Chapter 2: The Derivative Summary: Chapter 2 builds upon the ideas of limits and continuity discussed in the previous chapter. By using limits, the instantaneous rate at which a function changes with respect to its inputs can be investigated. This leads to the ability to find tangent lines to a function at a given point. These two concepts are encompassed by the derivative which is a new function that can be used to find the slope of a curve at a given point. A progression from continuous functions to those which have tangent lines to those which are differentiable can be seen. The concept of the derivative also provides some valuable tools for determining the shape of a given function and how a function behaves. These tools are a direct result of a derivative s ability to describe the slope of a function at a point. Many tools for finding the derivatives of functions are also discussed in this chapter so that derivatives of more complicated functions can be found. In the latter part of this chapter, it is discussed whether or not the slope of a general curve can be found if the curve does not have an eplicit function representation. The process of implicit differentiation allows dy/d to be found even when an epression for y = f() is not eplicitly known. Implicit differentiation can then be applied to find a relationship between the rate of change of variables in many situations. Finally, the ideas of tangent lines are more thoroughly described as linear approimations. Linear approimations then lead to the idea of differentials which allow for the estimation of a small change in a function given a small change in its inputs. OBJECTIVES: After reading and working through this chapter you should be able to do the following:. Understand the definition of the derivative and its implications ( 2.,2.2) 2. Be able to read and understand different notations for the derivative ( 2.2) 3. Find derivatives of simple functions like polynomials ( 2.3) or trigonometric functions ( 2.5) 4. Use the product and/or quotient rules to find derivatives of functions involving products or quotients of other functions ( 2.4) 5. Use the chain rule (either form) to find the derivatives of functions which involve the compositions of other functions ( 2.6) 29
30 6. Find the slope of a curve that is defined implicitly ( 2.7). 7. Use derivatives to relate the rates of different quantities that depend upon the same parameter such as time ( 2.8). 8. Approimate functions using linearizations or differentials ( 2.9). 2. Tangent Lines and Rates of Change PURPOSE: To use limits and continuity to investigate the slope of a function at a point. slope of a secant line instantaneous rate of change This section attempts to bridge the gap between limits and continuity and the slope of a graph. The process is one where the slope of a secant line can be seen as the average rate of change of a function. This becomes the instantaneous rate of change as the two points defining the secant line approach each other in the limit. The slope of the tangent line at a point 0 is defined as a limit: m tan = lim h 0 f( 0 + h) f( 0 ) h This slope can then be used to construct the tangent line to a function f() at the point ( 0, f( 0 )): y f( 0 ) = m tan ( 0 ) This is nothing more than the point-slope form of a line. In this case, the slope m tan is determined by a limiting process. The big idea in this section (and the net) is the equivalence of the following three things:. The slope of a tangent line to y = f() at the point = 0. 2. The instantaneous rate of change of y = f() at the point = 0. 3. The derivative of the function y = f() at the point = 0. In terms of rectilinear motion and velocity, this section shows the specific eam- ple where instantaneous velocity is the instantaneous rate of change of the position function of a particle. Be careful to make the distinction between displacement and distance (displacement implies a direction) and also the difference between velocity and speed (velocity implies a direction). rectilinear motion velocity Checklist of Key Ideas: finding the equation of a tangent line to a function speed or velocity of a particle
3 rectilinear motion graphing position versus time position function of a particle displacement and distance average velocity and average speed instantaneous velocity and speed rate of change of y with respect to average and instantaneous rate of change of y with respect to units of rate of change of y with respect to 2.2 The Derivative Function PURPOSE: To define the derivative, discuss its meaning and introduce some notation for the derivative. In this section, the limit that defines the slope of the tangent line at a point is referred to more generally as the derivative at a point. The derivative function is defined as: f () = lim h 0 f(+h) f() h The concepts of the previous section are etended in this section by the derivative. The instantaneous rate of change of a function at any point (where this limit eists) is the derivative. Similarly, the derivative function can be used to forecast the behavior of f() since it describes the slope of f() (or equivalently the slope of the tangent line) at any given point. definition of derivative The biggest ideas to come out of this section are the relationships between limits, continuity, tangent lines, and derivatives. IDEA: Continuity, differentiability, and tangent lines are related in the following ways:. If a function f() is differentiable at a point, then it must have a tangent line at that point. 2. If a function f() has a tangent line at a point, then it must be continuous at that point. 3. If a function f() is continuous at a point, then its limit must eist at that point. Notice that the relationships described above do not go the other direction. Here are some countereamples:
