Module M4.2 Basic differentiation

Transcription

1 F L E X I B L E L E A R N I N G A P P R O A C H T O P H Y S I C S Moule M4. Opening items. Moule introuction. Fast track questions.3 Reay to stuy? Variables, functions an erivatives. Functions an variables. Rates of change, graients an erivatives.3 Notation for erivatives 3 Derivatives of simple functions 3. Derivatives of basic functions 3. Derivative of a sum of functions 3.3 Derivative of a constant multiple of a function 3.4 Derivative of a prouct of functions 3.5 Derivative of a quotient of functions 4 More about logarithmic an exponential functions 4. Why exp(x) is special 4. Derivative of a x 4.3 Derivative of log a x 4.4 Resiste motion uner gravity:an example 5 Closing items 5. Moule summary 5. Achievements 5.3 Exit test Exit moule COPYRIGHT 998 S570 V.

2 Opening items. Moule introuction Being able to ifferentiate accurately an confiently is a necessary skill for any physicist. Differentiation is such a powerful tool with so many applications that it is worth some effort to ensure that you have mastere the techniques explaine in this moule. The main objective of this moule is to answer the following question: If we know how to ifferentiate two functions f (x) an g(x), how o we ifferentiate their sum, ifference, prouct an quotient? To answer this question we iscuss first the iea of a function an how functions can be combine to prouce new ones. Section examines the mathematical basis of ifferentiation, giving first a graphical interpretation an then a formal efinition. COPYRIGHT 998 S570 V.

3 Section 3 answers our question an is the core of the moule. It provies a list of erivatives of various stanar functions (such as power functions (xn) an trigonometric functions) an introuces the various rules that will enable you to ifferentiate a wie range of combinations of these stanar functions. In particular it will explain how to ifferentiate sums, constant multiples, proucts an quotients of the stanar functions. Sometimes the metho is straightforwar, as with the erivative of a sum, but it may also be quite complicate, as with the erivative of a quotient. We also iscuss the logarithmic an exponential functions, an, while loge x an ex are the most important of such functions, it is sometimes necessary to be able to ifferentiate loga x an a x where a is some positive number other than e. In Section 4 we consier why ex is so important an then a ax an loga x to the list of functions that can be ifferentiate. Stuy comment Having rea the introuction you may feel that you are alreay familiar with the material covere by this moule an that you o not nee to stuy it. If so, try the Fast track questions given in Subsection.. If not, procee irectly to Reay to stuy? in Subsection.3. COPYRIGHT 998 S570 V.

4 . Fast track questions Stuy comment Can you answer the following Fast track questions?. If you answer the questions successfully you nee only glance through the moule before looking at the Moule summary (Subsection 5.) an the Achievements liste in Subsection 5.. If you are sure that you can meet each of these achievements, try the Exit test in Subsection 5.3. If you have ifficulty with only one or two of the questions you shoul follow the guiance given in the answers an rea the relevant parts of the moule. However, if you have ifficulty with more than two of the Exit questions you are strongly avise to stuy the whole moule. COPYRIGHT 998 S570 V.

5 Question F Differentiate each of the following functions (it is not necessary to simplify your answers). (a) f (x) = x sin x + loge x (b) f (x) = (x3 6x)(x + 5x ) (use the prouct rule) (c) f (x) = sin x x + 4x f(x) = x +9 x tan x f(x) = 3x x + f(x) = + x + e f(x) = (e + e ) () (e) (f) (g) x x x COPYRIGHT 998 S570 V.

6 Question F If at time t the isplacement, sx(t), of a particle along the x-axis from some particular reference point is given by sx(t) = e αt [A cos (ω t) + B sin (ω t)] where A, B, α an ω are constants, fin expressions for v x(t), the velocity at time t, an ax(t) the acceleration at time t, an show that ax(t) + α vx(t) + (ω + α)sx(t) = 0 Stuy comment Having seen the Fast track questions you may feel that it woul be wiser to follow the normal route through the moule an to procee irectly to Reay to stuy? in Subsection.3. Alternatively, you may still be sufficiently comfortable with the material covere by the moule to procee irectly to the Closing items. COPYRIGHT 998 S570 V.

7 .3 Reay to stuy? Stuy comment To begin the stuy of this moule you will nee to be familiar with the following terms: base (of a logarithm), exponential function, inequality (in particular greater than (>) an less than (<)), integer, inverse function, logarithmic function, moulus (or absolute value, x ), prouct, quotient, raian, real number, reciprocal, set, square root, sum, trigonometric function an trigonometric ientities. (The trigonometric ientities that you require are repeate in this moule, as is the efinition of the moulus function.) In aition you will nee to have some familiarity with the concept of a function, an with relate terms such as argument, coomain, omain an variable (both epenent an inepenent), but the precise meaning of these terms is briefly reviewe in the moule. Similar comments apply to physics concepts use in this moule, such as acceleration, isplacement, force, spee, velocity an Newton s secon law. If you are uncertain of any of these items you can review them now by referring to the Glossary, which will inicate where in FLAP they are evelope. The following Reay to stuy questions will allow you to establish whether you nee to review some of the topics, or to improve your general algebraic skills, before embarking on this moule. Question R Write the following sums, proucts an quotients of logarithms an exponentials as compactly as you can: (a) log0 x + log0 y, (b) (log 4 log 4)3, (c) loge x loge y, () ex/ey, (e) (ex/ey), (f) loge x + loge x + loge x COPYRIGHT S570 V.

8 Question R If f (x) = cos x, g(x) = sin x, an h(x) = tan x, rewrite the following expressions in terms of trigonometric functions an simplify them where possible : (a) [ f (x) + g(x)] (b) f (x)h(x) + g(x)/h(x) (c) [ f (x) g(x)]/[ + h(x)] Question R3 Given that f (x) = x, fin the following values, f (), f (0), f ( ), f (0), f (0) f ( ) an sketch the graph 4 of y = f (x) COPYRIGHT 998 S570 V.

9 Question R4 3(b) exp(log (0 )),3(c) log ( ). Given that ax an loga x are inverse functions, simplify the following : (a) log e [e log0 ( x ) ], e x COPYRIGHT 998 S570 V.

