure 3.4. Does the graph of the erivative have the features you epect? 3 Figure 3.4: Graphs of an its erivative, 3

Justification of (n ) = n n, for n a Positive Integer 3. POWERS AND POLYNOMIALS 3 To calculate the erivatives of an 3, we ha to epan ( + h) an ( + h) 3. To calculate the erivative of n, we must epan ( + h) n. Let s look back at the previous epansions: ( + h) = + h + h, ( + h) 3 = 3 + 3 h + 3h + h 3, an multiply out a few more eamples: In general, we can say ( + h) 4 = 4 + 4 3 h + 6 h + 4h 3 + h 4, ( + h) 5 = 5 + 5 4 h + 0 3 h + 0 h 3 + 5h 4 + h 5. }{{} Terms involving h an higher powers of h ( + h) n = n + n n h + + h n. }{{} Terms involving h an higher powers of h We have just seen this is true for n =, 3, 4, 5. It can be prove in general using the Binomial Theorem (see the online theory supplement). Now to fin the erivative, ( + h) n n (n ) = lim h 0 h = lim h 0 ( n + n n h + + h n ) n h Terms involving h an higher powers of h {}}{ n n h + + h n = lim. h 0 h When we factor out h from terms involving h an higher powers of h, each term will still have an h in it. Factoring an iviing, we get: h(n n + + h n ) (n ) = lim = lim h 0 h But as h 0, all terms involving an h will go to 0, so Derivatives of Polynomials These terms go to 0 Terms involving h an higher powers of h {}}{ (n n + + h n ). h 0 (n ) = lim (n n + + h n ) = n h 0 }{{} n. Now that we know how to ifferentiate powers, constant multiples, an sums, we can ifferentiate any polynomial. Eample 3 Fin the erivatives of (a) 5 + 3 +, (b) 3 7 5 5 + π. (a) (5 + 3 + ) = 5 ( ) + 3 () + () = 5 + 3 + 0 (Since the erivative of a constant, (), is zero.) = 0 + 3.

4 Chapter Three SHORT-CUTS TO DIFFERENTIATION (b) ( ) 3 7 5 5 + π = 3 (7 ) 5 (5 ) + (π) = 3 7 6 5 54 + 0 (Since π is a constant, π/ = 0.) = 7 3 6 4. We can also use the rules we have seen so far to ifferentiate epressions which are not polynomials. Eample 4 Differentiate (a) 5 0 + (a) (b) ( 5 0 + ) (b) 0. 3 + = (5 / 0 + ) / = 5 / 0( ) 3 + = 5 + 0 3 4 3/. ( ) 3/ (0.3 + ) = 0. (3 ) + ( ) = 0.3 +. Eample 5 Fin the secon erivative an interpret its sign for (a) f() =, (b) g() = 3, (c) k() = /. (a) If f() =, then f () =, so f () = () =. Since f is always positive, f is concave up, as epecte for a parabola opening upwar. (See Figure 3.5.) (b) If g() = 3, then g () = 3, so g () = (3 ) = 3 ( ) = 3 = 6. This is positive for > 0 an negative for < 0, which means 3 is concave up for > 0 an concave own for < 0. (See Figure 3.6.) (c) If k() = /, then k () = (/) = /, so 4 k () = ( / ) = ( ) (/) = 4 3/. Now k an k are only efine on the omain of k, that is, 0. When > 0, we see that k () is negative, so k is concave own. (See Figure 3.7.) f > 0 Figure 3.5: f() = an f () = 8 g < 0 8 g > 0 Figure 3.6: g() = 3 an g () = 6 k < 0 Figure 3.7: k() = / an k () = 4 3/ 4 Eample 6 If the position of a boy, in meters, is given as a function of time t, in secons, by s = 4.9t + 5t + 6, fin the velocity an acceleration of the boy at time t.

3. POWERS AND POLYNOMIALS 5 The velocity, v, is the erivative of the position: v = s t = t ( 4.9t + 5t + 6) = 9.8t + 5, an the acceleration, a, is the erivative of the velocity: a = v t = ( 9.8t + 5) = 9.8. t Note that v is in meters/secon an a is in meters/secon. Eample 7 Figure 3.8 shows the graph of a cubic polynomial. Both graphically an algebraically, escribe the behavior of the erivative of this cubic. A B f A f C C B Figure 3.8: The cubic of Eample 7 Figure 3.9: Derivative of the cubic of Eample 7 Graphical approach: Suppose we move along the curve from left to right. To the left of A, the slope is positive; it starts very positive an ecreases until the curve reaches A, where the slope is 0. Between A an C the slope is negative. Between A an B the slope is ecreasing (getting more negative); it is most negative at B. Between B an C the slope is negative but increasing; at C the slope is zero. From C to the right, the slope is positive an increasing. The graph of the erivative function is shown in Figure 3.9. Algebraic approach: f is a cubic that goes to + as +, so with a > 0. Hence, f() = a 3 + b + c + f () = 3a + b + c, whose graph is a parabola opening upwar, as in Figure 3.9. Eercises an Problems for Section 3. Eercises. Let f() = 7. Using the efinition of the erivative, show that f () = 0 for all values of.. Let f() = 7+. Use the efinition of the erivative to calculate f (). For Eercises 3 44, fin the erivatives of the given functions. Assume that a, b, c, an k are constants. 3. y = 4. y = 5. y = 6. y = 7. y = 3. 8. y = 4/3 3. f(z) = z 6. 4. y = r 7/ 5. y = 6. f() = 4 7. h(θ) = 8. f() = 3 θ 3 9. f() = e 0. y = 4 3/ 5 /. f(t) = 3t 4t +. y = 7 + 4 / 3. y = z + z 4. f() = 5 4 + 5. h(w) = w 3 +3 w 6. y = 6 3 + 4 9. y = 3/4 0. y = 3/4 7. y = 3t 5 5 t + 7 t 8. y = 3t + t t. f() = 4. g(t) = t 5 9. y = ( + ) 30. y = t 3/ ( + t)

