4.2 First Differentiation Rules; Leibniz Notation

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1 .. FIRST DIFFERENTIATION RULES; LEIBNIZ NOTATION 307. First Differentiation Rules; Leibniz Notation In this section we erive rules which let us quickly compute the erivative function f (x) for any polynomial function f(x), an for sinx an cosx. Along the way we will erive a few general (though not comprehensive) rules for erivatives. We will also introuce the very powerful Leibniz notation for erivatives, an show how knowing the erivative helps us to further analyze a function. One consequence is that we can more accurately graph a function s behavior by han...1 Positive Integer Power Rule We will often be intereste in fining erivatives of functions f(x) = x n. Fortunately there is a simple rule which covers all such functions. It is usually calle the power rule, as state below. (Recall N = {1,, 3,, 5, 6, }.) Theorem..1 (f(x) = x n ) (n N) = f (x) = n x n. Note that implicit in this theorem is that the erivative of x n exists for every x R i.e., exists everywhere since it is equal to nx n, efine everywhere. Proof: The proof we give here epens upon the binomial expansion (.36), page 98. It is important to remember that x is a fixe number in the limit, an is the variable approaching zero as far as the limit is concerne. With that in min, f f(x + ) f(x) (x) 0 (x + ) n x n ( ) x n + nx n + n(n)xn () () n x n nx n + (n)(n) 1 x n () + + () n ( nx n + = nx n = nx n, q.e.. ) (n)(n 1) x n () () n 1 The only term which survives in the limit in the fifth line is the nx n term because the others have positive integer powers of, which is approaching zero. We will see later that this power rule is actually much more general. In fact, it can be use for n R but we nee some more avance methos to prove such generality. For now we will apply it only to n N. Example..1 Here we list the erivatives of some of the positive integer powers of x. The first case liste below (n = 1) oes follow from the proof, though we woul be reaing the statement of the theorem for that case f(x) = x 1 = f (x) = 1x 0 = 1. Again we o not really wish to say x 0 = 1 regarless of x, for several technical reasons (though it is fine as long as x > 0), but we see how the formula naively gives us what we want for n = 1. The rest of the table is more straightforwar: f(x) x x x 3 x x 100 f (x) 1 x 3x x 3 100x 99

2 308 CHAPTER. THE DERIVATIVE.. Leibniz Notation We will fin that other erivative rules will be unwiely to write with our present notation. Thus we will introuce the very powerful Leibniz 13 notation an use it except in a few settings where our present (prime) notation is simpler. Definition..1 f(x) = f (x). This is also written f(x), an sometimes shortene to f/ when it is clearly unerstoo that f is a function of x. The symbol is a ifferential operator which takes a function of x an returns the erivative with respect to x. (Thus takes a function as its input, an returns a function as its output.) When we are intereste in position an velocity, we can write s (t) = s(t) t, or when it is unerstoo that s = s(t), we might write v = s t. (.9) Notice that the notation resembles ifference quotients, because, with our efinition of erivatives, we have f(x) f(x) 0, s t t 0 s t. Similarly for any such relate quantities. With Leibniz notation our power rule becomes: (xn ) = nx n. (.10) If we woul like to compute f (a), i.e., the erivative at a particular point, in the Leibniz notation we use a vertical line which is rea, evaluate at, as in f (a) = f(x). x=a So, for example, 5 = 5x, an the slope at x = 1 of the function f(x) = x 5 is given by 1 f (1) = 5 1 = 5, or 5 = 5x x=1 = 5 1 = 5. x=1 ( Note how the Leibniz notation is often assume to act like a fraction: ) x 5 = 5. However the in the numerator, an (separately) the in the enominator are treate as inviolable; we o not ever break those terms up further. 13 Name for Gottfrie Wilhelm Leibniz (July 1, 166 November 1, 1716), a German mathematician an philosopher. Most creit him an English philosopher, mathematician, physicist an theologian Sir Isaac Newton, (December 5,16 March 0, 177) with inepenently iscovering calculus. Much is written about rivalries between the Newton camp an the Leibniz camp, regaring who iscovere what first. Newton s notation for erivative use a ot above the function, as in s/t = ṡ, a notation still use in some physics textbooks. 1 Often the x = is omitte when the variable is obvious, as in 5 = 5x 1 = 5 1 = 5. 1