32. If a function has a limit at a point, it does not have to be continuous at a point. Consider y = ( 2 4)/(+2) which has a hole at = 2 and so is not continuous there. 2. If a function is continuous, it does not have to have a tangent line. Consider y = at = 0 where it has a corner. 3. If a function has a tangent line, it does not have to be differentiable. Consider y = /3 which has a vertical tangent line at = 0. the derivative f () and the graph of f() Another key concept in this section is the relationship that eists between the graphs of the function f() and its derivative f (). As you continue on in this chapter, always try to be aware of how the graph of f () relates to the slope of the function f(). Thinking of f () as the slope of the function f() will establish a natural connection between the behavior of f() and f (). Checklist of Key Ideas: derivative of f with respect to f () and the slope of f() graphing f and f together f (t) and instantaneous velocity v(t) when a function does not have a derivative at a point differentiable at a point and on an interval relationship between derivatives, continuity, and tangent lines derivatives of piecewise functions and one-sided derivatives different notations for derivatives evaluating the derivative at a point derivatives and increment notation 2.3 Introduction to Techniques of Differentiation PURPOSE: To introduce basic rules for calculating derivatives. Evaluating derivatives by using the definition (i.e. the limit of the difference quotient from the previous two sections) can be tedious and may require some algebraic manipulation. Several basic derivatives are introduced in this section. First, the derivatives of constant functions and powers of are developed. Net, since the derivative is really a limit, it follows that derivatives can be added and subtracted and also multiplied by constants with ease. Unfortunately, the same
33 straightforward results do not hold when functions are multiplied and divided (see the net section). The most important result to come out of this section is the ability to calculate the derivative of a polynomial or the derivative of a sum of functions. Essentially, a sum of functions can be differentiated by taking the derivative of each function separately and then adding the derivatives back together. Lastly, the idea of higher order derivatives is introduced. If we can take a derivative of a function, then why not take the derivative of the derivative? In this way, so-called second order and higher derivatives are introduced. Order here simply refers to how many times a derivative is taken beginning with the original function. Pay close attention also to the different notations that are introduced for the derivative. There are (perhaps confusingly) several ways to say the same thing with derivatives. To this end, it may be useful to write these things down on a note card and have them close at hand as you work on problems in the tet. derivative of n derivative of c f() derivative of f() + g() write new derivative rules on notecards Checklist of Key Ideas: derivatives of constants, the power rule derivatives of constant multiples, sums, and differences derivatives of polynomials finding horizontal tangent lines higher derivatives and order of a derivative notation of higher order derivatives 2.4 The Product and Quotient Rules PURPOSE: To develop rules for differentiating products or quotients of functions. As indicated in the previous section, the derivatives of products of functions and quotients of functions are not straightforward. This is primarily because the limit definition involves a difference of functions. When these limits are multiplied together, it is not generally true that these differences will behave nicely. Some simple eamples can show that we may not simply multiply derivatives together. For eample, consider f() = and g() = 2 (see p.42-43 in the tet). Since y = f()g() = 3 we know that its derivative is y = 3 2. Notice that this is not equivalent to f ()g (). These same two functions can be used to show that if y = g()/ f() that the derivative is not simply g ()/ f (). Can you see why? Write the product and quotient rules on note cards for easy reference. They are used quite frequently.
34 This section introduces two rules that are frequently used throughout calculus: the product rule and the quotient rule for calculating derivatives of y = f()g() and y = f()/g() respectively. In words we have: Product Rule: (first) (derivative of second) + (second) (derivative of first) Quotient Rule: [ (bottom) (derivative of top) (top) (derivative of bottom) ]/ bottom 2 Product Rule: If y = f()g() then dy d = f()g ()+ f ()g() Quotient Rule: If y = f()/g() and g() 0 then dy d = g() f () f()g () (g()) 2 Checklist of Key Ideas: derivative of a product of functions, i.e., y = f()g() derivative of a quotient of functions, i.e., y = f()/g() 2.5 Derivatives of Trigonometric Functions PURPOSE: To give rules for calculating derivatives of the si basic trigonometric functions. Write these si derivative rules on a notecard. Memorize them in pairs sin() and cos() tan() and sec() cot() and csc() This section develops the derivatives of the si basic trigonometric functions starting with the derivatives of sin and cos. All trigonometric derivatives can be traced back to these two. d [sin] = cos d d [cos] = sin d From these two derivatives, the other four trigonometric ratios can be obtained using the quotient rule. Knowing these si derivatives is important. There are many patterns which can aide in the memorization of these derivatives. For eample, the graphs of y = sin and y = cos can help remind you what the derivatives of each of these functions are. Since cos begins at = 0 by decreasing, this indicates why the negative should be in front of its derivative. On the other hand, since sin starts by increasing with a positive slope, this indicates that positive cos should be its derivative. The derivatives of tan and sec are easy to remember since they are related to each other. d [sec] = sectan d d d [tan] = sec2