6 Chapter Three SHORT-CUTS TO DIFFERENTIATION 3. h(t) = 3 t + 4 t 3. y = θ ( θ + θ 33. y = + 34. f(z) = z + 3z 35. f(t) = t + t 3 t 4 36. y = θ θ 37. j() = 3 a + a b c 38. f() = a + b ) 39. h() = a + b c 4. 43. V r if V = 4 3 πr b 4. y if y = a + b + c 44. 40. g(t) = t( + t) t w q if w = 3ab q P t if P = a + b t Problems For Problems 45 50, etermine if the erivative rules from this section apply. If they o, fin the erivative. If they on t apply, inicate why. 45. y = ( + 3) / 46. y = 3 47. g() = π π 48. y = 3 + 4 49. y = 3 + 4 50. y = 3z + 4 5. If f(t) = t 3 4t + 3t, fin f (t) an f (t). 5. If f() = 3 8 + an f (r) = 4, fin r. 53. Fin the equation of the line tangent to the graph of f at (, ), where f is given by f() = 3 +. 54. (a) Fin the equation of the tangent line to f() = 3 at the point where =. (b) Graph the tangent line an the function on the same aes. If the tangent line is use to estimate values of the function, will the estimates be overestimates or unerestimates? 55. If f() = 4 3 + 6 3 + 7, fin the intervals on which f (). 56. Using a graph to help you, fin the equations of all lines through the origin tangent to the parabola y = + 4. Sketch the lines on the graph. 57. On what intervals is the function f() = 4 4 3 both ecreasing an concave up? 58. For what values of is the graph of y = 5 5 both increasing an concave up? 59. If f() = 3 6 5 + 0, fin analytically all values of for which f () = 0. Show your answers on a graph of f. 60. (a) Fin the eighth erivative of f() = 7 + 5 5 4 3 + 6 7. Think ahea! (The n th erivative, f (n) (), is the result of ifferentiating f() n times.) (b) Fin the seventh erivative of f(). 6. The height of a san une (in centimeters) is represente by f(t) = 700 3t, where t is measure in years since 995. Fin f(5) an f (5). Using units, eplain what each means in terms of the san une. 6. A ball is roppe from the top of the Empire State builing to the groun below. The height, y, of the ball above the groun (in feet) is given as a function of time, t, (in secons) by y = 50 6t. (a) Fin the velocity of the ball at time t. What is the sign of the velocity? Why is this to be epecte? (b) Show that the acceleration of the ball is a constant. What are the value an sign of this constant? (c) When oes the ball hit the groun, an how fast is it going at that time? Give your answer in feet per secon an in miles per hour ( ft/sec = 5/ mph). 63. At a time t secons after it is thrown up in the air, a tomato is at a height of f(t) = 4.9t + 5t + 3 meters. (a) What is the average velocity of the tomato uring the first secons? Give units. (b) Fin (eactly) the instantaneous velocity of the tomato at t =. Give units. (c) What is the acceleration at t =? () How high oes the tomato go? (e) How long is the tomato in the air? 64. The gravitational attraction, F, between the earth an a satellite of mass m at a istance r from the center of the earth is given by F = GMm r, where M is the mass of the earth, an G is a constant. Fin the rate of change of force with respect to istance. 65. The perio, T, of a penulum is given in terms of its length, l, by l T = π g, where g is the acceleration ue to gravity (a constant). (a) Fin T/l. (b) What is the sign of T/l? What oes this tell you about the perio of penulums?

3. THE EXPONENTIAL FUNCTION 7 66. (a) Use the formula for the area of a circle of raius r, A = πr, to fin A/r. (b) The result from part (a) shoul look familiar. What oes A/r represent geometrically? (c) Use the ifference quotient to eplain the observation you mae in part (b). 67. What is the formula for V, the volume of a sphere of raius r? Fin V /r. What is the geometrical meaning of V /r? 68. Show that for any power function f() = n, we have f () = n. 69. Given a power function of the form f() = a n, with f () = 3 an f (4) = 4, fin n an a. 70. Is there a value of n which makes y = n a solution to the equation 3(y/) = y? If so, what value? 7. Using the efinition of erivative, justify the formula ( n )/ = n n. (a) For n = ; for n = 3. (b) For any negative integer n. 3. THE EXPONENTIAL FUNCTION What o we epect the graph of the erivative of the eponential function f() = a to look like? The eponential function in Figure 3.0 increases slowly for < 0 an more rapily for > 0, so the values of f are small for < 0 an larger for > 0. Since the function is increasing for all values of, the graph of the erivative must lie above the -ais. It appears that the graph of f may resemble the graph of f itself. f() = a Figure 3.0: f() = a, with a > In this section we see that f () = k a, so in fact f () is proportional to f(). This property of eponential functions makes them particularly useful in moeling because many quantities have rates of change which are proportional to themselves. For eample, the simplest moel of population growth has this property. Derivatives of Eponential Functions an the Number e We start by calculating the erivative of g() =, which is given by ( g +h ) () = lim h 0 h ( h = lim h 0 h = lim h 0 ( h ) = lim h 0 ( h ) h h ). (Since an are fie uring this calculation). To fin lim h 0 ( h )/h, see Table 3. where we have substitute values of h near 0. The table suggests that the limit eists an has value 0.693. Let us call the limit k, so k = 0.693. Then ( ) = k = 0.693. So the erivative of is proportional to with constant of proportionality 0.693. A similar calculation shows that the erivative of f() = a is ( a f +h a ) ( a h ) () = lim = lim a. h 0 h h 0 h

8 Chapter Three SHORT-CUTS TO DIFFERENTIATION Table 3. h ( h )/h 0. 0.6697 0.0 0.6908 0.00 0.699 0.00 0.6934 0.0 0.6956 0. 0.777 Table 3.3 a a k = lim h h 0 h 0.693 3.099 4.396 5.609 6.79 7.946 Table 3.4 h ( + h) /h 0.00.7964 0.000.78478 0.0000.78954 0.0000.7868 0.000.78459 0.00.76939 The quantity lim h 0 (a h )/h is also a constant, although the value of the constant epens on a. Writing k = lim h 0 (a h )/h, we see that the erivative of f() = a is proportional to a : (a ) = k a. For particular values of a, we can estimate k by substituting values of h near 0 into the epression (a h )/h. Table 3.3 shows the results. Notice that for a =, the value of k is less than, while for a = 3, 4, 5,..., the values of k are greater than. The values of k appear to be increasing, so we guess that there is a value of a between an 3 for which k =. If so, we have foun a value of a with the remarkable property that the function a is equal to its own erivative. So let s look for such an a. This means we want to fin a such that a h lim =, or, for small h, h 0 h Solving for a, we can estimate a as follows: a h h. a h h, or a h + h, so a ( + h) /h. Taking small values of h, as in Table 3.4, we see a.78.... This is the number e introuce in Chapter. In fact, it can be shown that if e = lim ( + h) /h e h =.78... then lim =. h 0 h 0 h This means that e is its own erivative: (e ) = e. Figure 3. shows the graphs, 3, an e together with their erivatives. Notice that the erivative of is below the graph of, since k < there, an the graph of the erivative of 3 is above the graph of 3, since k > there. With e.78, the function e an its erivative are ientical. Note on Roun-Off Error an Limits If we try to evaluate ( + h) /h on a calculator by taking smaller an smaller values of h, the values of (+h) /h at first get closer to.78.... However, they will eventually move away again because of the roun-off error (i.e., errors introuce by the fact that the calculator can only hol a certain number of igits). As we try smaller an smaller values of h, how o we know when to stop? Unfortunately, there is no fie rule. A calculator can only suggest the value of a limit, but can never confirm that this