3 .. FIRST DIFFERENTIATION RULES; LEIBNIZ NOTATION 309 Note the flexibility of the Leibniz notation in the following: 3 = 3x, u 3 u = 3u, t 3 t = 3t. These are actually the same rule (with ifferent variables): that the cube of a quantity changes with respect to that quantity at the (instantaneous) rate of 3 times the square of the quantity, be it x, u or t. Put another way, if the horizontal axis is given by t, an we graph the height t 3 on the vertical axis, then the slope is always 3t. The Leibniz notation also keeps us from making the mistake of trying to use the erivative rules (such as the power rule) to compute, for example, u3. Since the variables (u an x) o not match, the power rule cannot be use irectly. 15 With the power rule (.10) an a few other results we can quickly calculate the erivatives of polynomials. Much of this chapter will be evote to calculating erivatives using known rules, which save an enormous amount of time when compare to calculating erivatives using limits of ifference quotients as in the previous section...3 Sum an Constant Derivative Rules Theorem.. (Sum Rule) Suppose f(x) an g(x) exist. Then (f(x) + g(x)) = f(x) + g(x). (.11) In other wors the erivative of a sum is the sum of the respective erivatives. Before we give the proof, it is worth mentioning that some texts write this using the prime notation: (f + g) = f + g. Proof: Assume that f(x) an g(x) exist at a particular x. Then (f(x + ) + g(x + )) (f(x) + g(x)) (f(x) + g(x)) 0 f(x + ) f(x) + g(x + ) g(x) 0 ( f(x + ) f(x) + 0 ) g(x + ) g(x) f(x + ) f(x) g(x + ) g(x) + lim 0 0 = f(x) + g(x), q.e.. The reason that we coul break this into two limits legitimately is because the two limits both existe an were finite by assumption. (See Theorem 3.9.1, page 66.) 15 Later in the text we will have the chain rule, which helps us get aroun the problem of computing u3, an similar erivatives where the variable in the numerator oes not match the variable of the enominator, i.e., the variable of the ifferential operator (here /). There we will see some of the true power of the Leibniz notation, as we compute for instance u 3 = u3 u u = 3u u. Notice how we apparently multiplie an ivie by u to achieve the secon expression.

4 310 CHAPTER. THE DERIVATIVE Example.. ( x 3 + x + x ) = = 3x + x + 1. With practice, one learns to skip the first step in the example above. Note how = 1, as one might hope. This reflects the fact that x an x change at the same rate (i.e., the ratio of their rates of change is always 1). Put another way, the slope of the line y = x is always 1. The next theorem is usually given separately for emphasis. Theorem..3 The erivative of a constant is zero; if a function is efine by f(x) = C for all x R, where C is some fixe constant, then f (x) = 0 for all x R. Written two ifferent ways, we thus have: f(x) = C = f (x) = 0; (.1) C = 0. (.13) There are several ways to see this. From the ifference quotient limit efinition (.3) (page 96), regarless of 0 the ifference quotient [f(x + ) f(x)]/ = [C C]/ = 0/ = 0, so it remains zero in the limit. From another perspective regaring what we know about lines we have that f(x) = C is a line of slope m = 0. From a qualitative stanpoint this theorem is reasonable since constants have rate of change zero (hence the term constant) with respect to x. Some texts write 16 (C) = 0. With this theorem we can write, for example, ( x ) = ( x 3 ) + ( ) 8 = 3x + 0 = 3x. ( With little or no practice one learns to write simply x ) = 3x. We nee just one more result before we can fin erivatives of arbitrary polynomials. This answers the question of what to o with the coefficients of a polynomial, an multiplicative constants in general. Theorem.. Multiplicative constants are preserve in the erivative. In other wors, (C f(x)) = C f(x). (.1) The proof is left as an exercise. It follows from the fact that multiplicative constants go along for the rie in limits as well. (See again Theorem 3.9.1, page 66.) For a simple example, we have ( 5x 7 ) = 5 ( x 7 ) = 5 7x 6 = 35x 6. Again, with very little practice one learns to compute such a erivative in one step. Note how the erivative operator treats aitive constants (which o not survive) ifferently from 16 One weakness of Taylor s prime notation is that we o not know what variable we are taking the erivative with respect to. For instance, in an earlier example we have fuel volume V as a function of position s, an so V /s measure the flow rate of fuel per mile. However, since s = s(t), we have V = V (s(t)), so ultimately V = V (t), i.e., V can be written as a (algebraically ifferent) function of t instea, in which case we can calculate V /t, measuring the flow rate with respect to time. So when aske to calculate V, or even V (5), there is this ambiguity which is not present in the Leibniz notation. If one wrote V (s), it woul probably be unerstoo to be V/s an not V/t. Similarly, V (5 secons) woul be unerstoo to mean V/t evaluate at t = 5 secons.