10 Variables, functions an erivatives. Functions an variables Stuy comment An unerstaning of functions is crucial to an unerstaning of ifferentiation, an it is vital that the notation an terminology use to escribe functions shoul be clear an unambiguous. For that reason, this subsection reviews the efinitions of terms such as function an variable even though it is assume that you have met these ieas before. If you are completely unfamiliar with these concepts you shoul consult the entry on functions in the Glossary. A function f is a rule that assigns a single value f (x) in a set calle the coomain to each value x in a set calle the omain. Functions are very often efine by formulae, for example f (x) = x, an in such cases we assume, unless we are tol otherwise, that the omain is the largest set of real values for which the formula makes sense. In the case of f (x) = x the omain of the function is the set of all real numbers. The function g( x ) = is not efine x when x =, since /0 has no meaning, an so we take the set of all real numbers x with x as its omain. The function f (x) = + x + x (a) is another example of a function that is efine for all values of x. COPYRIGHT 998 S570 V.

11 One may think of the function as a sort of machine with x as the input an + x + x as the output. The input x is known as the inepenent variable an, if we write y = f (x), the output y is known as the epenent variable since the function f (x) etermines the way in which y epens on x. It is important to note that the same function f coul equally well be efine using some other symbol, such as t, to represent the inepenent variable: f (t) = + t + t (b) This freeom to relabel the inepenent variable is often of great use, though it is vital that such changes are mae consistently throughout an equation. We may evaluate the function in Equation, whether we call it f (x) or f (t), for any value of the inepenent variable; for example, if we choose to use x to enote the inepenent variable, an set x =, we have f(π) = + π + π f(a) = + (a) + (a) = + a + 4a f () = + + = 3 Similarly, if x = π an, if x = a COPYRIGHT 998 S570 V.

12 We may even apply the function to x, x, or + x to obtain f( x) = + ( x) + ( x) = x + x f( + x) = + ( + x) + ( + x) = 3 + 3x + x When we write expressions such as f(π) or f(a), whatever appears within the brackets is calle the argument of the function. The value of f(x) is etermine by the value of its argument, irrespective of what we call the f (x) = + (x) + (x) = + x + 4x argument. 3 A special note about x Generally in FLAP we follow the convention that x may be positive or negative. Thus 4 = ±. A consequence of this convention is that f (x) = x oes not efine a function since it oes not associate a unique value of f (x) with each value of x. This woul be an exceptionally inconvenient convention to follow in this moule so instea we aopt the convention that x is positive. Of course, it remains true that the square roots of x may be positive or negative, x or x, so within this convention we shoul call x the positive square root of x, an not simply the square root. In a similar spirit x/ = x will also be a positive quantity. COPYRIGHT 998 S570 V.

13 3fin f(x) an If f (x) = cos x + sin x π f 4 3 Question T If f (x) = x + an g(x) = x, x (a) write own expressions for f ( x ) an g. (b) For which values of the inepenent variable x is f (x) = g(x)? (c) For which value of the inepenent variable x is the following true? f ( x + 0.) f ( x ) = COPYRIGHT 998 S570 V.

14 . Rates of change, graients an erivatives g(t) 0 t The graph of the function g(t) = is shown in Figure. It is clear from the graph that as the value of t increases from t = 0, the value of g(t) also increases from 0. In particular, note that as t increases from 0 to the value of g(t) changes from 0 to, an that as t increases from to the value of g(t) changes from to 4. Thus, the change in the value of g(t) that correspons to a change of one unit in the value of t epens on the initial value of t. For this particular function, if the initial value of t is large, then the change in g(t) will also be large; for example, the change in g(t) from t = 00 to t = 0 is t g(0) g(00) = 0 00 = 0 Figure The graph of g(t) = t. Clearly, as t increases the rate at which g(t) is changing is itself changing, becoming greater an greater. How are we to measure the rate at which a function changes? If you imagine yourself walking up a hill in the shape of the graph shown in Figure, starting at t = 0 an travelling to the right, your path woul become steeper an steeper. It woul become increasingly ifficult to make progress, because, with each step you took, the slope woul become greater an your walk woul become more an more of a climb. The slope of the graph is the key; once we can escribe the slope precisely for any value of t, we will be able to measure the rate of change of the function g(t) or of any other function. COPYRIGHT 998 S570 V.

15 f (x) f (x) run f (a + h) tangent Q rise f (a) f (a) P a (a) a+h a x 3The graph of an arbitrary function. Figure P x (b) The graph of an arbitrary function f (x) is shown in Figure a. Roughly speaking, the slope of this graph at the point P, where x = a an f (x) = f (a), is the same as the slope of the ashe straight line PQ where Q is a nearby point corresponing to x = a + h an f (x) = f (a + h). Of course, the two slopes are not exactly the same; the point P is fixe an its location etermines the slope of f (x) at P, whereas the slope of the line PQ will epen on the location of Q as well as that of P. Nonetheless, the two slopes are similar, so we can roughly escribe the slope of the curve in terms of the graient of the line PQ which is efine by rise f ( a + h) f ( a) f ( a + h) f ( a) graient of the line PQ = = = () run ( a + h) a h COPYRIGHT 998 S570 V.

16 f (x) f (x) run f (a + h) tangent Q rise f (a) f (a) P a (a) a+h P a x 3The graph of an arbitrary function. x (b) Figure Now, the line PQ (the ashe line in Figure a) that cuts the graph of f (x) at P an Q an is calle a chor, but if we let the point Q get closer an closer to P then this chor becomes more an more like the tangent in Figure b that just touches the graph at P. It is the graient of this tangent that really represents the slope of the curve at P, this is clear from the figure an from the fact that the graient of the tangent (unlike that of the chor) is etermine by the location of P alone. Fortunately, the graient of the tangent is fairly easy to work out, all we have to o is to consier what happens to the graient of the chor PQ as Q gets closer an closer to P. As Q approaches P, h gets smaller an the graient of the chor PQ approaches the graient of the tangent at P ever more closely. COPYRIGHT 998 S570 V.

17 Expresse more formally, the graient of the tangent at P is given by Equation in the limit as h tens to zero. graient of the line PQ = rise f ( a + h) f ( a) f ( a + h) f ( a) = = run ( a + h) a h (Eqn ) Thus f ( a + h) f ( a) graient of tangent at P = lim h 0 h (3) Since this graient represents the slope of the graph of f (x) at P, it makes sense to efine the graient of the graph of f (x) at P to be the graient of its tangent at P. We call this graient the erivative of f (x) at x = a an enote it by f (a) so that f ( a + h) f ( a) f ( a) = lim h 0 h COPYRIGHT 998 (4) S570 V.