35 The last two derivatives cot and csc are co -functions which follow the same patterns as tan and sec with their derivatives ecept that they have negative signs. d [csc] = csccot d d d [cot] = csc2 Knowing the graphs of these other four trigonometric ratios can also be helpful in remembering these definitions. In particular, matching where a derivative crosses the horizontal ais with places where a function has a horizontal asymptote is a useful strategy in identifying any function with its derivative. For eample, since tan(0) = 0, this indicates that sec will have a horizontal tangent at = 0 because the derivative of sec is sectan. All of the co -functions have a negative sign in their derivatives. Checklist of Key Ideas: derivatives of sin and cos derivatives of the other trigonometric functions (using the quotient rule) 2.6 The Chain Rule PURPOSE: To give a rule for calculating the derivative of a function that is a composition of one or more functions. In the previous sections, the derivatives of basic algebraic combinations of functions such as sums, products, and quotients have been described. Additionally, the derivatives of common functions such as polynomials, rational functions and trigonometric functions have been established. In this section, the goal is to establish the derivative of a composition of functions. The idea is very simple and yet easy to confuse. If a function can be written as a composition of one function inside of another, then the derivative of the composition can be found by a product of the derivatives. A composition of functions may be easiest to think of as an inside function and an outside function. The derivative of the one function inside of the other is essentially the derivative of the inside function times the derivative of the outside function. In other words if y = f(u) and u = g() (or y = f(g())) then dy d = f (u)g () = d f du du d derivative of outside derivative of inside This now allows the derivatives of simple looking functions such as sin(2) and (+3) 00 to be taken in a quick and simple fashion. Both of the functions mentioned here could be differentiated using previous methods, but the process may f (u) g ()
36 have been tedious or may not have been straightforward. For eample, the product rule can be used if it is seen that sin(2) = 2sincos. On the other hand, the derivative of (+3) 00 can be found by first epanding the polynomial ( foil ing 99 times!) and then taking the derivatives of each term. The chain rule makes this process much simpler. Checklist of Key Ideas: derivatives of compositions two versions of the chain rule inside and outside functions derivatives of the inside and outside functions generalized derivative formulas d d [ f(u)] = f (u) du d multiple applications of the chain rule 2.7 Implicit Differentiation PURPOSE: To give rules for epressing the derivative of a function that is written implicitly. implicit means that y = f() is not known Implicit Differentiation. take derivatives using the chain rule 2. solve for dy/d This section approaches the problem of being able to find the slope of a curve even if an eplicit formula for the curve is unknown. In other words, most curves are described by a formula of the form y = f(). However, it may not always be practical or possible to find such an epression for y (i.e., consider y+sin(y) = 3). In such cases, the chain rule can be used to find the slope dy/d. The basic idea in this section is to assume that y is a function of, that is y = f(), and then to differentiate the relationship describing the curve with respect to by using the chain rule. In the end, the derivative dy/d may be epressed in terms of and y. This brings up two questions. differentiate with the chain rule ( differentiate implicitly ) ( use the = chain rule. Can we find dy/d? 2. How is dy/d evaluated? ) The first question can be answered using the chain rule. In fact, the phrase differentiate implicity is just another way of saying use the chain rule. For eample, if we have a curve described by 3 + y+y 3 =, then it is assumed that y = f() is a function that corresponds to at least a portion of the graph of this curve. Then the relationship can be rewritten as follows. 3 + f()+[f()] 3 =
37 Now to differentiate, we take the derivative of each term with respect to on both sides of the equation. This leads us to the following. 3 2 + f()+f ()+3[ f()] 2 f () = 0 Here we have used the product rule to find d [ f()] and the chain rule to find d d ( [ f()] 3 ) as shown. But since y = f() and dy/d = f () we usually write d this statement as 3 2 + y+ dy dy + 3y2 d d = 0 Now we only have to solve a linear equation to find dy/d. This is a key feature of implicit differentiation. IDEA: After differentiating an implicit equation, dy/d is found by solving a linear equation. Once the equation of the curve has been differentiated, there will only remain a linear equation for dy/d to be solved. To solve for dy/d we put all of the dy/d terms on one side, factor and divide. In the eample above, this looks like the following: solve a linear equation for dy/d dy dy + 3y2 d d = 32 y Put dy/d terms on one side. dy ( +3y 2 ) = 3 2 y d Factor out dy/d. dy d = 32 y +3y 2 Solve for dy/d. The second key concern is how to evaluate the derivative dy/d. The answer is that you now need both an and a y value since dy/d will often depend upon both. However, arbitrary values of and y cannot be used. Instead, the values of and y must satisfy the initial relationship that implicitly defined y in terms of. In the eample above, this means that the ordered pair (,y) must satisfy the relationship 3 + y+y 3 = 3. One way to find a valid point is to pick an value (i.e., = 0 or some other value) and then solve the resulting equation for y. evaluating dy/d both a value for y and are needed to find dy/d [ 3 + y+y 3 = 3 ] =0 y 3 = 3 y = 3 (/3).4422 Once dy/d is known at a particular point ( 0,y 0 ) then it is a simple matter to find a tangent line using, for eample, the point-slope form of a line.