5 .. FIRST DIFFERENTIATION RULES; LEIBNIZ NOTATION 311 multiplicative constants (which o survive). We can combine the power rule (.10), (.13), an (.1) to quickly compute the erivative of any given polynomial (where n N): ( an x n + a n x n + + a x ) + a 1 x + a 0 = a n nx n + a n (n 1)x n + + a x + a 1. (.15) To be clear on the logic, note that we first use the sum rule to break this into a sum of erivatives of the a k x k, k = 1,, n an a 0, calculating the erivatives of the a k x k terms in turn, each time using the fact that the ( multiplicative constants a k are along for the rie in what are otherwise simple power rules: ak x k) = a k kx k. The final term a 0 is an aitive constant with erivative zero an thus oes not appear on the right han sie of (.15). Example..3 To see how (.15) can be carrie out quickly, we list a couple of brief examples: ( 5x + 9x + 13x + 7 ) = 5 x x = 0x x ( 9 6x + 5x 11 ) = 0 + ( 6) x 10 = x 10. One learns quickly to think but not necessarily write the first computational step in such problems. Note that the negative sign also goes along for the rie, since it is just a factor of. In fact we coul list a ifference rule, (f(x) g(x)) = f(x) g(x), but that woul be reunant given the sum rule, an how a multiplicative constant, even if negative, is preserve in the erivative. We nee to also point out that to use (.15), we nee to have the function written in the form of the left han sie of that equation, i.e., expane an not left factore. Consier for instance the following example. Example.. [ (x + 1) ] = [ x + x + 1 ] = x 3 + x. In the above we neee to multiply out the polynomial. Thus [(x + 1) ] (x + 1) 1, since we are taking the erivative with respect to x an not (x + 1). It is also important to note that erivatives o not allow variable quantities to go along for [ the rie. Thus ] [ x x 3 x ] x 3. Inee, x x3 = x 3x = 3x 3, while (x x3 ) = x = x 3. Finally we point out again that it shoul be clear from the previous examples that using (.15) for such computations is much simpler than using the original efinition of the erivative (letting 0 in a limit of ifference quotients as in (.3), page 96) to calculate erivatives of polynomials... Increasing an Decreasing Functions; Graphing Polynomials Recall that while f(x) gives the height of the graph y = f(x) at a particular value of x, the erivative f (x) gives the slope there. If the slope is positive the graph is sloping upwars; if negative the graph is sloping ownwars. Another way to speak of such things is to iscuss functions which are increasing or ecreasing on an interval, say (a, b).