3. THE EXPONENTIAL FUNCTION 9 value is correct. In this case, it looks like the limit is.78... because the values of (e h )/h approach this number for a while. To be sure this is correct, we have to fin the limit analytically. f() = e an its erivative g () (.)3 f () (0.69) g() = 3 Figure 3.: Graphs of the functions, e, an 3 an their erivatives A Formula for the Derivative of a To get a formula for the erivative of a, we must calculate f a +h a ( a h ) () = lim = lim a. h 0 h h 0 h }{{} k However, without knowing the value of a, we can t use a calculator to estimate k. We take a ifferent approach, rewriting a = e ln a, so a h (e ln a ) h e (ln a)h lim = lim = lim. h 0 h h 0 h h 0 h To evaluate this limit, we use a limit that we alreay know e h lim =. h 0 h In orer to use this limit, we substitute t = (ln a)h. Since t approaches 0 as h approaches 0, we have e (ln a)h e t ( ) ( lim = lim h 0 h t 0 (t/ ln a) = lim ln a et e t ) = ln a lim = (ln a) = ln a. t 0 t t 0 t Thus, we have f a +h a ( () = lim = lim h 0 h h 0 a h ) a = (ln a)a. h In Section 3.6 we obtain the same result by another metho. We conclue that: (a ) = (ln a)a. Thus, for any a, the erivative of a is proportional to a. The constant of proportionality is ln a. The erivative of a is equal to a if the constant of proportionality is, that is, if ln a =, then a = e. The fact that the constant of proportionality is when a = e makes e a particularly convenient base for eponential functions.

0 Chapter Three SHORT-CUTS TO DIFFERENTIATION Eample Differentiate 3 + 5e. ( 3 + 5e ) = (3 ) + 5 (e ) = ln 3 3 + 5e (.97)3 + 5e. Eercises an Problems for Section 3. Eercises Fin the erivatives of the functions in Eercises 6. Assume that a, b, c, an k are constants.. f() = e +. y = 5t + 4e t 3. y = 5 + 4. f() = e + 5. y = 5 + + 3 6. f() = + 3 7. y = 4 0 3 8. y = 3 4 9. y = + 3 0. y = 3 3 + 33. z = (ln 4)e. z = (ln 4)4 3. f(t) = (ln 3) t 4. y = 5 5 t + 6 6 t 5. h(z) = (ln ) z 6. f() = e + e 7. f() = 3 + 3 8. y = π + π 9. f() = e π + π 0. f() = π + π. f() = e k + k. f() = e + 3. f(t) = e t+ 4. y = e θ 5. y() = a + a. 6. f() = π + (π ) Which of the functions in Eercises 7 35 can be ifferentiate using the rules we have evelope so far? Differentiate if you can; otherwise, inicate why the rules iscusse so far o not apply. 7. y = + 8. y = ( ) 9. y = 30. y = 3. y = e +5 3. y = e 5 33. y = 4 ( ) 35. f(θ) = 4 θ 34. f(z) = ( 4) z Problems 36. With a yearly inflation rate of 5%, prices are given by P = P 0(.05) t, where P 0 is the price in ollars when t = 0 an t is time in years. Suppose P 0 =. How fast (in cents/year) are prices rising when t = 0? 37. Since January, 960, the population of Slim Chance has been escribe by the formula P = 35,000(0.98) t, where P is the population of the city t years after the start of 960. At what rate was the population changing on January, 983? 38. The population of the worl in billions of people can be moele by the function f(t) = 5.3(.08) t, where t is years since 990. Fin f(0) an f (0). Fin f(30) an f (30). Using units, eplain what each answer tells you about the population of the worl. 39. Certain pieces of antique furniture increase very rapily in price in the 970s an 980s. For eample, the value of a particular rocking chair is well approimate by V = 75(.35) t, where V is in ollars an t is the number of years since 975. Fin the rate, in ollars per year, at which the price is increasing. 40. The value of a certain automobile purchase in 997 can be approimate by the function V (t) = 5(0.85) t, where t is the time, in years, from the ate of purchase, an V is the value, in thousans of ollars. (a) Evaluate an interpret V (4). (b) Fin an epression for V (t), incluing units. (c) Evaluate an interpret V (4). () Use V (t), V (t), an any other consierations you think are relevant to write a paragraph in support of or in opposition to the following statement: From a monetary point of view, it is best to keep this vehicle as long as possible.

3.3 THE PRODUCT AND QUOTIENT RULES 4. (a) Fin the slope of the graph of f() = e at the point where it crosses the -ais. (b) Fin the equation of the tangent line to the curve at this point. (c) Fin the equation of the line perpenicular to the tangent line at this point. (This is the normal line.) 4. Fin the value of c in Figure 3., where the line l tangent to the graph of y = at (0, ) intersects the -ais. y y = l 43. Fin the quaratic polynomial g() = a + b + c which best fits the function f() = e at = 0, in the sense that g(0) = f(0), an g (0) = f (0), an g (0) = f (0). Using a computer or calculator, sketch graphs of f an g on the same aes. What o you notice? 44. Using the equation of the tangent line to the graph of e at = 0, show that e + for all values of. A sketch may be helpful. 45. Fin all solutions of the equation c Figure 3. =. How o you know that you foun all solutions? 46. For what value(s) of a are y = a an y = + tangent at = 0? Eplain. 3.3 THE PRODUCT AND QUOTIENT RULES The Prouct Rule We now know how to fin erivatives of powers an eponentials, an of sums an constant multiples of functions. This section shows how to fin the erivatives of proucts an quotients. Using Notation To epress the ifference quotients of general functions, some aitional notation is helpful. We write f, rea elta f, for a small change in the value of f, f = f( + h) f(). In this notation, the erivative is the limit of the ratio f/h: f f () = lim h 0 h. Suppose we know the erivatives of f() an g() an want to calculate the erivative of the prouct, f()g(). The erivative of the prouct is calculate by taking the limit, namely, [f()g()] f( + h)g( + h) f()g() = lim. h 0 h To picture the quantity f( + h)g( + h) f()g(), imagine the rectangle with sies f( + h) an g( + h) in Figure 3.3, where f = f( + h) f() an g = g( + h) g(). Then Now ivie by h: f( + h)g( + h) f()g() = (Area of whole rectangle) (Unshae area) = Area of the three shae rectangles = f g() + f() g + f g. f( + h)g( + h) f()g() h = f h g g() + f() h f g +. h