38 Because of the way that dy/d is solved for, the derivative is often written as some sort of fraction. This information may be used to find where the curve has either horizontal or vertical tangents. To find horizontal tangents, the top of the fraction is set equal to zero. To find vertical tangents, the bottom of the fraction is set equal to zero. It is not quite as simple as this as in some cases, the top and bottom of the fraction may be zero simultaneously. This may be indicative of a vertical tangent and it may not. In the above eample, to find horizontal tangents to the curve 3 + y+y 3 = 3, we set the top of the derivative equal to zero. 3 2 y = 0 Then the curve may have a horizontal tangent where y = 3 2. This does not give very specific information about and y but it is enough. This relationship may be substituted back into the original curve to find the coordinates of a point with a horizontal tangent. Checklist of Key Ideas: y defined eplicitly and implicitly finding the slope of an implicitly defined curve implicit differentiation epressing dy/d in terms of and y finding tangent lines to implicitly defined curves 2.8 Related Rates PURPOSE: To use derivatives to relate the rates of change of multiple functions. In this section, derivatives are applied to problems that may involve more than one function of an independent variable. Most often, the independent variable that is considered is time, t. The main idea here is that a real situation may eist such that two or more quantities can be related to each other using algebraic equations (such as area, a, and radius, r of a circle). Then both of these quantities are assumed to be functions of time, t. The most confusing aspect of this is that t does not show up eplicitly in the equations and definitions for the quantities are not known (i.e., a(t) =? and r(t) =? in the previous circle mentioned). As it turns out, all that is needed is the information that these quantities are functions of t. Then the equation that relates them can be differentiated. The result is an equation that relates their rates or relates their derivatives. It may be helpful to write π r(t) 2 as For eample, if a = πr 2 in a circle then we assume that area and radius are func- π [r(t)] 2
39 tions of time, t. a(t) = πr(t) 2 Then we differentiate this equation. a (t) = π2r(t)r (t) On the right hand side of the last equation, the derivative of r(t) 2 was obtained using the chain rule: let y = u 2 and y = r(t) and then apply the chain rule. dy dt = dy du du dt = 2u r (t) = 2r(t)r (t) Formulas are still not known for a(t), r(t) or their derivatives. But if numerical information is known about them at a particular time, then frequently we can discover the value of how one of the quantities is changing with respect to time. Checklist of Key Ideas: relating the rates of change of different variables strategy for solving related rates problem identifying equations to relate the variables identifying variables as functions of time, t 2.9 Local Linear Approimations; Differentials PURPOSE: To use a tangent line to approimate a function. This section builds upon the notion at the beginning of this chapter that we can use the derivative of a function to find the equation of a tangent line to the function. The tangent line can then be used to approimate the values of the function nearby the point of tangency. This is the idea of a local linear approimation. The tangent line of a function is this local linear approimation. In some cases, it may be simpler to obtain the tangent line of a function and then use this in place of actual function values. For eample, if the value of a function and its derivative are known at a point but a definition of the actual function is not known, then a tangent line approimation can be used. IDEA: The derivative f () is the slope of the tangent line. Differentials take the approimation idea a step further. The notion of a differential for the function y = f() given by dy = f ()d
40 is just another way of using a local linear approimation. The derivative f () is also the slope of the tangent line. But near a point of tangency, the slope should be equal to a small change in y divided by a small change in. Then dy = f ()d says that a small change in y equals the slope of the tangent line times a small change in. This would be eact for a line but this is only an approimation for a general function. In this section and y are used to denote actual (eact) changes in -values and function values. Usually, d is assumed to be equal to a desired. But then dy becomes an approimation to y. Sometimes y is called the error in y and so dy would be an approimation to this error. On the other hand, y gives a percentage y error or relative error in the value of y. Then dy would be approimation of this y relative error. Checklist of Key Ideas: differentiable functions are locally linear general equation of tangent line at = 0 using tangent line as a linear approimation differentials and differential form of derivatives actual change, y, and estimated change, dy error, relative error and percentage error general differential formulas