6 31 CHAPTER. THE DERIVATIVE Definition.. Consier a function f(x) with an interval (a, b) containe within its omain. 1. We say f(x) is increasing on (a, b) if an only if ( x, y (a, b))(x < y f(x) < f(y)).. We say f(x) is ecreasing on (a, b) if an only if ( x, y (a, b))(x < y f(x) > f(y)). (Note that it is possible that a function is not consistently increasing or consistently ecreasing on a given interval.) Clearly, for an increasing function on (a, b), the height increases as x increases through the interval. Similarly, for a ecreasing function on (a, b), the height ecreases as x increases through the interval. If we know exactly where a function is increasing, an where it is ecreasing, that information can be of great help in plotting or analyzing the function. To see what this has to o with erivatives we state the following theorem. Its proof relies on the Mean Value Theorem which will be introuce in a later section. However, it shoul alreay have the ring of truth given what we know of erivatives an slopes. Theorem..5 Suppose f(x) is efine for x (a, b), an f (x) exists for x (a, b). Then 1. ( x (a, b))(f (x) > 0) = f(x) is increasing on (a, b);. ( x (a, b))(f (x) < 0) = f(x) is ecreasing on (a, b). (Again, if f (x) changes sign on (a, b), then neither of these hol.) Example..5 To see how we might use this to graph polynomials, consier the graph of the function f(x) = x 3 3x. This function is continuous on all of R = (, ). Also notice that lim f(x) [x (1 3 3x )] x x lim f(x) [x (1 3 3x )] x x 1, 1. If we raw a sign chart for f(x), showing where the function is positive an where it is negative, we can get some iea of what the graph looks like. To construct a sign chart for any function we look at all the possible points where the function can change signs. Recall that the Intermeiate Value Theorem (Corollary 3.3., page 196) implies a function f(x) can only change signs, as we increase x, by either passing through zero height or having a iscontinuity. Since our particular f(x) here is continuous on all of R, we look to where f(x) = 0 to ivie R into intervals of constant sign. Now f(x) = x 3 3x = x(x 3) is zero for x = 0, ± 3. This gives us four intervals on which f(x) oes not change signs. We can test for the sign of f(x) at a single point in each interval to get the sign of f(x) on that interval. Doing so as we i in Section 3.3, we construct the sign chart for f(x): Function: f(x) = x(x 3) Test x = 0 1 Sign Factors: 10 Sign f(x): 3 0 3

7 .. FIRST DIFFERENTIATION RULES; LEIBNIZ NOTATION 313 local maximum? local minimum? Figure.5: Rough graph of f(x) = x 3 3x base upon its sign chart an behavior as x ±. In particular we o not know the exact locations of the local maximum(s) or minimum(s) without investigating the erivative of f(x). From the sign chart an the behavior as x ± we can get some iea of what the graph of f(x) looks like. That information is reflecte, however imprecisely, in Figure.5. A serious rawback to such a graph is that we know from the Extreme Value Theorem (Corollary 3.3.1, page 196) that there will be a value in [ 3, 0 ] which is a local maximum, an another in [ 0, 3 ] which is a local minimum, but we o not know exactly where these are from the sign chart of the function (we will formally efine the bolface terms shortly). However, a sign chart for the erivative of f(x) can possibly give us this information. Since f(x) = x 3 3x, it follows quickly that f (x) = 3x 3. Recall that on intervals where f > 0, the function f is increasing, while on those intervals on which f < 0, the function is ecreasing. Since f (x) is also an easily factore polynomial, constructing its sign chart is easy. Note f (x) = 3x 3 = 3(x 1) = 3(x + 1)(x 1) is zero exactly where x = ±1. Test x = Sign f (x) = f (x) = 3(x + 1)(x 1) 0 10 Sign f (x): Behavior of f(x): 1 INC DEC INC ր ց ր Here we use INC to abbreviate increasing, which we also signifie by the arrow pointing upwars (ր), an we use DEC an (ց) to signify ecreasing. From this we euce that we get a local maximum at (, f()) = (, ), an a local minimum at (1, f(1)) = (1, ). These two bits of information allow us to raw a more accurate sketch of the graph of f(x) = x 3 3x, as illustrate in Figure.6. That graph is computer-generate, but we can get a very accurate picture of the function s general behavior by plotting the information we have gathere: the sign