Chapter Three SHORT-CUTS TO DIFFERENTIATION Area = f g() Area = f g f f( + h) f() Area = f() g() Area = f() g g() g g( + h) Figure 3.3: Illustration for the prouct rule (with f, g positive) To evaluate the limit as h 0, we eamine the three terms on the right separately. Notice that f lim h 0 h g() = f ()g() an lim f() g h 0 h = f()g (). In the thir term we multiply the top an bottom by h to get f h f g f lim = lim h 0 h h 0 h g h h = lim f h 0 h Therefore, we conclue that f( + h)g( + h) f()g() lim = lim h 0 h h 0 Thus we have prove the following rule: ( f h g h h. Then, lim g h 0 h lim h = f () g () 0 = 0. h 0 g f g g() + f() + h h f g = lim g() + lim f() h 0 h h 0 h + lim h 0 = f ()g() + f()g (). ) f g h Theorem 3.3: The Prouct Rule If u = f() an v = g() are ifferentiable, then (fg) = f g + fg. The prouct rule can also be written (uv) = u v + u v. In wors: The erivative of a prouct is the erivative of the first times the secon plus the first times the erivative of the secon. Another justification of the prouct rule is given in Problem 6 on page 55. Eample Differentiate (a) e, (b) (3 + 5)e, (c) e.

(a) ( e ) ( ( ) = ) e + (e ) = e + e = ( + )e. (b) ((3 + 5)e ( ) (3 ) + 5) = e + (3 + 5) (e ) = (6 + 5)e + (3 + 5)e = (3 + + 5)e. 3.3 THE PRODUCT AND QUOTIENT RULES 3 (c) First we must write e as the prouct e : ( ) e = ( e ( ) ( ) ) = e + (e ) = 3 e + e = ( 3 + )e. The Quotient Rule Suppose we want to ifferentiate a function of the form Q() = f()/g(). (Of course, we have to avoi points where g() = 0.) We want a formula for Q in terms of f an g. Assuming that Q() is ifferentiable, we can use the prouct rule on f() = Q()g(): Solving for Q () gives f () = Q ()g() + Q()g () = Q ()g() + f() g() g (). f () f() Q g() g () () =. g() Multiplying the top an bottom by g() to simplify gives ( ) f() = f ()g() f()g () g() (g()). So we have the following rule: Theorem 3.4: The Quotient Rule If u = f() an v = g() are ifferentiable, then or equivalently, ( ) f = f g fg g g, ( u ) = v u v u v v. In wors: The erivative of a quotient is the erivative of the numerator times the enominator minus the numerator times the erivative of the enominator, all over the enominator square. The metho in Eample 6 on page 30 can be use to eplain why Q() must be ifferentiable.

4 Chapter Three SHORT-CUTS TO DIFFERENTIATION Eample Differentiate (a) (a) 5 3 +, (b) + e, (c) e. ( ) ( ) 5 (5 ) ( 3 + ) 5 (3 + ) 3 = + ( 3 + ) = 0(3 + ) 5 (3 ) ( 3 + ) = 54 + 0 ( 3 + ). (b) ( ) ( ) () ( + e ) ( + e ) + e = ( + e ) = 0( + e ) (0 + e ) ( + e ) = e ( + e ). (c) This is the same as part (c) of Eample, but this time we o it by the quotient rule. ( (e ( ) ) ( ) ( e e ) ) = ( ) = e e () 4 ( = e ) ( ) 4 = e 3. This is, in fact, the same answer as before, although it looks ifferent. Can you show that it is the same? Eercises an Problems for Section 3.3 Eercises. If f() = ( 3 + 5), fin f () two ways: by using the prouct rule an by multiplying out before taking the erivative. Do you get the same result? Shoul you?. If f() = 3, fin f () two ways: by using the prouct rule an by using the fact that 3 = 6. Do you get the same result? For Eercises 3 30, fin the erivative. It may be to your avantage to simplify first. Assume that a, b, c, an k are constants. 3. f() = e 4. y = 5. y = 6. y = (t + 3)e t 5. q(r) = 3r 5r + 7. z = 3t + 5t + 9. z = t + 3t + t +. w = y3 6y + 7y y 6. g(t) = t 4 t + 4 8. z = t + 5t + t + 3 0. f() = + 3 t. y = t + 3. f(z) = z + 4 4. g(t) = z 3 + t 7. f() = ( )3 8. z = (s s)(s + s) 9. f(y) = 4 y ( y ) 0. y = (t 3 7t + )e t 5. h(r) = r r + 6. f(z) = 3z 5z + 7z. f() = e. g() = 5 3. y = t + t 4. g(w) = w3. 5 w e 7. w() = 7e 8. h(p) = + p 3 + p 9. f() = + 30. f() = a + b + 3 + 4 c + k