18 The erivative at x = a efine in this way also represents the rate of change of f (x) with respect to x when x = a. This last point is even easier to appreciate if we introuce a epenent variable y such that y = f (x), for we can the represent the vertical rise in Figure a by y = f ( a + h) f (a), an the horizontal run by x = h. With these efinitions it follows from Equation 4 f ( a + h) f ( a) (Eqn 4) f ( a) = lim h 0 h that the erivative at x = a is given by y f ( a) = lim x 0 x Even greater emphasis can be given to the iea that the erivative is a rate of change with respect to x by y ( a). replacing f (a) by the alternative symbol Thus, y f ( a + h) f ( a) y ( a) = lim = lim x 0 x h 0 h COPYRIGHT 998 S570 V. (5)

19 The efinition of the erivative at x = a represente by Equations 4 an 5 f ( a + h) f ( a) (Eqn 4) f ( a) = lim h 0 h y y f ( a + h) f ( a) Thus, (Eqn 5) ( a) = lim = lim x 0 x h 0 h can be applie at any point on the graph provie that y an x can be efine at that point an that their quotient y/ x has a unique limit as x 0 at that point. Thus, provie the erivative of f (x) exists at every point within some omain (i.e. some set of x values) it is possible to efine a new function on that omain that associates any given value of x with the graient of f (x) at that point. This new function is calle the erive f function or erivative of f (x) an is written f (x) or ( x ). If y = f (x), the erive function may also be written y y (x) or just. Thus, provie unique limits exist f ( x + h) f ( x ) y = lim h 0 h COPYRIGHT 998 (6) S570 V.

20 When using this formula f ( x + h) f ( x ) y (Eqn 6) = lim h 0 h it is important to remember that y/ is not a quotient of two quantities y an even though it may look like one. At a given value of x, the erivative y/ represents the graient of the graph of y = f (x) at that value of x. Although the graphical interpretation of the erivative is important, it oes not len itself to calculation since rawing tangents can only be approximate, an in any case it woul be impossible to o it for every point on the graph. However, in simple cases it is not ifficult to etermine the erivative from the formula for the function an the above efinition (Equation 6). Consier the case of f (x) = x, for which f (x + h) f (x) (x + lim f (x) = lim = h 0 h 0 h h i.e. f (x) = lim ( x + h ) = x h) x xh + h lim = h 0 h h 0 so that f (x) = x The function f (x) = x has the set of all real numbers as its omain, an its erivative f (x) = x has the same omain. COPYRIGHT 998 S570 V.

21 As a further example, we can use the efinition to obtain the erivative of the function f (x) = /x. In this case f (x + h) = x+h giving us x ( x + h) f ( x + h) f ( x ) = = = ( x + h) x h ( x + h) x h h x + h x As h tens to zero the expression So, approaches. ( x + h) x x f (x) = lim = x h 0 ( x + h) x (7) The function f (x) = /x has the set of all non-zero real numbers as its omain, an its erivative f (x) = /x has the same omain. COPYRIGHT 998 S570 V.

22 Question T (a) If f (x) = /(ω x) where ω is a constant, use the efinition of the erivative to fin f (x). (b) If g(t) = /(ω t) use the answer to part (a) to write g (t) an g (t). 3 Don t be misle into thinking that if a function exists then its erivative must also exist. The function f (x) = x ( the moulus of x ) illustrates the situation where the erivative has a smaller omain than the function (see Figure 3). The moulus of x is efine by x = x if x > 0 x = x if x < 0 an f (x) x For positive x the graph of f (x) = x is a straight line with graient, so while for negative x the graph is a straight line with graient, so f (x) = if x < 0 COPYRIGHT 998 3The graph of f(x) = x. Figure 3 f (x) = if x > 0 S570 V.

23 However, when x = 0 no unique limit exists since we get ifferent values for the limit epening on how we calculate it. In particular, if we approach x = 0 from the positive sie, where x > 0 an h is positive f (0 + h) f (0) h 0 h = lim = lim = lim h 0 + h 0 + h h h h 0 + whereas, if we approach x = 0 from the negative sie, where x < 0 an h is negative f (0 + h) f (0) h 0 h lim = lim = lim = h 0 h 0 h 0 h h h So, if h approaches zero through positive values the limit is, but if h approaches zero through negative values the limit is. A unique limit exists only if we obtain the same result no matter how h approaches 0. Since ifferent answers have been obtaine in this case there is no unique limit an hence no erivative at x = 0. Although the original function is efine for all real values of x the erivative is only efine for non-zero values of x. On the whole, the omains of functions are of more concern to mathematicians than to physicists. Nonetheless, it is important that you shoul be aware of their significance, an prepare to investigate them if necessary. COPYRIGHT 998 S570 V.

24 Question T3 (a) Use your calculator to convince yourself that exp (h) h is very nearly equal to when h is very small. (b) Let f (x) = exp (x). What oes part (a) tell you about f (0)? 3 exp (h) (c) Use the fact that lim =, an the efinition of the erivative to show that f (x) = exp (x). h 0 h.3 Notation for erivatives Various notations are in common use for the erivative of a given function. Two such notations have alreay been introuce in this moule an you will probably meet a thir elsewhere. Here we provie a brief summary. Function notation This is the notation we have mainly use so far, in which a function is represente by f or f (x) or some similar symbol an its erivative (the erive function) is represente by f or f (x). This is a neat an compact notation but care is neee in hanwritten work to make sure that the all-important prime ( ) is clearly visible. COPYRIGHT 998 S570 V.

25 Leibniz notation This is the y/ notation that is especially popular among physicists. We will make much use of it in what follows. y f ( x ) or ( x ). Sometimes these are abbreviate to y (in which case it is assume that we are regaring the variable y as a function y(x) of x), or If we let y = f (x), then in Leibniz notation the erivative is written f (in which case it is assume that we are iscussing a function f (x)). An avantage of Leibniz notation is that it is also possible to write the erivative as [ f ( x )] so that for a specific function, f (x) = x say, we coul write ( x ) = x COPYRIGHT 998 S570 V.

26 Newtonian notation A ifferent notation that is especially common in mechanics, but which will not be use in this moule, uses a ot to enote erivatives. Thus, if x(t) represents the x coorinate of a moving particle at time t, then x (t ) represents the rate of change of x with respect to time, i.e. the x component of the particle s velocity. Given that y = f (x) = /x, an recalling Equation 7, etermine the following: (a) y (), 3(b) f ( a),3(c) t t,3() f ( + 3x). COPYRIGHT 998 S570 V.

27 3 Derivatives of simple functions Now that the efinitions are out of the way we can get own to the real business of oing calculus. The proceure by which erivatives are etermine is calle ifferentiation. In practice everyboy who uses ifferentiation regularly, knows the erivatives of various stanar functions (x, sin x, exp (x), etc.) an knows some simple rules for fining the erivatives of various combinations (sums, ifferences, proucts, quotients an reciprocals) of those stanar erivatives. The formal efinition of a erivative is rarely use in practice. This section introuces the stanar erivatives an the basic rules for combining them. COPYRIGHT 998 S570 V.