8 31 CHAPTER. THE DERIVATIVE (true) local maximum (true) local minimum Figure.6: Partial graph of f(x) = x 3 3x showing the sign of f(x), the limiting behavior as x ±, an the sign of f (x) (which inicates also the locations of local extrema). The x- intercepts (where f(x) = 0), the local maximum an local minimum points are also illustrate, as are the facts that x = f(x), an x = f(x). of f, incluing the x-intercepts (where f(x) = 0), the limiting behavior of f(x) as x ±, an where f(x) is increasing/ecreasing, incluing any local maximum an minimum points. 17 It is important to istinguish the meanings of a sign chart for f(x), an one for f (x). The former just tells us where the function is below or above the x-axis; the latter tells us where the function is increasing an where the function is ecreasing. In the above we use the following terms, which we now efine: Definition..3 Given a function f(x). 1. We call a point x 0 a local maximum of f(x) if an only if ( (a, b) x 0 )( x (a, b))(f(x) f(x 0 )). (.16). We call a point x 0 a local minimum of f(x) ( (a, b) x 0 )( x (a, b))(f(x) f(x 0 )). (.17) In other wors, x 0 is a local maximum of f(x) if there is an open interval containing x 0 in which the function is never greater than x 0 on that interval. Local minimum is efine analogously. If f(x) is continuous in an open interval aroun x 0, an f exists in that interval, then a change of signs of f at x 0 inicates one of these local extrema. If, for instance, f > 0 to the left of x 0 an f < 0 to the right, then f increases before an ecreases after x 0, making x 0 a local maximum. This can be seen in the erivative sign chart an graph above for our example function f(x) = x 3 3x. Example..6 Use the erivative to etermine where the graph of f(x) = x 6x + 8x is increasing, an where it is ecreasing. Use this information to sketch a graph of y = f(x). 17 It is also worth noticing that f (x) = 3(x + 1)(x 1) for both x an x, an so the slope of f(x) grows larger as x ±. This is not the case with all graphs (see Figure.3, page 300 for example), but it is a nice feature to notice when plotting a graph such as Figure.6 above.

9 .. FIRST DIFFERENTIATION RULES; LEIBNIZ NOTATION 315 Solution: We wish to know where f (x) > 0 an f (x) < 0, so we compute this erivative an construct its sign chart. f (x) = [ x 6x + 8x ] = f (x) = x 3 1x + 8 = f (x) = (x 3 3x + ). To construct the sign chart we nee to solve x 3 3x + = 0. While solving a thir-egree polynomial equation can be quite ifficult, in this case we are somewhat fortunate that x = 1 is one solution: (1) 3 3(1) + = = 0, an so (x 1) is one factor. From this we can use either synthetic ivision or regular long ivision to fin the quotient when x 3 3x+ is ivie by x 1: x + x Hence, we have x 1 ) x 3 3x + x 3 + x x 3x x + x x + x f (x) = (x 1)(x + x ) f (x) = (x 1)(x + )(x 1) f (x) = (x 1) (x + ). From this we get f (x) = 0 x {, 1}. We use this for our sign chart: 0 Function: f (x) = (x 1) (x + ) Test x = 0 0 Sign f (x) = 10 Sign f (x): Behavior of f(x): 1 DEC INC INC ց ր ր The graph of f(x) will clearly have a local minimum at x =. At x = 1 we have a curious situation where the graph is increasing on the intervals (, 1) an (1, ), but has slope zero at x = 1. This is illustrate in Figure.7, showing that the curve briefly levels off at x = 1. In fact, f(x) here can be sai to be increasing on [, ) an ecreasing on (, ]. Whether a function is increasing or ecreasing (or neither) is technically a escription of its behavior on an interval, not at a particular point. (See Definition.., page 31.) Note also that [ lim f(x) x 6x + 8x ] x ± x ± x ± [x (1 6x + 8x 3 )] 1.

10 316 CHAPTER. THE DERIVATIVE (1, 3) (, ) Figure.7: Partial graph of f(x) = x 6x +8x. From the erivative s sign chart an form f (x) = x 3 1x + 8 = (x 1) (x + ), it is clear that f (x) = 0 at x =, 1, but while the function has a local (inee global ) mimimum at x =, the graph only levels out at x = 1, an continues to increase on all of [ ). We can easily see one x-intercept occurs at x = 0 (since f(0) = 0 clearly), an from our computergenerate graph or from experimentation an the Intermeiate Value Theorem, another occurs just to the right of x = Computing also we get the graph in Figure.7. f( ) = ( ) 6( ) + 8( ) = =, f(1) = (1) 6(1) + 8(1) = = 3, It is important to note that having f (x) = 0 oes not imply that there is a local maximum or minimum point there. We shoul always consier the full sign chart of f (x). Local extrema o exist at points of continuity x = a on the graph if f (x) changes sign at x = a. In the above example, x = 1 is a zero of multiplicity for f (x), an so f oes not change sign there, an there is no local extremum at x = 1. At x = we have a zero of multiplicity 1 for f (x), an so f oes change sign there, giving us a local extremum (a minimum in that case)...5 Derivatives of Sine an Cosine In this subsection we show how sin x an cos x are both ifferentiable, compute their erivatives, an apply them to functions involving the chain rule. We will prove the following theorem. Theorem..6 The functions sin x an cos x are ifferentiable for all x R, an when x is 18 In Section 5. we will see how to fin x-intercepts of most functions to as much accuracy as we esire.