3.3 THE PRODUCT AND QUOTIENT RULES 5 Problems In Problems 3 33, use Figure 3.4 to estimate the erivative, or state that it oes not eist. The graph of f() has a sharp corner at =. 4 3 y f() g() 3 4 Figure 3.4 3. Let h() = f() g(). Fin: (a) h () (b) h () (c) h (3) 3. Let k() = (f())/(g()). Fin: (a) k () (b) k () (c) k (3) 33. Let j() = (g())/(f()). Fin: (a) j () (b) j () (c) j (3) For Problems 34 39, let h() = f() g(), an k() = f()/g(), an l() = g()/f(). Use Figure 3.5 to estimate the erivatives. 44. For what intervals is f() = e concave up? 45. For what intervals is g() = concave own? + 46. Fin the equation of the tangent line to the graph of f() = 5 at the point at which = 0. + 47. (a) Differentiate y = e, y = e e, an y =. 3 (b) What o you anticipate the erivative of y = e n will be? Confirm your guess. 48. Using the prouct rule an the fact that () that ( ) = an (3 ) = 3. 49. Use the prouct rule to show that [Hint: Write = / /.] =, show (/ ) =. / 50. Suppose f an g are ifferentiable functions with the values shown in the following table. For each of the following functions h, fin h (). (a) h() = f() + g() (b) h() = f()g() (c) h() = f() g() 3 3 3 3 3 f() g() 3 3 3 Figure 3.5 f() g() f () g () 3 4 5 5. If H(3) =, H (3) = 3, F (3) = 5, F (3) = 4, fin: (a) G (3) if G(z) = F (z) H(z) (b) G (3) if G(w) = F (w)/h(w) 5. Fin a possible formula for a function y = f() such that f () = 0 9 e + 0 e. 34. h () 35. k () 36. h () 37. k () 38. l () 39. l () 40. If f() = (3 + 8)( 5), fin f () an f (). 4. Differentiate f(t) = e t by writing it as f(t) = e t. 4. Differentiate f() = e by writing it as f() = e e. 43. Differentiate f() = e 3 by writing it as f() = e e an using the result of Problem 4. 53. The quantity, q, of a certain skateboar sol epens on the selling price, p, in ollars, so we write q = f(p). You are given that f(40) = 5,000 an f (40) = 00. (a) What o f(40) = 5,000 an f (40) = 00 tell you about the sales of skateboars? (b) The total revenue, R, earne by the sale of skateboars is given by R = pq. Fin R p. p=40 (c) What is the sign of R? If the skateboars p p=40 are currently selling for $40, what happens to revenue if the price is increase to $4?

6 Chapter Three SHORT-CUTS TO DIFFERENTIATION 54. When an electric current passes through two resistors with resistance r an r, connecte in parallel, the combine resistance, R, can be calculate from the equation R = r + r. Fin the rate at which the combine resistance changes with respect to changes in r. Assume that r is constant. 55. A museum has ecie to sell one of its paintings an to invest the procees. If the picture is sol between the years 000 an 00 an the money from the sale is investe in a bank account earning 5% interest per year compoune annually, then B(t), the balance in the year 00, epens on the year, t, in which the painting is sol an the sale price P (t). If t is measure from the year 000 so that 0 < t < 0 then B(t) = P (t)(.05) 0 t. (a) Eplain why B(t) is given by this formula. (b) Show that the formula for B(t) is equivalent to B(t) = (.05) 0 P (t) (.05). t (c) Fin B (0), given that P (0) = 50,000 an P (0) = 5000. 56. Let f(v) be the gas consumption (in liters/km) of a car going at velocity v (in km/hr). In other wors, f(v) tells you how many liters of gas the car uses to go one kilometer, if it is going at velocity v. You are tol that f(80) = 0.05 an f (80) = 0.0005. (a) Let g(v) be the istance the same car goes on one liter of gas at velocity v. What is the relationship between f(v) an g(v)? Fin g(80) an g (80). (b) Let h(v) be the gas consumption in liters per hour. In other wors, h(v) tells you how many liters of gas the car uses in one hour if it is going at velocity v. What is the relationship between h(v) an f(v)? Fin h(80) an h (80). (c) How woul you eplain the practical meaning of the values of these functions an their erivatives to a river who knows no calculus? 57. The function f() = e has the properties f () = f() an f(0) =. Eplain why f() is the only function with both these properties. [Hint: Assume g () = g(), an g(0) =, for some function g(). Define h() = g()/e, an compute h (). Then use the fact that a function with a erivative of 0 must be a constant function.] 58. Fin f () for the following functions with the prouct rule, rather than by multiplying out. (a) f() = ( )( ). (b) f() = ( )( )( 3). (c) f() = ( )( )( 3)( 4). 59. Use the answer from Problem 58 to guess f () for the following function: f() = ( r )( r )( r 3) ( r n) where r, r,..., r n are any real numbers. 60. (a) Provie a three imensional analogue for the geometrical emonstration of the formula for the erivative of a prouct, given in Figure 3.3 on page. In other wors, fin a formula for the erivative of F () G() H() using Figure 3.6. (b) Confirm your results by writing F () G() H() as [F () G()] H() an using the prouct rule twice. (c) Generalize your result to n functions: what is the erivative of f () f () f 3() f n()? Figure 3.6: A graphical representation of the 3-imensional prouct rule 6. If P () = ( a) Q(), where Q() is a polynomial, we call = a a ouble zero of the polynomial P (). (a) If = a is a ouble zero of a polynomial P (), show that P (a) = P (a) = 0. (b) If P () is a polynomial an P (a) = P (a) = 0, show that = a is a ouble zero of P ().

3.4 THE CHAIN RULE 7 3.4 THE CHAIN RULE Composite functions such as sin(3t) or e occur frequently in practice. In this section we see how to ifferentiate such functions. The Derivative of a Composition of Functions Suppose f(g()) is a composite function, with f being the outsie function an g being the insie. Let us write z = g() an y = f(z), so y = f(g()). Then a small change in, calle, generates a small change in z, calle z. In turn, z generates a small change in y calle y. Provie an z are not zero, we can say: Since y = we have: lim 0 The Chain Rule y = y z z. y, this suggests that in the limit as, y, an z get smaller an smaller, y = y z z. Since y z = f (z) an z = g (), we can also write f(g()) = f (z) g (). Substituting z = g(), we can rewrite this as follows: Theorem 3.5: The Chain Rule If f an g are ifferentiable functions, then f(g()) = f (g()) g (). In wors: The erivative of a composite function is the prouct of the erivatives of the outsie an insie functions. The erivative of the outsie function must be evaluate at the insie function. A justification of the chain rule is given in Problem 7 on page 55. Eample The length, L, in cm, of a steel bar epens on the air temperature, H C, an the temperature H epens on time, t, measure in hours. If the length increases by cm for every egree increase in temperature, an the temperature is increasing at 3 C per hour, how fast is the length of the bar increasing? What are the units for your answer? We epect the rate at which the length is increasing to be in cm/hr. We are tol that Rate length increasing with respect to temperature = L H = cm/ C Rate temperature increasing with respect to time = H t = 3 C/hr.