28 Table 3. Derivatives of basic functions Tables a an b list a number of functions with which you shoul alreay be familiar along with their erivatives. Each of these erivatives can be euce from the efinition given in the last section, though the proof is not always easy. If you are going to use calculus frequently you will nee to know these erivatives or at least know where you can look them up quickly. 3Some stanar erivatives. (a) (b) f (x) f (x) f (x) f (x) k (constant) 0 0 kxn nkxn sin kx k cos kx tankx coseckx seckx cotkx exp(kx) log (kx) cos kx e COPYRIGHT 998 ksec kx kcosec kxcot kx kseckx tankx kcosec kx kexp (kx) k sin kx /x S570 V. xn nxn sin x cos x cos x sin x tanx cosec x exp(x) log (x) sec x cot x e sec x secxtan x cosec x exp(x) cosec x cot x /x

29 When using the tables it is important to remember the following points: Table 3Some stanar erivatives. (a) (b) o n an k are constants. f (x) f (x) f (x) f (x) o The functions in Table b are special cases of those in Table a, corresponing to k =. k (constant) 0 0 kxn nkxn sin kx k cos kx o In each of the trigonometric functions x must be an angle in raians or a imensionless real variable. tankx coseckx seckx cotkx exp(kx) log (kx) cos kx e COPYRIGHT 998 ksec kx kcosec kxcot kx kseckx tankx kcosec kx kexp (kx) k sin kx /x S570 V. xn nxn sin x cos x cos x sin x tanx cosec x exp(x) log (x) sec x cot x e sec x secxtan x cosec x exp(x) cosec x cot x /x

30 Table Question T4 Use the efinition of the erivative to show that if f (x) = xn an n is a positive integer, then f (x) = nxn. 3 The best way to get to know the stanar erivatives is to use them frequently. Here are a few questions to start the process of familiarization. 3Some stanar erivatives. (a) (b) f (x) f (x) f (x) f (x) k (constant) 0 0 kxn nkxn sin kx k cos kx tankx coseckx seckx cotkx exp(kx) log (kx) cos kx e COPYRIGHT 998 ksec kx kcosec kxcot kx kseckx tankx kcosec kx kexp (kx) k sin kx /x S570 V. xn nxn sin x cos x cos x sin x tanx cosec x exp(x) log (x) sec x cot x e sec x secxtan x cosec x exp(x) cosec x cot x /x

31 Table Fin the erivatives (using Table ) of the following functions: (a) f (x) = x3, 3 (b) g(t) = /t3, (c) h(x) = 5. x Which function in Table is its own erivative? 3Some stanar erivatives. (a) (b) f (x) f (x) f (x) f (x) k (constant) 0 0 kxn nkxn sin kx k cos kx tankx coseckx seckx cotkx exp(kx) log (kx) cos kx e COPYRIGHT 998 ksec kx kcosec kxcot kx kseckx tankx kcosec kx kexp (kx) k sin kx /x S570 V. xn nxn sin x cos x cos x sin x tanx cosec x exp(x) log (x) sec x cot x e sec x secxtan x cosec x exp(x) cosec x cot x /x

32 Question T5 Fin where the graients of the tangents to the graph of y = sin x have the following values: (a) 0, (b), (c) +. 3 Question T6 (a) Use your calculator (making sure that it is in raian moe) to investigate the limits lim h 0 cos (h) h 3an3 lim sinh h h 0 Using successively smaller values of h (e.g. h = ±0., ±0.0, etc.) try to estimate the limit in each case. (b) Use the limits obtaine in part (a) to fin ( sin x ) from the general efinition of the erive function. 3 (Hint: Use the trigonometric ientity sin (A + B) = sin A cos B + cos A sin B.) COPYRIGHT 998 S570 V.

33 Table Question T7 Which functions in Table b have erivatives which are: (a) always positive; (b) always negative; (c) can be positive or negative? 3 3Some stanar erivatives. (a) (b) f (x) f (x) f (x) f (x) k (constant) 0 0 kxn nkxn sin kx k cos kx tankx coseckx seckx cotkx exp(kx) log (kx) cos kx e COPYRIGHT 998 ksec kx kcosec kxcot kx kseckx tankx kcosec kx kexp (kx) k sin kx /x S570 V. xn nxn sin x cos x cos x sin x tanx cosec x exp(x) log (x) sec x cot x e sec x secxtan x cosec x exp(x) cosec x cot x /x

34 3. Derivative of a sum of functions While it is possible, in theory, to fin the erivative of any given function from the efinition, this woul in fact be an aruous process. In practice, those using calculus employ a set of simple rules which can be applie to combinations of functions to fin the erivatives of a wie variety of functions with relative ease. The first of these simple rules is calle the sum rule, which states: The erivative of a sum of functions is the sum of the erivatives of the iniviual functions. If f (x) an g(x) are two functions, this can be written as: sum rule [ f (x) + g(x)] = [ f (x)] + [g(x)] (8) Alternatively we may write this rule in the form 3or just3(f + g) = f + g [ f (x) + g(x)] = f (x) + g (x) This rule enables us to ifferentiate functions such as y = x/ + loge x. To o so we first note that it is the sum of two stanar functions f (x ) = x/ an g(x) = loge x, we then ifferentiate each of these functions an finally a the erivatives to obtain the require answer. COPYRIGHT 998 S570 V.