4 Chapter 2 Sample Tests Section 2.. Find the average rate of change of y with respect to over the interval [,5] if y = f() = 2. (a) 0.24 0.24 0.48 0.48 2. Find the average rate of change of y with respect to over the interval [,4] if y = f() = 3. (a) 2 2 3.5 3.5 3. Find the instantaneous rate of change of y = 4 with respect to at 0 = 3. (a) 08 27 54 3.5 4. Find the instantaneous rate of change of y = with respect to at 0 = 2. (a) 0.25 0.5 0.25 0.5 5. Find the instantaneous rate of change of y = 2 3 with respect to at a general point 0. (a) 6 2 0 4 2 0 2 2 0 3 2 0 6. Find the instantaneous rate of change of y = 2 with respect to at a general point 0. (a) 3 0 3 0 2 3 2 0 2 2 0 7. Find the slope of the tangent to the graph of f() = 3 2 at a general point 0. (a) 2 0 2 3 2 0 2 2 0 3 2 0 8. Answer true or false. The slope of the tangent line to the graph of f() = 2 4 at 0 = 3 is 2. 9. Answer true or false. Use a graphing utility to graph y = 3 on [0,5]. If this graph represents a position versus time curve for a particle, then the instantaneous velocity of the particle is increasing over the graphed domain. 0. Use a graphing utility to graph y = 2 6+4 on [0,0]. If this graph represents a position versus time curve for a particle, then the instantaneous velocity of the particle is zero at what time? Assume time is in seconds. (a) 0 s 3 s 5 s 0 s. A rock is dropped from a height of 64 feet and falls toward the earth in a straight line. What is the instantaneous velocity downward when it hits the ground? (a) 64 ft/s 32 ft/s 2 ft/s 6 ft/s 2. Answer true or false. The magnitude of the instantaneous velocity is never less than the magnitude of the average velocity. 3. Answer true or false. If a rock is thrown straight upward from the ground, when it returns to earth its average velocity will be zero. 4. Answer true or false. If an object is thrown straight upward with a positive instantaneous velocity, its instantaneous velocity when it returns to the ground will be negative. 5. An object moves in a straight line so that after t s its distance in mm from its original position is given by s = t 3 + t. Its instantaneous velocity at t = 5 s is (a) 28 mm/s
42 28 mm/s 27 mm/s 76 mm/s 6. A particle moving along an -ais with a constant velocity is at the point = 3 when t = and = 8 when t = 2. The velocity of the particle if is in meters and t is in seconds is (a) 5 m/s 2 m/s 8 3 m/s 3 8 m/s 7. A family travels north along a highway at 60 mi/hr, then turns back and travels south at 65 mi/hr until returning to the starting point. Their average velocity is (a) 62.5 mi/hr 25 mi/hr 5 mi/hr 0 mi/hr 8. Find the equation of the tangent line to y = f() = 4 2 at = 2. (a) y = 8 y = 6 6 y = 8 6 y = 6+6 Section 2.2. Find the equation of the tangent line to y = f() = +2 at = 7. (a) y = 6 + 6 y = 6 4 y = 3 + 6 y = 3 4 2. If y = 2 then dy/d = (a) 2 2 2 2 2 3. If y = then dy/d = (a) 2 2 2 4. Answer true or false. y 5-5 5-5 5-5 The derivative of the function graphed on the left is graphed on the right. 5. Answer true or false. Use a graphing utility to help in obtaining the graph of y = f() =. The derivative f () is not defined at = 0. 6. Find f (t) if f(t) = 4t 3 2. (a) 2t 3 2 3t 2 4t 2 2t 2 (4+h) 3 64 7. lim represents the derivative of f() = 3 at h 0 h = a. Find a. (a) 4 4 64 64 3 27+h 3 8. lim represents the derivative of f() = 3 at h 0 h = a. Find a. (a) 27 3 3 27 9. Find an equation for the tangent line to the curve y = 3 2 + + at (,2). y 5-5
43 (a) y = 2 y = 2 y = 2 y = 2 ( 0. Let f() = cos. Estimate f π ) by using a graphing utility. 4 (a) 4 2 2 2 π 4. An air source constantly increases the air supply rate of a balloon. The volume V in cubic feet is given by V(t) = t 2 for 0 t 5, where t is time in seconds. How fast is the volume of the balloon increasing at t = 3 s? (a) 6 ft 3 /s 9 ft 3 /s 8 ft 3 /s 3 ft 3 /s 2. Answer true or false. Using a graphing utility it can be shown that f() = 5 is differentiable everywhere on [ 0,0]. 3. Answer true or false. { A graphing utility can be used to determine that f() = 3, 2 is differentiable at =., > 4. Answer true or false. { A graphing utility can be used to determine that f() = 3, 0 5 is differentiable at = 0., > 0 5. Answer true or false. If a function y = f() has a vertical tangent at = a then y = f() is differentiable at = a. 6. Answer true or false. The function y = is continuously differentiable. Section 2.3. Find dy/d if y = 7 9. (a) 6 9 63 9 6 8 63 8 2. Find dy/d if y = e 3. (a) 3e 2 2e 2 2e 3 0 3. Find dy/d if y = 3( 2 2+4). (a) 6 6 3 2 6+4 9 3 2 2 + 4 2 2 4. Answer true or false. If f() = + 2, then f () = 2 + 2. 5. Find d 2 y/d 2 if y = 9 2 + 4+6. (a) 9 8+4 8 9 6. Find y if y = 6 + 3. (a) 20 4 + 6 336 9 + 6 20 4 + 6 336 9 + 6 7. Answer true or false. The equation y = y + 2y 48 3 is satisfied by the function y = 3 + 6 2 + 2. 8. Use a graphing utility to locate all horizontal tangent lines to the curve y = 3 + 6 2 + 2. (a) = 0,4 = 4,0 = 0 = 4 9. Find the -coordinate of the point on the graph of y = 4 where the tangent line is parallel to the secant line that cuts the curve at = 0 and at =. (a) 3 4 4 2 0. The position of a moving particle is given by s(t) = 3t 2 + 2t + where t is time in seconds. The velocity in m/s is given by ds/dt. Find the velocity at t = 2 s. (a) 6 m/s 8 m/s
44 6 m/s 4 m/s {. If f() = 2, 2 then what value of b makes 4+b, < 2 f() differentiable everywhere? (a) b = 4 b = 0 b = 4 b can be any real number 2. If y = 2 3 and y = c 5 then the value of c is (a) 20 24 40 any real number 3. If y = 3 + + (a) 3 2 + 2 2 2 2 3 2 then dy/d = 4. If y = (2+3) 2 then y = (a) 4+6 4 2 + 2+9 8+6 8+2 5. Let h() = 3 f() 2g(). If f (3) = 3 and h() has a horizontal tangent line at = 3, then g (3) = (a) 0 3 3 9/2 6. If y = 4 /2 and y = c then c = (a) 2 4 8 any real number Section 2.4. Answer true or false. If y = 4+2 then y () = 4. 2. If y = 2 dy then 4 d = = (a) 8 3 8 3 8 9 8 9 3. Let y = 4 +5. Then y (0) = (a) 0 4 5 4 25 4 5 4. Suppose that g() = 2 f(). Find g (2) given that f(2) = 4 and f (2) = 8. (a) 48 6 6 32 5. Answer true or false. If f, g, and h are differentiable functions, ( ) and h 0 anywhere on its domain, then f g = h f g f gh h h 2. 6. Let h() = f()g(). If f () = g() and g () = f() then which of the following epressions could represent h ()? (a) 0 f()g() 2 f()g() 4 f()g() 7. If y = ( 2) f() has a tangent line with a slope of 3 at = 3 then find the value of f (3) when f(3) = 7. (a) f (3) = 4 f (3) = 7/2 f (3) = 3/2 f (3) can be any real number 8. A line of the form y = +b could be tangent to the graph of y = at which of the following points?