8 Chapter Three SHORT-CUTS TO DIFFERENTIATION We want to calculate the rate at which the length is increasing with respect to time, or L/t. We think of L as a function of H, an H as a function of t. By the chain rule we know that L t = L H H ( t = Thus, the length is increasing at 6 cm/hr. cm C ) ( ) C 3 = 6 cm/hr. hr Eample shows us how to interpret the chain rule in practical terms. The net eamples show how the chain rule is use to compute erivatives of functions given by formulas. Eample Fin the erivatives of the following functions: (a) ( + ) 00 (b) 3 + 5 (c) + 4 () e 3 (e) e (a) Here z = g() = + is the insie function; f(z) = z 00 is the outsie function. Now g () = an f (z) = 00z 99, so (( + ) 00 ) = 00z 99 = 00( + ) 99 = 00( + ) 99. (b) Here z = g() = 3 + 5 an f(z) = z, so g () = 6 + 5 an f (z) = z. Hence ( 3 + 5 ) = z (6 + 5) = (6 + 5). 3 + 5 (c) Let z = g() = + 4 an f(z) = /z, so g () = + 4 3 an f (z) = z = z. Then ( + 4 ) = z ( + 43 ) = + 43 ( + 4 ). We coul have one this problem using the quotient rule. Try it an see that you get the same answer! () Let z = g() = 3 an f(z) = e z. Then g () = 3 an f (z) = e z, so ( e 3 ) = e z 3 = 3e 3. (e) To figure out which is the insie function an which is the outsie, notice that to evaluate e we first evaluate an then take e to that power. This tells us that the insie function is z = g() = an the outsie function is f(z) = e z. Therefore, g () =, an f (z) = e z, giving (e ) = e z = e = e. To ifferentiate a complicate function, we may have to use the chain rule more than once, as in the following eample. Eample 3 Differentiate: (a) e /7 + 5 (b) ( e t ) 9

3.4 THE CHAIN RULE 9 (a) Let z = g() = e /7 + 5 be the insie function; let f(z) = z be the outsie function. Now f (z) = z, but we nee the chain rule to fin g (). We choose new insie an outsie functions whose composition is g(). Let u = h() = /7 an k(u) = e u +5 so g() = k(h()) = e /7 +5. Then h () = /7 an k (u) = e u, so ( g () = e u ) = 7 7 e /7. Using the chain rule to combine the erivatives of f(z) an g(), we have ( e /7 + 5) = ( 7 ) e /7 z e /7 = 4 e /7 + 5. (b) Let z = g(t) = e t be the insie function an f(z) = z 9 be the outsie function. Then f (z) = 9z 8 but we nee the chain rule to ifferentiate g(t). Now we choose u = h(t) = t an k(u) = e u, so g(t) = k(h(t)). Then h (t) = t / = t an k (u) = e u, so t g (t) = e u = e. t t Using the chain rule to combine the erivatives of f(z) an g(t), we have ) t t ( t e ) 9 = 9z ( 8 e = 9 e ( e t ) 8. t t It is often faster to use the chain rule without introucing new variables, as in the following eamples. Eample 4 Differentiate sin(e 3+ ). The chain rule is neee four times: ( sin(e ) 3+ ) = cos(e 3+ ) (e 3+ ) = cos(e 3+ ) e 3+ ( 3 + ) = cos(e 3+ ) e 3+ (3 + ) / = cos(e 3+ ) e 3+ (3 + ) /. ( 3 + ) Eample 5 Fin the erivative of e by the chain rule an by the prouct rule. Using the chain rule, we have (e ) = e () = e = e. Using the prouct rule, we write e = e e. Then (e ) = ( ) ( ) (e e ) = (e ) e + e (e ) = e e + e e = e.

30 Chapter Three SHORT-CUTS TO DIFFERENTIATION Using the Prouct an Chain Rules to Differentiate a Quotient If you prefer, you can ifferentiate a quotient by the prouct an chain rules, instea of by the quotient rule. The resulting formulas may look ifferent, but they will be equivalent. Eample 6 Fin k () if k() = +. One way is to use the quotient rule: k () = ( + ) () ( + ) = ( + ). Alternatively, we can write the original function as a prouct, an use the prouct rule: k() = + = ( + ), k () = ( + ) + Now use the chain rule to ifferentiate ( + ), giving Therefore, [ ( + ) ]. [ ( + ) ] = ( + ) = ( + ). k () = + + ( + ) = + ( + ). Putting these two fractions over a common enominator gives the same answer as the quotient rule. Eercises an Problems for Section 3.4 Eercises Fin the erivatives of the functions in Eercises 5. Assume that a, b, c, an k are constants.. f() = ( + ) 99. w = (t + ) 00 3. w = (t 3 + ) 00 4. (4 + ) 7 5. f() = 6. e + 7. w = ( t + ) 00 8. h(w) = (w 4 w) 5 9. w(r) = r 4 + 0. k() = ( 3 + e ) 4. f() = e ( + 5 ). f(t) = e 3t 3. g() = e π 4. f(θ) = θ 5. y = π (+) 6. g() = 3 (+7) 7. f(t) = te 5 t 8. p(t) = e 4t+ 9. v(t) = t e ct 0. g(t) = e (+3t). y = e 3w/. y = e 4t 3. y = s 3 + 4. w = e s 5. y = te t 6. f(z) = ze z 7. z() = 3 + 5 8. z = 5t 3 9. w = ( 5 ) 3 30. f(y) = 0 (5 y) 3. f(z) = z z e z 3. y = z

