Chapter : The Derivative Summary: Chapter 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. 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 (.,.). Be able to read and understand different notations for the derivative (.) 3. Find derivatives of simple functions like polynomials (.3) or trigonometric functions (.5) 4. Use the product and/or quotient rules to find derivatives of functions involving products or quotients of other functions (.4) 5. Use the chain rule (either form) to find the derivatives of functions which involve the compositions of other functions (.6). Tangent Lines and Rates of Change PURPOSE: To use limits and continuity to investigate the slope of a function at a point. 33
34 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.. 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 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
35. 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.. 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:. If a function has a limit at a point, it does not have to be continuous at a point. Consider y = ( 4)/(+) which has a hole at = and so is not continuous there.. 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. 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 (). the derivative f () and the graph of f()
36 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.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 straightforward results do not hold when functions are multiplied and divided (see the net section). derivative of n derivative of c f() derivative of f() + g() write new derivative rules on notecards 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.
37 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.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() = (see p.63-64 in the tet). Since y = f()g() = 3 we know that its derivative is y = 3. 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. 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. 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()) Checklist of Key Ideas: 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 derivative of a product of functions, i.e., y = f()g() derivative of a quotient of functions, i.e., y = f()/g()
38.5 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() All of the co -functions have a negative sign in their derivatives. 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] = sec 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] = csc 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. Checklist of Key Ideas: derivatives of sin and cos derivatives of the other trigonometric functions (using the quotient rule)
39.6 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() 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 have been tedious or may not have been straightforward. For eample, the product rule can be used if it is seen that sin() = sincos. 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. f (u) g () 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
40 Chapter Sample Tests Section.. Find the average rate of change of y with respect to over the interval [,5] if y = f() =. (a) 0.4 (b) 0.4 0.48 (d) 0.48. Find the average rate of change of y with respect to over the interval [,4] if y = f() = 3. (a) (b) 3.5 (d) 3.5 3. Find the instantaneous rate of change of y = 4 with respect to at 0 = 3. (a) 08 (b) 7 54 (d) 3.5 4. Find the instantaneous rate of change of y = with respect to at 0 =. (a) 0.5 (b) 0.5 0.5 (d) 0.5 5. Find the instantaneous rate of change of y = 3 with respect to at a general point 0. (a) 6 0 (b) 4 0 0 3 (d) 0 6. Find the instantaneous rate of change of y = with respect to at a general point 0. (a) 3 0 (b) 3 0 3 0 (d) 0 7. Find the slope of the tangent to the graph of f() = 3 at a general point 0. (a) 0 (b) 3 0 0 (d) 3 0 8. Answer true or false. The slope of the tangent line to the graph of f() = 4 at 0 = 3 is. 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 = 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 (b) 3 s 5 s (d) 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 (b) 3 ft/s ft/s (d) 6 ft/s. 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) 8 mm/s