45 (a) ( 2, /3) (0,) (2,) No such points eist. 9. If y = 2 then y has a horizontal tangent line at + (a) = 0 = or = only = y has no horizontal tangents. 0. If y = 2 then y has a horizontal tangent line at + (a) = 0 only = or = =,0, or y has no horizontal tangents.. If y = f()g() then d2 y d 2 = (a) f ()g () f ()g ()+ f ()g () f()g ()+ f ()g() f ()g()+2f ()g ()+ f()g () 2. Answer true or false. If y = f()g() and f (2) = 0 then y has a horizontal tangent line at = 2. Section 2.5. Find f () if f() = 4 sin. (a) 4 3 cos 4 3 cos 4 3 sin+ 4 cos 4 3 sin 4 cos 2. Find f () if f() = sintan. (a) cos (sin)(+sec 2 ) (sin)( sec 2 ) cos 3. Find f () if f() = sin 2 +cos. (a) sin 2sincos sin 2sin cos 3sin 4. Find d 2 y/d 2 if y = sin. (a) sin+2cos 0 cos cos 5. Answer true or false. If y = sec then d 2 y/d 2 = sec. 6. Find the equation of the line tangent to the graph of y = sin at the point where = 0. (a) y = y = y = y = 7. Find the -coordinates of all points in the interval [ 2π, 2π] at which the graph of f() = csc has a horizontal tangent line. (a) 3π/2, π/2, π/2, 3π/2 π, π π, 0,π 3π/2, 0, 3π/2 8. Find d 93 (cos)/d 93. (a) cos cos sin sin 9. Find all -values on (0,2π) where f() is not differentiable if f() = tancos. (a) π/2, 3π/2 π π/2, π, 3π/2 None 0. Answer true or false. If is given in radians, the derivative formula for y = tan in degrees is y = π 80 sec2.. A rock at an elevation angle of θ is falling in a straight line. If at a given instant it has an angle of elevation of θ = π/4 and has a horizontal distance s from an observer, find the rate at which the rock is falling with respect to θ. ( π ) (a) sec 4 ( s sec 2 π ) 4
46 sec(π/4) s ( sπ ) sec 4 2. Answer true or false. If f() = tancos cotsin, then f () = cos sin. 3. Answer true or false. If f() = tan then f () = csc 2. 4. Answer true or false. The function f() = sin is differentiable cos everywhere. 5. If y = 3 sin then find d 2 y/d 2. (a) 6 sin 6sin+6 2 cos+ 3 sin 6sin+6 2 cos 3 sin 6sin 3 sin Section 2.6. Let f() = 2 4+3. f () = (a) 2 2 4+3 4 2 4+3 2 2 4+3 2 4 2. Let f() = ( 5 2) 20. f () = (a) 20( 5 2) 9 00 4 ( 5 2) 9 00 5 ( 2) 20 00 4 3. Let f() = sin(3). f () = (a) cos(3) 3cos(3) cos(3) 3cos(3) 4. Answer true or false. If f() = f cos () = sin 2 +3. 5. Let f() = 3 2 2. f () = (a) 3 2 2 2 + 32 2 2 sin 2 +3 then 3 2 2 2 3 4 2 2 4 2 2 + 32 2 2 6. y = sin(cos). Find dy/d. (a) cos(cos) sin cos(cos ) sin sin(cos) cos sin(cos) cos 7. y = 4 tan(6). Find dy/d. (a) 24 3 sec 2 (6) 24 3 sec(6)tan(6) 6 4 sec 2 (6)+4 3 tan(6) 6 4 sec(6)tan(6)+4 3 tan(6) ( +sin 2 ) 8. y =. Find dy/d. cos (a) 2sin+sincos 2 cos 2 sin+2sincos 2 sin 3 2cos sin 2cos sin cos 2 9. Answer true or false. If y = cos(5 3 ) then d 2 y/d 2 = cos(5 3 ). 0. Answer true or false. If y = sin 3 cos 2 then y = 6cos 3 9 4 sin 3 + 2sin 2 + 4 2 cos 2.. Find an equation for the tangent line to the graph of y = tan at = π/4. (a) y π ( π )( 4 = 2 + π ) 4 y = π 4 y π 4 = y π 2 ( 4 = π ) 2 4 2. y = sin 3 (π 2θ). Find dy/dθ. (a) 6sin 2 (π 2θ)cos(π 2θ) 6sin 2 (π 2θ) 6sin 3 (π 2θ) 3sin 2 (π 2θ)cos(π 2θ)
47 3. Use a graphing utility to obtain the graph of f() = ( + 2) 3. Determine the slope of the tangent line to the graph at =. (a) 54 27 40.5 4. Find the value of the constant A so that y = Acos3t satisfies the equation d 2 y/dt 2 + 3y = cos3t. (a) 2 6 6 9 2 5. Answer true or false. Suppose that f () = +2 and g() = 3. If F() = f(g()) then F () = 3 2 3 + 2. Section 2.7. Answer true or false. If y = 5 4+2 then dy d = 4 5 4+2. 2. Answer true or false. If y 3 = then dy d = 3y 2. 3. Find dy/d if 3 y cos = 2. (a) dy/d = 3y 2/3 sin dy/d = 3y 2/3 sin dy/d = 6y 2/3 sin dy/d = 6y 2/3 sin 4. Find dy/d if 2 + y 2 = 25. (a) 25 y y y 25 y 5. Answer true or false. If y 2 + 2y = 5 then dy d = 5 2y+2. 6. Suppose that 2 2 + 3y 2 = 9. Find d 2 y/d 2. (a) d 2 y d 2 = 62 + 4y 36y 3 d 2 y d 2 = 62 + 4y 6y 3 d 2 y d 2 = 62 4y 36y 3 d 2 y d 2 = 6y2 4 2 9y 3 7. Find the slope of the tangent line to 2 + y 2 = 5 at the point (,2). (a) 2 2 5 2 5 2 8. Find the slope of the tangent line to y 2 = 4 at the point (,2). (a) 4 4 9. Find dy/d if y 4 = 3. (a) 3 4y 4 3 2 y 4 4y 3 32 4y 4 3 2 + y 4 4y 3 0. Find dy/d if = cos(y). (a) sin(y) sin(y) +ysin(y) sin(y) +ysin(y) sin(y). Answer true or false. If cos = siny then dy/d = tan. 2. Answer true or false. If tan(y) = 4 then dy d = ysec2 (y). 3. y 2 = 3 2 has a tangent line parallel to the -ais at (a) (,)