33. y = ( + 3 35. y = e + 37. h(z) = ) 34. h() = + 9 36. y = e 3 + ( b a + z ) 4 38. f(z) = 39. w = (t +3t)( e t ) 40. h() = e3 + 3 (e z + ) 3.4 THE CHAIN RULE 3 43. f(w) = (5w + 3)e w 44. f(θ) = (e θ + e θ ) 45. y = e 3t + 5 46. z = (te 3t + e 5t ) 9 47. f(y) = e e(y ) 48. f(t) = e et 49. f() = (a + b) 3 50. f(t) = ae bt 5. f() = ae b 5. g(α) = e αe α 4. f() = 6e 5 + e 4. f() = e ( ) Problems In Problems 53 56, use Figure 3.4 to estimate the erivative, or state it oes not eist. The graph of f() has a sharp corner at =. 4 y f() 6. Fin the equation of the tangent line to f() = ( ) 3 at the point where =. 6. Fin the equation of the line tangent to y = f() at =, where f() is the function in Problem 4. 3 g() 3 4 Figure 3.7 63. For what values of is the graph of y = e concave own? 64. For what intervals is f() = e concave own? 65. Suppose f() = ( + ) 0 (3 ) 7. Fin a formula for f (). Decie on a reasonable way to simplify your result, an fin a formula for f (). 53. Let h() = f(g()). Fin: (a) h () (b) h () (c) h (3) 54. Let u() = g(f()). Fin: (a) u () (b) u () (c) u (3) 55. Let v() = f(f()). Fin: (a) v () (b) v () (c) v (3) 56. Let w() = g(g()). Fin: (a) w () (b) w () (c) w (3) In Problems 57 60, use Figure 3.8 to evaluate the erivative. 80 80 g() f() 66. At any time, t, a population, P (t), is growing at a rate proportional to the population at that moment. (a) Using erivatives, write an equation representing the growth of the population. Let k be the constant of proportionality. (b) Show that the function P (t) = Ae kt satisfies the equation in part (a) for any constant A. 67. Fin the mean an variance of the normal istribution of statistics using the information in parts (a) an (b) with m(t) = e µt+σ t /. (a) Mean = m (0) (b) Variance = m (0) (m (0)) 57. 59. 0 80 0 Figure 3.8 80 f(g()) =30 58. f(g()) =70 g(f()) =30 60. g(f()) =70 68. If the erivative of y = k() equals when =, what is the erivative of (a) k() when =? (b) k( ( + ) ) when = 0? (c) k 4 when = 4?

3 Chapter Three SHORT-CUTS TO DIFFERENTIATION 69. Is = 3 t + 5 a solution to the equation 3 t =? Why or why not? 70. Fin a possible formula for a function m() such that m () = 5 e (6). 7. Given F () =, F () = 5, F (4) = 3, F (4) = 7 an G(4) =, G (4) = 6, G(3) = 4, G (3) = 8, fin: (a) H(4) if H() = F (G()) (b) H (4) if H() = F (G()) (c) H(4) if H() = G(F ()) () H (4) if H() = G(F ()) (e) H (4) if H() = F ()/G() 7. Given y = f() with f() = 4 an f () = 3, fin 79. If you invest P ollars in a bank account at an annual interest rate of r%, then after t years you will have B ollars, where ( B = P + r ) t. 00 (a) Fin B/t, assuming P an r are constant. In terms of money, what oes B/t represent? (b) Fin B/r, assuming P an t are constant. In terms of money, what oes B/r represent? 80. The theory of relativity preicts that an object whose mass is m 0 when it is at rest will appear heavier when moving at spees near the spee of light. When the object is moving at spee v, its mass m is given by (a) g () if g() = f(). (b) h () if h() = f( ). m = m 0 (v /c ), where c is the spee of light. In Problems 73 77, use Figures 3.9 an 3.0 an h() = f(g()). b c b a a b c b Figure 3.9 b c b a a b c b 73. Evaluate h(0) an h (0). Figure 3.0 f() g() 74. At = c, is h positive, negative, or zero? Increasing or ecreasing? 75. At = a, is h increasing or ecreasing? 76. What are the signs of h() an h ()? 77. How oes the value of h() change on the interval < < b? 78. The worl s population is about f(t) = 6e 0.03t billion, where t is time in years since 999. Fin f(0), f (0), f(0), an f (0). Using units, interpret your answers in terms of population. (a) Fin m/v. (b) In terms of physics, what oes m/v tell you? 8. The charge, Q, on a capacitor which starts ischarging at time t = 0 is given by { Q0 for t 0 Q = Q 0e t/rc for t > 0, where R an C are positive constants epening on the circuit an Q 0 is the charge at t = 0, where Q 0 0. The current, I, flowing in the circuit is given by I = Q/t. (a) Fin the current I for t < 0 an for t > 0. (b) Is it possible to efine I at t = 0? (c) Is the function Q ifferentiable at t = 0? 8. A particle is moving on the -ais, where is in centimeters. Its velocity, v, in cm/sec, when it is at the point with coorinate is given by v = + 3. Fin the acceleration of the particle when it is at the point =. Give units in your answer. 83. A function f is sai to have a zero of multiplicity m at = a if f() = ( a) m h(), with h(a) 0. Eplain why a function having a zero of multiplicity m at = a satisfies f (p) (a) = 0, for p =,,... m. [Note: f (p) is the p th erivative.] 84. If f( + ) = +, what is f ()? 85. If t G(a bt) = H(a bt), what is G (t)? (Give your answer in terms of H(t).) 86. If t (tf(t)) = + f(t), what is f (t)? 87. If f(e ) = e, what is f()?

3.5 THE TRIGONOMETRIC FUNCTIONS 33 3.5 THE TRIGONOMETRIC FUNCTIONS Derivatives of the Sine an Cosine Since the sine an cosine functions are perioic, their erivatives must be perioic also. (Why?) Let s look at the graph of f() = sin in Figure 3. an estimate the erivative function graphically. f() = sin π π π π 3π 4π Figure 3.: The sine function First we might ask where the erivative is zero. (At = ±π/, ±3π/, ±5π/, etc.) Then ask where the erivative is positive an where it is negative. (Positive for π/ < < π/; negative for π/ < < 3π/, etc.) Since the largest positive slopes are at = 0, π, an so on, an the largest negative slopes are at = π, 3π, an so on, we get something like the graph in Figure 3.. f () π π π π 3π 4π Figure 3.: Derivative of f() = sin The graph of the erivative in Figure 3. looks suspiciously like the graph of the cosine function. This might lea us to conjecture, quite correctly, that the erivative of the sine is the cosine. Of course, we cannot be sure, just from the graphs, that the erivative of the sine really is the cosine. However, for now we ll assume that the erivative of the sine is the cosine an confirm the result at the en of the section. One thing we can o now is to check that the erivative function in Figure 3. has amplitue (as it ought to if it is the cosine). That means we have to convince ourselves that the erivative of f() = sin is when = 0. The net eample suggests that this is true when is in raians. Eample Using a calculator set in raians, estimate the erivative of f() = sin at = 0. Since f() = sin, Table 3.5 f sin(0 + h) sin 0 sin h (0) = lim = lim h 0 h h 0 h. Table 3.5 contains values of (sin h)/h which suggest that this limit is, so we estimate f (0) = lim h 0 sin h h =. h (raians) 0. 0.0 0.00 0.000 0.000 0.00 0.0 0. (sin h)/h 0.99833 0.99998.0000.0000.0000.0000 0.99998 0.99833 Warning: It is important to notice that in the previous eample h was in raians; any conclusions we have rawn about the erivative of sin are vali only when is in raians. If you fin the erivative with h in egrees, you get a ifferent result. Eample Starting with the graph of the cosine function, sketch a graph of its erivative.