4 (b) 8 mm/s 7 mm/s (d) 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 =. The velocity of the particle if is in meters and t is in seconds is (a) 5 m/s (b) m/s 8 3 m/s 3 (d) 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) 6.5 mi/hr (b) 5 mi/hr 5 mi/hr (d) 0 mi/hr 8. Find the equation of the tangent line to y = f() = 4 at =. (a) y = 8 (b) y = 6 6 y = 8 6 (d) y = 6+6 Section.. Find the equation of the tangent line to y = f() = + at = 7. (a) y = 6 + 6 (b) y = 6 4 y = 3 + 6 (d) y = 3 4. If y = then dy/d = (a) (b) (d) 3. If y = then dy/d = (a) (b) (d) 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. (a) t 3 (b) 3t 4t (d) t (4+h) 3 64 7. lim represents the derivative of f() = 3 at h 0 h = a. Find a. (a) 4 (b) 4 64 (d) 64 3 7+h 3 8. lim represents the derivative of f() = 3 at h 0 h = a. Find a. (a) 7 (b) 3 3 (d) 7 9. Find an equation for the tangent line to the curve y = 3 + + at (,). y 5-5
4 (a) y = (b) y = y = (d) y = ( 0. Let f() = cos. Estimate f π ) by using a graphing utility. 4 (a) 4 (b) (d) 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 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 (b) 9 ft 3 /s 8 ft 3 /s (d) 3 ft 3 /s. 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, 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.3. Find dy/d if y = 7 9. (a) 6 9 (b) 63 9 6 8 (d) 63 8. Find dy/d if y = e 3. (a) 3e (b) e e 3 (d) 0 3. Find dy/d if y = 3( +4). (a) 6 6 (b) 3 6+4 9 3 + 4 (d) 4. Answer true or false. If f() = +, then f () = +. 5. Find d y/d if y = 9 + 4+6. (a) 9 (b) 8+4 8 (d) 9 6. Find y if y = 6 + 3. (a) 0 4 + 6 (b) 336 9 + 6 0 4 + 6 (d) 336 9 + 6 7. Answer true or false. The equation y = y + y 48 3 is satisfied by the function y = 3 + 6 +. 8. Use a graphing utility to locate all horizontal tangent lines to the curve y = 3 + 6 +. (a) = 0,4 (b) = 4,0 = 0 (d) = 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 (b) 4 (d) 0. The position of a moving particle is given by s(t) = 3t + t + where t is time in seconds. The velocity in m/s is given by ds/dt. Find the velocity at t = s. (a) 6 m/s (b) 8 m/s
43 6 m/s (d) 4 m/s {. If f() =, then what value of b makes 4+b, < f() differentiable everywhere? (a) b = 4 (b) b = 0 b = 4 (d) b can be any real number. If y = 3 and y = c 5 then the value of c is (a) 0 (b) 4 40 (d) any real number 3. If y = 3 + + (a) 3 + (b) (d) 3 then dy/d = 4. If y = (+3) then y = (a) 4+6 (b) 4 + +9 8+6 (d) 8+ 5. Let h() = 3 f() g(). If f (3) = 3 and h() has a horizontal tangent line at = 3, then g (3) = (a) 0 (b) 3 3 (d) 9/ 6. If y = 4 / and y = c then c = (a) (b) 4 8 (d) any real number Section.4. Answer true or false. If y = 4+ then y () = 4.. If y = dy then 4 d = = (a) 8 3 8 (b) 3 8 9 8 (d) 9 3. Let y = 4 +5. Then y (0) = (a) 0 4 (b) 5 4 5 (d) 4 5 4. Suppose that g() = f(). Find g () given that f() = 4 and f () = 8. (a) 48 (b) 6 6 (d) 3 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. 6. Let h() = f()g(). If f () = g() and g () = f() then which of the following epressions could represent h ()? (a) 0 (b) f()g() f()g() (d) 4 f()g() 7. If y = ( ) 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 (b) f (3) = 7/ f (3) = 3/ (d) 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?
44 (a) (, /3) (b) (0,) (,) (d) No such points eist. 9. If y = then y has a horizontal tangent line at + (a) = 0 (b) = or = only = (d) y has no horizontal tangents. 0. If y = then y has a horizontal tangent line at + (a) = 0 (b) only = or = =,0, or (d) y has no horizontal tangents.. If y = f()g() then d y d = (a) f ()g () (b) f ()g ()+ f ()g () f()g ()+ f ()g() (d) f ()g()+f ()g ()+ f()g (). Answer true or false. If y = f()g() and f () = 0 then y has a horizontal tangent line at =. Section.5. Find f () if f() = 4 sin. (a) 4 3 cos (b) 4 3 cos 4 3 sin+ 4 cos (d) 4 3 sin 4 cos. Find f () if f() = sintan. (a) cos (b) (sin)(+sec ) (sin)( sec ) (d) cos 3. Find f () if f() = sin +cos. (a) sin (b) sincos sin sin cos (d) 3sin 4. Find d y/d if y = sin. (a) sin+cos (b) 0 cos (d) cos 5. Answer true or false. If y = sec then d y/d = sec. 6. Find the equation of the line tangent to the graph of y = sin at the point where = 0. (a) y = (b) y = y = (d) y = 7. Find the -coordinates of all points in the interval [ π, π] at which the graph of f() = csc has a horizontal tangent line. (a) 3π/, π/, π/, 3π/ (b) π, π π, 0,π (d) 3π/, 0, 3π/ 8. Find d 93 (cos)/d 93. (a) cos (b) cos sin (d) sin 9. Find all -values on (0,π) where f() is not differentiable if f() = tancos. (a) π/, 3π/ (b) π π/, π, 3π/ (d) None 0. Answer true or false. If is given in radians, the derivative formula for y = tan in degrees is y = π 80 sec.. 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 ( (b) s sec π ) 4