48 (0,0) (2,2) (,2) 4. 2 + y 2 = 6 has tangent lines parallel to the y-ais at (a) (0, 6) and (0, 6) (0, 4) and (0, 4) ( 4, 0) and (4, 0) ( 6, 0) and (6, 0) 5. If y = 3, find the formula for y. (a) y = 3 + 3 y = 3 + y = 3 3 (+ ) 2 3 3 2 y = 3 3 (+ ) 2 6. If y = /3 + then dy d = (a) 6 7/6 + 2 /2 6 7/6 + 2 /2 2 ( ) 3 2/3 + ( 3 2/3 + ) ( ) + 2 /2 /3 + 7. Find the equation of the tangent line to the function y = ( 2 + 4) /3 at = 2. (a) y = ( 2)+2 2 y = ( 2)+8 2 y = ( 2)+2 3 y = ( 2)+8 3 Section 2.8. The volume of a sphere is given by V = 4 3 πr3. Find dv dt terms of dr dt. (a) dv dt = 4πr 2 dr dt in dv = 4 dr πr3 dt 3 dt dv = 4 dr πr2 dt 3 dt dv = 3r 2 dr dt dt 2. A cylinder of length 2 m and radius m is epanding such that dl/dt = 0.0 m/sec and dr/dt = 0.02 m/sec. Find dv/dt. (a) 0.003 m/s 3 0.2827 m/s 3 0.05 m/s 3 0.03 m/s 3 3. A 0-ft ladder rests against a wall. If it were to slip so that when the bottom of the ladder is 6 feet from the wall it will be moving at 0.02 ft/s, how fast would the ladder be moving down the wall? (a) 0.02 ft/s 0.0025 ft/s 0.05 ft/s 0.2 ft/s 4. A plane is approaching an observer with a horizontal speed of 200 ft/s and is currently 0,000 ft from being directly overhead at an altitude of 20,000 ft. Find the rate at which the angle of elevation, θ, is changing with respect to time, dθ/dt. (a) 0.020 rad/s 0.00 rad/s 0.08 rad/s 0.009 rad/s 5. Answer true or false. Suppose that z = y 5 3. Then dz/dt = (dy/dt) 5 +(d/dt) 3. 6. Suppose that z = 2 + y 2. Find dz/dt. (a) 2 d dy + 2y dt dt d dt + dy dt ( ) d 2 ( ) dy 2 + dt dt d dt + y dy dt 2 + y 2 7. The power in watts for a certain circuit is given by P = I 2 R. How fast is the power changing if the resistance, R, of the circuit is,000 Ω, the current, I is 2 A, and the current is decreasing with respect to time at a rate of 0.03 A/s. (a) 0.09 w/s 60 w/s
49 20 w/s.8 w/s 8. Gravitational force is inversely proportional to the distance between two objects squared. If F = 5 N at a distance d2 d = 3 m, how fast is the force diminishing if the objects are moving away from each other at 2 m/s? (a) 2 N/s 0.74 N/s 6.7 N/s. N/s 9. A point P is moving along a curve whose equation is given by y = 4 + 9. When P = (2,5), y is increasing at a rate of 2 units/s. How fast is changing? (a) 2 units/s 5 8 units/s 64 units/s 5 6 units/s 0. Water is running out of an inverted conical tank at a rate of 3 ft 3 /s. How fast is the height of the water in the tank changing if the height is currently 5 ft and the radius is 5 ft? (a) 0.038 ft/s 0.377 ft/s.3 ft/s 9.425 ft/s. Answer true or false. If z = lny then dz dt = 2. Answer true or false. If sinθ = 3y then dθ dt ( d dt )( ) dy. dt dy d = 3 +3y dt dt. 3. Answer true or false. If V = 3 2 y then dy dt = dv d 6 dt dt. 4. Answer true or false. If V = 0 3 then d dt = 30 2 dv dt. Section 2.9. If y = 4, find the formula for y. (a) y = (+ ) 4 y = 4 3 y = 4( ) 3 y = (+ ) 4 4 2. If y = cos, find the formula for y. (a) y = cos(+ ) cos y = cos(+ ) y = sin y = +cos 3. Answer true or false. The formula for dy is obtained from the formula for y by replacing with d. 4. Let y = 5. Find the formula for dy. (a) dy = (+d) 5 dy = (+d) 5 5 dy = 5 + 5 dy = 5 4 d 5. Let y = tan. Find the formula for dy. (a) dy = (sec 2 )d dy = (sec tan)d dy = tan(+d) tan dy = tan(+d) 6. Let y = 3 sin. Find the formula for dy. (a) dy = (3 2 sin+ 3 cos)d dy = 3(+d) 3 sin(+d) dy = (3 2 cos)d dy = (+d) 3 sin(+d) 3 sin 7. Let y =. Find dy at = if d = 0.0. 2 (a) 0.02 0.02 0.00 0.00 8. Let y = 5. Find dy at = if d = 0.0. (a) 0.00000005 0.00000005 0.05 0.05 9. Let y =. Find y at = 3 if =. (a) 0.268 0.268 0.289 0.250 0. Use dy to approimate 3.96 starting at = 4. (a) 2.0.99