35 Table So, if f (x) = x / 3an3g(x) = log x e then, from Table f = x an 3 3 g = x so, y f g = + = x + x an if 3 then 3 y = x + loge x y = + x x It is usually a goo iea to think before applying the rule, as in the following question. 3Some stanar erivatives. (a) (b) f (x) f (x) f (x) f (x) k (constant) 0 0 kxn nkxn sin kx k cos kx tankx coseckx seckx cotkx exp(kx) log (kx) cos kx e COPYRIGHT 998 ksec kx kcosec kxcot kx kseckx tankx kcosec kx kexp (kx) k sin kx /x S570 V. xn nxn sin x cos x cos x sin x tanx cosec x exp(x) log (x) sec x cot x e sec x secxtan x cosec x exp(x) cosec x cot x /x

36 Table Fin the erivative of cos x sin x h(x) = +. sin x cos x The rule erivative of a sum equals the sum of the erivatives applies equally well if more than two functions are ae together. This allows us to obtain the erivatives of functions such as those in the following question. 3Some stanar erivatives. (a) (b) f (x) f (x) f (x) f (x) k (constant) 0 0 kxn nkxn sin kx k cos kx tankx coseckx seckx cotkx exp(kx) log (kx) cos kx e COPYRIGHT 998 ksec kx kcosec kxcot kx kseckx tankx kcosec kx kexp (kx) k sin kx /x S570 V. xn nxn sin x cos x cos x sin x tanx cosec x exp(x) log (x) sec x cot x e sec x secxtan x cosec x exp(x) cosec x cot x /x

37 Table Question T8 Use the sum rule to fin y/ in each of the following cases: (a) y = + 3 x + ex (b) y = ( + x ) (c) y = loge (xex) () y = cot x sin x + tan x cos x. (Hint: Write each function as a sum of functions appearing in Table.) 3 3Some stanar erivatives. (a) (b) f (x) f (x) f (x) f (x) k (constant) 0 0 kxn nkxn sin kx k cos kx tankx coseckx seckx cotkx exp(kx) log (kx) cos kx e COPYRIGHT 998 ksec kx kcosec kxcot kx kseckx tankx kcosec kx kexp (kx) k sin kx /x S570 V. xn nxn sin x cos x cos x sin x tanx cosec x exp(x) log (x) sec x cot x e sec x secxtan x cosec x exp(x) cosec x cot x /x

38 Question T9 If you were to use the efinition of the erive function (Equation 6) f ( x + h) f ( x ) y = lim h 0 h 3 (Eqn 6) to show that the erivative of the sum is the sum of the erivatives, what assumption must you make about the limit of a sum? COPYRIGHT 998 S570 V.

39 3.3 Derivative of a constant multiple of a function Table gives the erivative of sin x, for example, but what about sin x? In other wors, what is the erivative of a constant times a function? The answer is given by the constant multiple rule, which states: The erivative of a constant times a function is equal to the constant times the erivative of the function. Using k for the constant an f (x) for the function, this result can be written as: constant multiple rule i.e. 3 [kf ( x )] = k [ f ( x )] [k0f (x)] = k0f (x) or just 3(k0f) = k0f Therefore, for the above example, ( sin x) = (sin x) = cos x similarly, (πe x ) = π (e x ) = π e x = πe x COPYRIGHT 998 S570 V. (9)

40 Differentiate h(x) = x + x 3 Use the constant multiple rule to show that [ f (x) g(x)] = f (x) g (x). This last result shows that the erivative of the ifference of two functions is the ifference of the erivatives, i.e. [ f (x) g(x)] = [ f (x)] [g(x)] COPYRIGHT 998 (0) S570 V.

41 Question T0 Fin the erivatives of the following functions: 4 3 (a) x (b) loge x x x x (c) sin + cos () (π x) π (f) tan x + ( tan x) 4 (e) sin (x + ) tan A + tan B Hint: tan ( A + B) = tan A tan B Question T 3 If k is a constant an f (x) a function, write own the efinition of f (x) an [kf (x)]. What assumption must you make about limits to justify the conclusion [kf (x)] = kf (x)?4 COPYRIGHT 998 S570 V.

42 3.4 Derivative of a prouct of functions This subsection introuces yet another way of combining functions an obtaining the erivative of the result. This is the prouct of two functions, f (x)g(x). Our aim is to express the erivative of such a prouct in terms of the erivatives of the functions f (x) an g(x) that are multiplie together to form the prouct, but the answer is not as obvious as it was for the sum or ifference of functions. The obvious answer that the erivative of a prouct is the prouct of the erivatives is wrong. You cannot simply multiply erivatives. COPYRIGHT 998 S570 V.

43 To convince yourself of this consier f (x) = x an g(x) = x3. Then f (x)g(x) = x x3 = x5. We can ifferentiate each of these functions using Table : implies3f (x) = x thus, f (x) = x 3Some stanar erivatives. (a) 3g (x) = 3x an f(x)g(x) = x implies3[f (x)g(x)] = 5x Clearly, [f (x)g(x)] f (x) g (x) 5 4 (b) f (x) f (x) f (x) f (x) k (constant) 0 0 kxn nkxn sin kx k cos kx tankx coseckx seckx cotkx exp(kx) log (kx) cos kx g(x) = x3 implies Table e COPYRIGHT 998 ksec kx kcosec kxcot kx kseckx tankx kcosec kx kexp (kx) k sin kx /x S570 V. xn nxn sin x cos x cos x sin x tanx cosec x exp(x) log (x) sec x cot x e sec x secxtan x cosec x exp(x) cosec x cot x /x

44 3 3g(x) = 4x Fin f (x), g (x), [f (x)g(x)] an show that [f (x)g(x)] f (x) g (x). Let f (x) = x an The correct etermination of [ f (x)g(x)] involves not only f (x) an g (x) but also f (x) an g(x). It is given by the prouct rule an is easier to express in symbols than in wors: prouct rule f g [ f (x)g(x)] = g(x) + f (x) 3 i.e. [ 0f (x)g(x)] = f (x)g(x) + f (x)g (x) or just () 3(0fg) = f g + fg The expression of this rule in terms of wors is probably not very helpful: The erivative of f times g equals the erivative of f times g plus f times the erivative of g. COPYRIGHT 998 S570 V.

45 You may fin Figure 4 a more useful memory ai. f (x) = x3an3g (x) = 3x In the case of f (x) = x an g(x) = x 3 (iscusse earlier) where g (x) f (x) the application of the prouct rule to f (x)g(x) f g [ f (x)g(x)] = g(x) + f (x) f (x) (Eqn ) an 5x4 is what we expect for the erivative of f (x)g(x) = xx3 = x5. If f (x) = x an g(x) = apply the prouct rule to fin the erivative of f (x)g(x). 4x COPYRIGHT f (x) g (x) Figure 4 Differentiating a prouct of functions. [ f (x)g(x)] = f (x)g(x) + f (x)g (x) = x x3 + x 3x = x4 + 3x4 = 5x4 3 gives g (x) S570 V.