45 sec(π/4) s ( sπ ) (d) sec 4. Answer true or false. If f() = tancos cotsin, then f () = cos sin. 3. Answer true or false. If f() = tan then f () = csc. 4. Answer true or false. The function f() = sin is differentiable cos everywhere. 5. If y = 3 sin then find d y/d. (a) 6 sin (b) 6sin+6 cos+ 3 sin 6sin+6 cos 3 sin (d) 6sin 3 sin Section.6. Let f() = 4+3. f () = (a) 4+3 (b) 4 4+3 4+3 (d) 4. Let f() = ( 5 ) 0. f () = (a) 0( 5 ) 9 (b) 00 4 ( 5 ) 9 00 5 ( ) 0 (d) 00 4 3. Let f() = sin(3). f () = (a) cos(3) (b) 3cos(3) cos(3) (d) 3cos(3) 4. Answer true or false. If f() = f cos () = sin +3. 5. Let f() = 3. f () = (a) 3 + 3 sin +3 then (b) 3 (d) 3 4 4 + 3 6. y = sin(cos). Find dy/d. (a) cos(cos) sin (b) cos(cos ) sin sin(cos) cos (d) sin(cos) cos 7. y = 4 tan(6). Find dy/d. (a) 4 3 sec (6) (b) 4 3 sec(6)tan(6) 6 4 sec (6)+4 3 tan(6) (d) 6 4 sec(6)tan(6)+4 3 tan(6) ( +sin ) 8. y =. Find dy/d. cos (a) (b) sin+sincos cos sin+sincos sin 3 cos sin (d) cos sin cos 9. Answer true or false. If y = cos(5 3 ) then d y/d = cos(5 3 ). 0. Answer true or false. If y = sin 3 cos then y = 6cos 3 9 4 sin 3 + sin + 4 cos.. Find an equation for the tangent line to the graph of y = tan at = π/4. (a) y π ( π )( 4 = + π ) 4 (b) y = π 4 y π 4 = (d) y π ( 4 = π ) 4. y = sin 3 (π θ). Find dy/dθ. (a) 6sin (π θ)cos(π θ) (b) 6sin (π θ) 6sin 3 (π θ) (d) 3sin (π θ)cos(π θ)
46 3. Use a graphing utility to obtain the graph of f() = ( + ) 3. Determine the slope of the tangent line to the graph at =. (a) (b) 54 7 (d) 40.5 4. Find the value of the constant A so that y = Acos3t satisfies the equation d y/dt + 3y = cos3t. (a) (b) 6 6 (d) 9 5. Answer true or false. Suppose that f () = + and g() = 3. If F() = f(g()) then F () = 3 3 +. (a) y = 7 60 (b) y = 7+60 y = 8+6 (d) y = 8 6 5. If y = 8 then dy/d = (a) 8 7 (b) 8 8 7 7 (d) 7 8 6. Answer true or false. y 5-5 5 y -3 3 Chapter Test. Find the average rate of change of y with respect to over the interval [,] if y = f() = 3. (a) 9 (b) 9 (d). Find the instantaneous rate of change of y = 4 with respect to at 0 = 3. (a) 6 (b) 54 08 (d) 7 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 t. Its instantaneous velocity at t = 4 s is (a) 56 (b) 46 4 (d).5 4. Find the equation of the tangent line to y = f() = 3 3 at = 3. -5 The derivative of the function graphed on the left is graphed on the right. (6+h) 36 7. lim represents the derivative of f() = () h 0 h at = (a) 6 (b) 3 0 (d) 3 8. Let f() = tan. Estimate f (5π/4) by using a graphing utility. (a) (b) 0 (d) 9. Find dy/d if y = π 5. (a) 5π 4 (b) π 5 0 (d) 4π 5 0. Answer true or false. If f() = 3 + 3, then f () = 3 + 3. -
47. If y = 5 (a) 0 (b) 0 8 (d) 8 dy then d = =. Let g() = f(). Find g (4) given that f(4) = 4 and f (4) = 6. (a) 4 (b).5 3 (d) 4 3. Find f () if f() = 3 tan. (a) 3 sec (b) 3 sectan 3 tan+ 3 sec (d) 3 tan+ 3 sectan 4. Find d y/d if y = (sin)(cos). (a) 8(cos)(sin) (b) 8(cos)(sin) (cos)(sin) (d) (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/, 3π/4 (b) π/4, 3π/4 π/ (d) None eist. 6. Answer true or false. d 05 [sin] = cos. d05 7. Answer true or false. If f() = 3 then f () = 3 4 3. 8. If f() = sin(6) then f () = (a) 6cos(6) (b) 6cos(6) cos(6) (d) cos(6) { sin(3), 0 9. Let y = 3. Find the value of a so + a+3, < 0 that y is differentiable everywhere. (a) a = 6 (b) a = 3 a = 3 (d) No such value of a eists. 0. If y = ( f()) 3 then dy d = (a) 3[ f()] f () (b) 3[ f()] [ f ()] 3 (d) 3[ f ()]
48 Chapter : Answers to Sample Tests Section.. b. a 3. a 4. c 5. a 6. d 7. d 8. false 9. true 0. b. a. false 3. true 4. true 5. d 6. a 7. d 8. b Section.. a. c 3. a 4. true 5. true 6. d 7. a 8. a 9. d 0. b. a. false 3. false 4. true 5. false 6. false Section.3. d. d 3. a 4. false 5. c 6. b 7. false 8. b 9. a 0. d. a. c 3. c 4. d 5. d 6. c Section.4. false. c 3. c 4. a 5. false 6. d 7. a 8. b 9. b 0. a. d. false Section.5. c. b 3. b 4. a 5. false 6. c 7. a 8. d 9. a 0. false. b. false 3. true 4. false 5. c Section.6. a. b 3. b 4. false 5. d 6. a 7. c 8. a 9. false 0. true. a. a 3. d 4. c 5. true Chapter Test. a. a 3. b 4. d 5. a 6. false 7. b 8. d 9. c 0. false. b. c 3. c 4. a 5. d 6. true 7. false 8. a 9. d 0. a