50 4.0 3.99. Answer true or false. A circular spill is spreading so that when its radius r is 2 m, dr = 0.05 m. The corresponding change in the area covered by the spill, A, is 0.63 m 2 (to the nearest hundredth). 2. A small suspended droplet of radius 0 microns is evaporating. If dr = 0.00 micron, find the change in the volume, dv, to the nearest thousandth of a cubic micron. (a) 420.237 cubic microns.257 cubic microns 0.49 cubic microns 4.89 cubic microns 3. Answer true or false. A cube is epanding as temperature increases. If the length of the cube is changing at a rate of = 2 mm when is m, the volume is eperiencing a corresponding change of 0.006 mm 3 (to the nearest thousandth). 4. Answer true or false. The radius of the base of a cylinder is 2 mm with a possible error of ±0.0 mm. The height of the cylinder is eactly 4 m. Using differentials to estimate the maimum error of the volume, it is found to be 502.65 mm 3 (to the nearest hundredth)..5 4. Find the equation of the tangent line to y = f() = 3 3 at = 3. (a) y = 27 60 y = 27+60 y = 8+62 y = 8 62 5. If y = 8 then dy/d = (a) 8 7 8 8 7 7 7 8 6. Answer true or false. y 5-5 5 y 2-3 3 Chapter 2 Test. Find the average rate of change of y with respect to over the interval [,2] if y = f() = 3 2. (a) 9 9 2. Find the instantaneous rate of change of y = 2 4 with respect to at 0 = 3. (a) 26 54 08 27 3. An object moves in a straight line so that after t s its distance from its original position is given by s = t 3 2t. Its instantaneous velocity at t = 4 s is (a) 56 46 4-5 The derivative of the function graphed on the left is graphed on the right. (6+2h) 2 36 7. lim represents the derivative of f() = (2) 2 h 0 h at = (a) 6 3 0 3 8. Let f() = tan. Estimate f (5π/4) by using a graphing utility. (a) 0 2 9. Find dy/d if y = π 5. (a) 5π 4 π 5 0 4π 5-2
5 0. Answer true or false. If f() = 3 + 3, then f () = 2 + 3 32.. If y = 5 2 (a) 0 0 8 8 dy then d = = 2. Let g() = f(). Find g (4) given that f(4) = 4 and f (4) = 6. (a) 24.5 3 4 3. Find f () if f() = 3 tan. (a) 3 2 sec 2 3 2 sectan 3 2 tan+ 3 sec 2 3 2 tan+ 3 sectan 4. Find d 2 y/d 2 if y = 2(sin)(cos). (a) 8(cos)(sin) 8(cos)(sin) 2(cos)(sin) 2(cos)(sin) 5. Find the -coordinates of all the points over the interval (0, π) where the graph of f() = cot has a horizontal tangent line. (a) π/4, π/2, 3π/4 π/4, 3π/4 π/2 None eist. 6. Answer true or false. d 05 [sin] = cos. d05 7. Answer true or false. If f() = 3 2 2 then f () = 3 2 4 3 2 2. 8. If f() = sin(6) then f () = (a) 6cos(6) 6cos(6) cos(6) cos(6) { sin(3), 0 9. Let y = 3 2. Find the value of a so + a+3, < 0 that y is differentiable everywhere. (a) a = 6 a = 3 a = 3 No such value of a eists. 20. If y = ( f()) 3 then dy d = (a) 3[ f()] 2 f () 3[ f()] 2 [ f ()] 3 3[ f ()] 2 2. If y = 9 then find the formula for y. (a) y = 9 + 9 y = 9 + 9 y = 9 9 8 y = 9 + + 9 22. Answer true or false. If 2 + y 2 = 4 then y = 4y 3. 23. If y = 2 tan, find the formula for dy. (a) dy = (2 tan)d dy = ( 2 sec 2 )d dy = (2tan+ 2 sec 2 )d dy = (2tan 2 sec 2 )d 24. Answer true or false. If y = 5 and = 2 then dy = 5 6 d. 25. Answer true or false. A spherical balloon is deflating. The rate at which the volume is changing when r = 2 m is given by dv = 6πdr m 3 /s.
52 Chapter 2: Answers to Sample Tests Section 2.. b 2. a 3. a 4. c 5. a 6. d 7. d 8. false 9. true 0. b. a 2. false 3. true 4. true 5. d 6. a 7. d 8. b Section 2.2. a 2. c 3. a 4. true 5. true 6. d 7. a 8. a 9. d 0. b. a 2. false 3. false 4. true 5. false 6. false Section 2.3. d 2. d 3. a 4. false 5. c 6. b 7. false 8. b 9. a 0. d. a 2. c 3. c 4. d 5. d 6. c Section 2.4. false 2. c 3. c 4. a 5. false 6. d 7. a 8. b 9. b 0. a. d 2. false Section 2.5. c 2. b 3. b 4. a 5. false 6. c 7. a 8. d 9. a 0. false. b 2. false 3. true 4. false 5. c Section 2.6. a 2. b 3. b 4. false 5. d 6. a 7. c 8. a 9. false 0. true. a 2. a 3. d 4. c 5. true Section 2.7. false 2. true 3. a 4. c 5. false 6. d 7. b 8. d 9. b 0. d. false 2. false 3. a 4. c 5. a 6. a 7. c Section 2.8. a 2. b 3. c 4. c 5. false 6. d 7. c 8. b 9. b 0. a. false 2. false 3. false 4. true Section 2.9. d 2. a 3. false 4. d 5. a 6. a 7. b 8. d 9. b 0. b. true 2. b 3. true 4. true Chapter 2 Test. a 2. a 3. b 4. d 5. a 6. false 7. b 8. d 9. c 0. false. b 2. c 3. c 4. a 5. d 6. true 7. false 8. a 9. d 20. a 2. b 22. true 23. c 24. false 25. true