46 Illustrations of the prouct rule With the ai of the prouct rule we can now ifferentiate many more functions. The important step is to express the function that is to be ifferentiate in terms of the sum, ifference or prouct of functions with known erivatives. Here are some examples. Example Differentiate sin x cos x. f(x) = sin x3an3g(x) = cosx for which f (x) = cosx3an3g (x) = sin x Solution sin x cos x may be written as a prouct of the functions Using the prouct rule f g [ f (x)g(x)] = g(x) + f (x) (Eqn ) we can therefore write ( sin x cos x) = ( cos x)(cos x) + ( sin x)( sin x) COPYRIGHT 998 S570 V. = (cos x sin x)4

47 Differentiate xex + sec x. Example xex sec x Solution + is the sum of two functions xex an sec x, each of which is a prouct. So we ifferentiate x xe first an then sec x (using the prouct rule for each) an then a the answers using the sum rule. First, xex may be written as the prouct of the functions 3an3g(x) = e f (x) = 3an 3g (x) = e f (x) = x x x for which then, using the prouct rule f g [ f (x)g(x)] = g(x) + f (x) we can write x (xe x ) = e x (x) + x (e ) = e x + x e x = ( + x)e x Secon, sec x = sec x sec x, therefore giving us f(x) = sec x 3an3g(x) = secx f (x) = secx tanx 3an3g (x) = sec xtanx COPYRIGHT 998 S570 V. (Eqn )

48 Then, again using the prouct rule f g [ f (x)g(x)] = g(x) + f (x) (Eqn ) we fin (sec x) = (sec x tan x) (sec x) + (sec x) (sec x tan x) = sec x tan x Thus, combining these results, an using the sum rule we fin (xex + sec x) = (xex) + (sec x) = ( + x)ex + sec x tan x 3 The kinetic energy of a particle of mass m moving with a spee v(t) that varies with time t, is m[v(t)] /. Use the prouct rule f g [ f (x)g(x)] = g(x) + f (x) (Eqn ) v to show that the rate of change of m[v(t)] / with time is mv(t ) (t ). t COPYRIGHT 998 S570 V.

49 Question T Fin the erivative of each of the following functions: (a) x loge x, (b) (sin x cos x), (c) ( x )( + 3 x ), () e x, (e) sec x cot x, (f) x3f(x), where F(x) is an arbitrary function Question T3 The instantaneous motion of a particle moving along a straight line (call it the x-axis) can be escribe, at time t, in terms of its isplacement sx(t) from a fixe reference position, its velocity vx(t) an its acceleration ax(t). These quantities are efine in such a way that vx s ax(t) = (t ) an v x(t) = x (t ) t t 3 3 If the isplacement is given as a function of time by 3where ω an α are constants a (t) = αv (t) (ω + α )s (t)333 sx(t) = [cos (ω t) sin (ω t)]exp(α t) show that x x x COPYRIGHT 998 S570 V.

50 Question T4 This question extens the prouct rule to three functions. f g [ f (x)g(x)] = g(x) + f (x) (Eqn ) If f (x), g(x), h(x) are any three functions, show that [ f ( x )g( x )h( x ) ] = f ( x )g( x )h( x ) + f ( x ) g ( x )h( x ) + f ( x )g( x ) h ( x ) Question T5 Use the result evelope in Question T4 to fin x (e loge x sin x ) 3 COPYRIGHT 998 S570 V. 3

51 3.5 Derivative of a quotient of functions It is frequently the case that we nee to fin the erivative of the quotient f (x) of two functions f (x) an g(x). g(x) The erivative of a quotient, like that of a prouct, epens not only on the values of f (x) an g (x), but also on f (x) an g(x). The quotient rule states: quotient rule i.e. g f f (x) g(x) f (x) g(x) = [g(x)] f (x) f (x)g(x) f (x)g (x) g(x) = [g(x)] () 3or just3 gf = f gg fg COPYRIGHT 998 S570 V.

52 You may fin Figure 5 a useful memory ai. f (x) To apply this result g f f (x) g(x) f (x) g(x) = [g(x)] to the function y = take so that Then (Eqn ) sin x x 3an3g(x) = x f (x) = cosx3an3g (x) = f (x) f (x) f (x) = sin x g (x) 3The quotient rule. Figure 5 y (cos x ) x (sin x ) x cos x sin x = = x x COPYRIGHT 998 g (x) 3 S570 V. g (x) [g (x)]

53 Write y = x3 as x5 /x an use the quotient rule g f f (x) g(x) f (x) = [g(x)] g(x) (Eqn ) to fin y/. Use the quotient rule to fin F( x ) in terms of F (x) an F(x). You may fin this useful, but there is no point in remembering it if you know the quotient rule since it is just a special case of that more general rule. COPYRIGHT 998 S570 V.

54 Question T6 Fin y/ in each of the following cases: (a) y = x ex 3(b) y = cos xx 3(c) y = x 5 x4 + x3 x ( x + ) log x 3() y = x tan x e Question T7 If f (x) = (x ) an g(x) = xex, fin the following: (a) g( x ) f (x) () f ( x ) g( x ) ] (b) f ( x )g( x ) ] (c) [ [ f ( x ) g( x ) 3 The aim of the next question is to erive the quotient rule from the prouct rule. It begins by showing how to obtain the reciprocal rule without using the quotient rule. COPYRIGHT 998 S570 V.

55 Question T8 (a) Let f (x) be an arbitrary function an let h(x) = /f (x) so that f (x)h(x) = Differentiate both sies of this equation an hence obtain a formula for f ( x ) (similar to that given in Equation 3). the reciprocal rule F (x) F(x) = [F(x)] (Eqn 3) f (x) = f (x), then use the answer to part (a) an g( x ) g( x ) f g the prouct rule [ f (x)g(x)] = g(x) + f (x) (Eqn ) f g g(x) f (x) f (x) = to erive the quotient rule (Eqn ) [g(x)] g(x) (b) Let f (x) an g(x) be any two functions, an write 3 COPYRIGHT 998 S570 V.

56 Question T9 Use (sin x ) = cos x, (cos x ) = sin x, an the quotient rule f g g(x) f (x) f (x) = [g(x)] g(x) 3 to ifferentiate (a) tan x, an (b) cot x. COPYRIGHT 998 S570 V. (Eqn )

57 Question T0 (a) Use the reciprocal rule to fin F (x) F(x) = [F(x)] (Eqn 3) x (e ). x (e + e x ) is sometimes known as the hyperbolic cosine of x (an is enote by cosh x ). (b) The function f (x) = Fin f an show that [ f ( x ) ] f =. COPYRIGHT S570 V.

58 Question T Let F(x) = G 3 Mm where G, M, m an a are constants. Fin F (x). ( x a) Question T (sin x) = cos x an (cos x) = sin x, along with the reciprocal rule, F (x) = (Eqn 3) F(x) [F(x)] Using 3 fin the erivatives of (a) sec x an (b) cosec x. COPYRIGHT 998 S570 V.