34 Chapter Three SHORT-CUTS TO DIFFERENTIATION The graph of g() = cos is in Figure 3.3(a). Its erivative is 0 at = 0, ±π, ±π, an so on; it is positive for π < < 0, π < < π, an so on; an it is negative for 0 < < π, π < < 3π, an so on. The erivative is in Figure 3.3(b). (a) g() = cos (b) π π π π 3π 4π π π g () π π 3π 4π Figure 3.3: g() = cos an its erivative, g () As we i with the sine, we use the graphs to make a conjecture. The erivative of the cosine in Figure 3.3(b) looks eactly like the graph of sine, ecept reflecte about the -ais. But how can we be sure that the erivative is sin? Eample 3 Use the relation (sin ) = cos to show that (cos ) = sin. Since the cosine function is the sine function shifte to the left by π/ (that is, cos = sin ( + π/)), we epect the erivative of the cosine to be the erivative of the sine, shifte to the left by π/. Differentiating using the chain rule: ( ( (cos ) = sin + π )) ( = cos + π ). But cos( + π/) is the cosine shifte to the left by π/, which gives a sine curve reflecte about the -ais. So we have ( (cos ) = cos + π ) = sin. At the en of this section an in Problems 5 an 5, we show that our conjectures for the erivatives of sin an cos are correct. Thus, we have: For in raians, (sin ) = cos an (cos ) = sin. Eample 4 Differentiate (a) sin(3θ), (b) cos, (c) cos( ), () e sin t. Use the chain rule: (a) (b) (c) () θ ( sin(3θ)) = θ (sin(3θ)) = (cos(3θ)) (3θ) = (cos(3θ))3 = 6 cos(3θ). θ (cos ) = ( (cos ) ) = (cos ) ( cos( ) ) = sin( ) ( ) = sin( ). t (e sin t ) = e sin t t ( sin t) = (cos t)e sin t. (cos ) = (cos )( sin ) = cos sin.

Derivative of the Tangent Function 3.5 THE TRIGONOMETRIC FUNCTIONS 35 Since tan = sin / cos, we ifferentiate tan using the quotient rule. Writing (sin ) for (sin )/, we have: (tan ) = ( sin cos ) = (sin ) (cos ) (sin )(cos ) cos = cos + sin cos = cos. For in raians, (tan ) = cos. The graphs of f() = tan an f () = / cos are in Figure 3.4. Is it reasonable that f is always positive? Are the asymptotes of f where we epect? π f () = cos π sin( + h) h A Q + h P sin f() = tan 0 cos Figure 3.4: The function tan an its erivative Figure 3.5: Unit circle showing sin( + h) an sin Eample 5 Differentiate (a) tan(3t), (b) tan( θ), (c) + tan t tan t. (a) Use the chain rule: (b) Use the chain rule: (c) Use the quotient rule: t t ( tan(3t)) = cos (3t) θ (tan( θ)) = cos ( θ) ( ) + tan t = tan t = = ( ( + tan t) t t (3t) = 6 cos (3t). ( θ) = θ cos ( θ). ) ( tan t) ( tan t) ( + tan t) t ( tan t) cos ( tan t) ( + tan t) t ( tan t) cos t ( tan t). ( cos t )

36 Chapter Three SHORT-CUTS TO DIFFERENTIATION Informal Justification of (sin ) = cos Consier the unit circle in Figure 3.5. To fin the erivative of sin, we nee to estimate sin( + h) sin. h In Figure 3.5, the quantity sin( + h) sin is represente by the length QA. The arc QP is of length h, so sin( + h) sin h = QA Arc QP. Now, if h is small, QAP is approimately a right triangle because the arc QP is almost a straight line. Furthermore, using geometry, we can show that angle AQP = + h. For small h, we have sin( + h) sin h As h 0, the approimation gets better, so = QA Arc QP cos( + h). sin( + h) sin (sin ) = lim = cos. h 0 h Other erivations of this result are given in Problems 5 an 5 on page 37. Eercises an Problems for Section 3.5 Eercises. Construct a table of values for cos, = 0, 0., 0.,..., 0.6. Using the ifference quotient, estimate the erivative at these points (use h = 0.00), an compare it with ( sin ). Fin the erivatives of the functions in Eercises 39. Assume a is a constant.. r(θ) = sin θ + cos θ 3. s(θ) = cos θ sin θ 4. z = cos(4θ) 5. f() = sin(3) 6. g() = sin( 3). 7. R() = 0 3 cos(π) 8. g(θ) = sin (θ) πθ 9. f() = cos 0. k() = (sin()) 3. f() = sin(3). y = e θ sin(θ) 3. f() = e sin 4. z = sin t 5. y = sin 5 θ 6. g(z) = tan(e z ) 7. z = tan(e 3θ ) 8. w = e sin θ 9. h(t) = t cos t + tan t 30. f(α) = cos α + 3 sin α 3. k(α) = sin 5 α cos 3 α 3. f(θ) = θ 3 cos θ 33. y = cos w + cos(w ) 34. y = sin(sin + cos ) 35. y = sin() sin(3) 0. w = sin(e t ). f() = e cos. f(y) = e sin y 3. z = θe cos θ 4. R(θ) = e sin(3θ) 5. g(θ) = sin(tan θ) 6. w() = tan( ) 7. f() = cos 8. f() = cos(sin ) 9. f() = tan(sin ) 36. t(θ) = cos θ sin θ 38. r(y) = y cos y + a 37. f() = sin cos 39. G() = sin + cos + Problems 40. Is the graph of y = sin( 4 ) increasing or ecreasing when = 0? Is it concave up or concave own? 4. Fin the fiftieth erivative of y = cos. 4. Fin a possible formula for the function q() such that q () = e sin e cos (sin ).