59 4 More about logarithmic an exponential functions 4. Why exp(x) is consiere to be special Of all the functions liste in Table, f (x) = exp (x) = ex is the only one which is its own erivative. Far from being a technical curiosity this fact turns out to be of great significance an ultimately explains why the exponential function plays such an important role in physics. COPYRIGHT 998 S570 V.

60 It also has an interesting interpretation in terms of the graph of y = ex, shown in Figure 6. Since f (x) = ex the graient of the tangent to the graph at the point A which has coorinates x = a an y = ea is f (a) = ea which is the same as the height of A above C. y y = ex A 4 So the rate of change of f (x) when x = a is simply f (a), that is, the rate of change of the exponential function at any point is equal to the value of the function itself at that point. In Figure 6 the point B has coorinates (a, 0). Since A is the point (a, ea), the line AB has a graient of CA e a = = ea BC (a, e a ) 3 an is therefore the tangent to the graph at A. So, to raw a tangent to y = ex at (a, ea ), simply join (a, ea ) to (a, 0) an the resulting line is boun to have a graient of ea. B 0 C (a, 0) (a, 0) 3The graph of y = e. Figure 6 COPYRIGHT 998 S570 V. x x

61 Is f (x) = ex the only function which is its own erivative? (Think about the rules of ifferentiation that you alreay know.) Using the prouct rule f g [ f (x)g(x)] = g(x) + f (x) (Eqn ) fin the erivatives of (a) ex, (b) e3x, (c) e4x. Deuce the formula for the erivative of e nx where n is any positive integer. A similar result hols true even when n is not a positive integer. This was given in Table, but it eserves to be emphasize again here. kx (e ) = kekx for any constant k COPYRIGHT 998 (4) S570 V.

62 Question T3 If y = Ce x y for some constant C, show that = y( y). x + Ce Question T4 Differentiate f (x) = (ex + e x)(ex e x). 3 COPYRIGHT 998 S570 V.

63 4. Derivative of ax The previous subsection consiere the erivative of ex. However, functions such as x an πx sometimes arise in physical applications, an so it is necessary to know how to ifferentiate any function of the form ax, where the positive number a being raise to the power x is not necessarily e. As an example we will consier the ifferentiation of x, but the metho is of general applicability. First note that exp is the inverse function of loge, this means that exp reverses the effect of loge so that e loge = exp (log e ) = since = exp (log e ) So, x = [exp (log e )]x = exp(x log e ) x ( ) = [exp(x log e )] COPYRIGHT 998 S570 V.

64 Now, from Equation 4 (or Table ) kx (e ) = kekx for any constant k (Eqn 4) we know that kx (e ) = kekx for any constant k an, if we choose k to be loge, this gives us x ( ) = [exp(x log e )] = (log e )[exp(x log e )] = (log e ) x A similar argument applies with any positive number, a say, in place of. So we have the result: (ax) = (loge a)ax (a > 0) (5) Rather than trying to remember this result you woul probably be wiser to try to remember the metho that was use to obtain it. Notice the way that the properties of exponentials an logarithms have been use to express a function of the form ax in terms of ekx, an how the simple properties of ekx have then been exploite. COPYRIGHT 998 S570 V.

65 Fin (πx). Apply the formula for ifferentiating ax (Equation 4) kx (e ) = kekx for any constant k (Eqn 4) when a = e. Question T5 Differentiate each of the following functions: (a) f (x) = 3x, 3(b) f(x) = 3,3(c) f(x) = ( + 3),3() f(x) = e.3 x COPYRIGHT 998 x x x S570 V.

66 4.3 Derivative of loga x As state in Table : (loge x) = x From this it is easy to euce the erivative of loge kx, where k is a constant (log (kx)) = (log k) + (log x) = x loge (kx) = loge k + loge x so e e e COPYRIGHT 998 S570 V.

67 If we ha been compelle to ifferentiate log 0 x or log x the problem woul have been a little more ifficult, but still not intractable. The general metho for ifferentiating logs to an arbitrary base, such as loga x is remarkably similar to that for ealing with a x: it involves expressing loga x in terms of loge x which can be ifferentiate easily. As an example, let us consier log0 x. To express it in terms of loge x we write y = log0 x x = 00y then an, taking logarithms to the base e of both sies of this equation, we obtain an recalling that y = log x this can be rearrange to give log x y = log x = log 0 loge x = loge (00y) = y loge 0 0 e 0 e COPYRIGHT 998 S570 V.

68 We may now easily ifferentiate log0 x because /loge 0 is a constant, so log e x (log0 x) = log e 0 (log e x) = = = log e 0 log e 0 x x log e 0 A similar argument applies for any positive base a giving the general result: ( log a x ) = x log e a (6) If a = e, check that the above formula gives our previous result for the erivative of loge x. COPYRIGHT 998 S570 V.

69 Differentiate logπ x. Question T6 Differentiate each of the following functions: (a) f (x) = log x (b) f (x) = log0 x log x (c) f (x) = [Hint: loga b = (loga c)(logc b)] log0 x () 3 f(x) = a log x3where a is a positive constant. 3 x a COPYRIGHT 998 S570 V.

70 4.4 Resiste motion uner gravity: an example Imagine a parachutist falling to Earth with velocity, vx(t) ownwars, at time t. For convenience we will choose the positive x-axis to be the ownwar vertical, with the origin at the point where the parachute opens. The parachutist is subject to two forces, gravity acting ownwars an the resistance of the air on his parachute acting upwars. Suppose that the resistive force is proportional to [vx(t)] with a constant of proportionality k (which has to be etermine experimentally). It follows from Newton s secon law that the equation governing the motion is m [v x(t)] = mg k[vx(t)] (7) t where m is the mass of the parachutist an g is the magnitue of the acceleration ue to gravity. The forces on the right-han sie have opposite sign because they act in opposite irections. Now, suppose you want to check that the following expression for vx(t) satisfies Equation 7 vx(t) = mg k + B exp ( t gk m ) B exp ( t gk m ) (8) where B is a constant (etermine by the spee of the parachutist at time t = 0). It appears that we must first ifferentiate v x(t), which is a rather complicate multiple of the quotient of two functions. COPYRIGHT 998 S570 V.

71 We coul certainly procee irectly, an just ifferentiate the function as it stans, but a little thought shoul tell you that the algebra is going to get very nasty; so it is worth trying to simplify the expression. To simplify the notation let A = mg k an α = gk m, an let us write vx rather than vx(t). So we have a simplifie version of Equation 8 vx(t) = mg k + B exp ( t gk m ) B exp ( t gk m ) (Eqn 8) + B exp( α t) vx = A B exp( α t) COPYRIGHT 998 (9) S570 V.

72 Now, using the quotient rule f g g(x) f (x) f (x) = [g(x)] g(x) (Eqn ) to ifferentiate this equation + B exp( α t) vx = A B exp( α t) (Eqn 9) with respect to t, we fin ( + Be α t ) ( Be α t ) ( Be α t ) ( + Be α t ) vx t t = A α t ) t ( Be ( Bα e α t )( Be α t ) (Bα e α t )( + Be α t ) vx = A t ( Be α t ) COPYRIGHT 998 S570 V.

73 ( Bα e α t + B α e α t ) (Bα e α t + B α e α t ) vx = A t ( Be α t ) Bα e α t vx = A α t t ( Be ) Substituting this expression for vx(t)/t into the left-han sie of Equation 7 m [v x(t)] = mg k[vx(t)] (Eqn 7) t an using the fact that Aα = g we obtain Bα e α t 4mgBe α t = ma α t α t ( Be ) ( Be ) COPYRIGHT 998 S570 V.

74 an substituting the same expression for vx(t)/t Bα e α t vx = A α t t ( Be ) into the right-han sie of Equation 7 m [v x(t)] = mg k[vx(t)] t (Eqn 7) an using the fact that ka = mg we obtain ( Be α t ) ( + Be α t ) + Be α t mg mg = mg α t Be ( Be α t ) α t 4mgBe = ( Be α t ) COPYRIGHT 998 S570 V.

75 Thus, the two sies of Equation 7 m [v x(t)] = mg k[vx(t)] t (Eqn 7) are inee equal if vx has the form given in Equation 8. vx(t) = mg k + B exp ( t gk m ) B exp ( t gk m ) (Eqn 8) This really completes the ifferentiation an subsequent manipulation, but having introuce an expression (Equation 8) for the ownwar velocity uner gravity in the presence of a resistive force it is worth noting at least one of its mathematical properties. What is the behaviour of v x(t) as t becomes large? In other wors, what is lim [ vx (t ) ]? t COPYRIGHT 998 S570 V.

76 What is the terminal velocity if m = 80 kg, k = 30 kg m an g = 9.8 m s? Question T7 Show that y(x) = log e x satisfies the equation log e + x y (x) loge + [ + y(x)] 3 COPYRIGHT 998 S570 V.

77 5 Closing items 5. Moule summary 3 A function, f say, is a rule that assigns a single value of the epenent variable, y say, (in the coomain) to each value of the inepenent variable x in the omain of the function, such that y = f (x). The rate of change, or erivative of a function f (x) at x = a, can be interprete as the graient of the graph of the function at the point x = a. Some stanar erivatives are given in Table a (repeate here for reference). The erivative is efine more formally in terms of a limit, an may be represente in a variety of ways. If y = f (x) y f ( x + h) f ( x ) y f ( x ) = = lim = lim x 0 x h 0 h COPYRIGHT 998 S570 V.

78 4 The sum, constant multiple, prouct an quotient rules for ifferentiating combinations of functions are as follows: sum rule (f + g) = f + g constant multiple rule (k0f ) = kf prouct rule (fg) = f g + fg f f g fg = g g quotient rule The erivatives of the stanar functions are given in Table. The erivatives of ax an loga x are given by (ax) = (loge a)ax (Eqn 5) an ( log a x ) = (Eqn 6) x log e a You shoul be aware that logarithms to ifferent bases are relate by loga b = (loga c)(logc b) COPYRIGHT 998 S570 V. Table (a) Some stanar erivatives. f (x) f (x) k (constant) 0 kxn nkxn sin kx k cos kx coskx tankx coseckx seckx cotkx exp(kx) log (kx) e ksec kx kcosec kxcot kx kseckx tankx kcosec kx kexp (kx) k sin kx /x

79 5. Achievements Having complete the moule, you shoul be able to: A Define the terms that are embolene an flagge in the margins of the moule. A Define the erivative of a function (at a point) as a limit an give the efinition a graphical interpretation. A3 Use the efinition of the erivative as a limit to calculate the erivative of simple functions. A4 Know the erivatives of a range of stanar functions. A5 Use aition, subtraction, multiplication an ivision to express a given function in terms of simpler ones in appropriate cases. A6 Differentiate a constant multiple of a function. A7 Apply the rule for ifferentiating the sum (an ifference) of functions. A8 Apply the prouct rule of ifferentiation. A9 Apply the quotient rule of ifferentiation. A0 Use the properties of logarithms an exponentials to etermine the erivatives of loga x an ax. COPYRIGHT 998 S570 V.

80 Stuy comment You may now wish to take the Exit test for this moule which tests these Achievements. If you prefer to stuy the moule further before taking this test then return to the Moule contents to review some of the topics. COPYRIGHT 998 S570 V.

81 5.3 Exit test Stuy comment Having complete this moule, you shoul be able to answer the following questions, each of which tests one or more of the Achievements. Question E 3 (A, A an A4) Write own the erivative of f (x) = loge x. As x increases from zero how oes the graient of the tangent to the graph change? What is the behaviour of loge x as x approaches infinity? Question E (A3) 3Using the efinition of the erivative as a limit, ifferentiate f(x) = /x. COPYRIGHT 998 S570 V.

82 3 Question E3 (A4, A5 an A6) Fin y/ for each of the following functions: (a) y = x(x )(x + ) (b) y = sin (x + π/4) (Hint: sin (A + B) = sin A cos B + cos A sin B) (c) y = loge x COPYRIGHT 998 S570 V.

83 Question E4 3 f(x) = x e x + x + x f(x) = x x sin x f(x) = log x exp ( x + log x ) f(x) = x f(x) = a x (A4 to A9) Fin f (x) for each of the following functions: (a) f (x) = x cos x + x sin x (b) x (c) () e (e) (f) e x a COPYRIGHT 998 S570 V.

84 Question E5 3 (A4, A5, A6, A8 an A9) If n is a positive integer, the series x n + + x + x + x3 + + xn has the sum x 3 3 show that the sum of the series 3 3 nx + x + 3x + + nxn is n + ( n + ) x n + ( x ) COPYRIGHT 998 S570 V.

85 Question E6 (A7 an A8) 3If f an g are functions show that + = f (x)g(x) f (x) g(x) f (x) g(x) Stuy comment This is the final Exit test question. When you have complete the Exit test go back to Subsection. an try the Fast track questions if you have not alreay one so. If you have complete both the Fast track questions an the Exit test, then you have finishe the moule an may leave it here. COPYRIGHT 998 S570 V.