11 .. FIRST DIFFERENTIATION RULES; LEIBNIZ NOTATION y = 0 y = sin x y = = 3 y = 0 1 y = y = cosx y = 0 = y = y = cosx 1 y = sinx 3 3 Figure.8: Partial graphs of y = sin x, y = cos x, an their respective erivative functions graphe below them. measure in raians their erivatives are given by: sin x = cosx, (.18) cosx = sinx. (.19) These shoul seem reasonable given the respective graphs of Figure.8. For instance, for the sine curve we have the following ata: x = 0,, 3, y = sin x = 0, 1, 0,, 0 y = cos x = 1, 0,, 0, 1 Looking at the graph of sin x as rawn in Figure.8, the slopes at these points an the values for cosx seem at least compatible. Similarly for the cosine curve: x = 0,, 3, y = cos x = 1, 0,, 0, 1 y = sin x = 0,, 0, 1, 0 We will prove the erivative formula for sinx, an leave the erivative of cosx as an exercise. (The two computations are very similar.) Proof:(.18): The proof is base upon the following: sin(α + β) = sin α cosβ + cosαsin β, sin θ 1 cosθ lim = 1, lim = 0, θ 0 θ θ 0 θ

12 318 CHAPTER. THE DERIVATIVE which are, respectively, a trigonometric ientity, (3.81) from page 70, an (3.86) from page 7. We will use the limit-efinition of the erivative, expan using the formula for sin(α+β), an rearrange the terms so we can use the trigonometric limits above. f f(x + ) f(x) (x) 0 sin(x + ) sinx sin xcos + cosxsin sin x sin x(cos 1) + cosxsin [ sin x cos 1 + cosx sin = sin x 0 + cosx 1 = cos x, q.e.. ] Now we combine what we know into other examples. Example..7 Fin f (x) if f(x) = x + sin x 3 cosx. Solution: f (x) = [ x + sinx 3 cosx ] = x + cosx 3( sinx) = x + cosx + 3 sin x. We can also use these erivatives to fin where functions involving sinx an cosx are increasing/ecreasing, an thus fin any local extrema (that is, local maxima an minima). Example..8 Consier the function f(x) = sin x cos x. Fin where f(x) is increasing an where f(x) is ecreasing, an use this information to plot f(x). Solution: Here f (x) = cosx ( sinx) = cosx + sinx. Since this is efine an continuous everywhere, we will check where it is zero to etect where it (f (x) here) possibly changes signs. The technique below works anytime we are intereste in solving a sinx + b cosx = 0, where a, b 0: cosx + sin x = 0 sin x = cosx sinx cosx = tan x =. The reason we can ivie by cosx is because there are no solutions where cosx = 0, because such solutions woul require also sin x = 0, an these cannot be zero simultaneously because (recall) sin x + cos x = 1. So we are looking for x R such that tan x =. This occurs in the secon quarant (if x represents an angle in stanar position) an in the fourth quarant, with reference angles /:

13 .. FIRST DIFFERENTIATION RULES; LEIBNIZ NOTATION 319 3/ / 7/ Thus we are looking for angles x = 3 +n, where n = 0, ±1, ±, ±3,. Now f(x) = cosx+sin x is -perioic, so we can analyze one perio to see what the graph shoul look like. We will use the points x = /, 3/, 7/ for our sign chart, an eclare the pattern from there: f (x) = cosx + sinx Test x= 0 f (x) = Sign f : Behavior of f: ր ց 3 7 Because this behavior continues, we see a local maximum at ( 3, f ( )) ( 3 = 3, ), since f(3/) = sin 3 cos 3 = + = = =. This local maximum height then repeats every in both (left an right) irections. Similarly, because of the sign chart an the fact that this function (an its erivative an its erivative s sign chart) repeats every, we have a local minimum at, for instance, ( 7, f ( )) ( 7 = 7, ), which also repeats every in both irections. This function is graphe in Figure.9. Example..9 Let f(x) = x + sin x. Fin where f(x) is increasing an where f(x) is ecreasing. Solution: Here f (x) = 1 + cosx = 0 when cosx =, which is at x = ±, ±3, ±5,. A partial sign chart is given below:

14 30 CHAPTER. THE DERIVATIVE Figure.9: Partial graph of f(x) = sin x cos x, showing for instance the local minima at x = 7/ an x = /, an the local maxima at x = 3/ an x = 5/. Each local extremum is repeate every. See Example..8. f (x) = 1 + cosx Test x = 0 sign f (x): 3 ր ր ր ր So this function is actually always increasing, only briefly having zero slope at the o multiples of. Note that these points occur at (, ), (3, 3), (7, 7), etc., an (, ), ( 3, 3), ( 7, 7), etc. This function is graphe in Figure.10, showing this behavior...6 Absolute Values an Piece-wise Define Functions Here we begin by exploring how one woul compute erivatives for piece-wise efine functions such as x. This is inee a piece-wise efine function, as we have x = { x if x 0, i.e., x [0, ) x if x < 0, i.e., x (, 0). = x { 1 if x > 0, i.e., x (0, ) = if x < 0, i.e., x (, 0). If we are safely within one of the pieces, then the erivatives are easily compute using other rules. For instance, if f(x) = x, then f (5) = (x) f ( 3) = ( x) = 1, x=5 =. x= 3 We can o this, because when we look at the efinition of the erivative, with small enough, x + will still be within the interior of the interval [0, ) or (, 0). However, f (0) will not

15 .. FIRST DIFFERENTIATION RULES; LEIBNIZ NOTATION Figure.10: Partial graph of f(x) = x+sin x. The erivative being f (x) = 1+cos x, which is positive except at x = ±, ±3, ±5,, the function is always increasing, momentarily leveling off at these points where f (x) = 0. exist: f(0 + ) f(0) lim f(0 + ) f(0) lim 0 0 >0 <0 lim = 1, lim 0 () =. 0 For f (0) to exist, these two limits woul have to be equal (an thus equal to f (0)), but these limits are ifferent an so f (0) oes not exist. However, the above process is somewhat more involve than necessary. We can simply note that the left-sie limit ( 0 ) woul yiel the same as ( x) = an the right-sie limit woul yiel the same as (x) = 1, which are not equal an so the Example..10 Fin f (x) if f(x) = 1, if x /, sinx, if / < x < /, x, if x /. Solution: Note first that f(x) is continuous at x = /, so there is a possibility of a erivative there. It is not continuous at x = /, as a quick check of left an right limits there woul show they o not match. We can therefore say that f (/) oes not exist. That sai, we o at least have x (/, ) = f (x) = (1) = 0, x ( /, /) = f (x) = (sin x) = cosx, x (, /) = f (x) = ( x) =.

16 3 CHAPTER. THE DERIVATIVE What is left to check is f (/). This is simpler than one might think at first, because when we look at the efinition of f (/) using limits, when computing this with 0 + it woul be exactly the same computation as if the function in question were the (constant) function 1, an so that limit woul be 0. When taking 0, that limit computation woul be exactly the same as if the function were sin x, an we know its erivative is cos x, an cos(/) = 0. Thus the left an right limits can be foun using the functions in the branches which efine f(x), assuming that the function is continuous there. (Otherwise some ifficulty woul be foun when trying to take the whole limit as 0 ±.) Summarizing, f (x) = an oes not exist otherwise (i.e., at x = /). 0, if x /, cosx, if / < x < /,, if x < /, Functions such as these inicate why it is important to recall the original efinition of f (x), namely f f(x + ) f(x) (x), 0 To make the argument above more formally, one coul write, f(/ + ) f(/) 1 1 lim = 0, f(/ + ) f(/) sin(/ + ) sin(/) lim = cos 0 0 = 0. The final computation came from what we know of the erivative of sinx, which is a two-sie limit an so efinitely suffices for the one-sie limit neee there. We were allowe to use sin(/) where we i since this equals 1, an so oes f(/). We were allowe to use sin(/+) because for < 0 (but small), we are on that mile branch of the function s omain where the function returns the sine of the input. These two limits above being the same, we can eclare the limit as 0 to exist an to be that value, i.e., f (/) = 0. Similar but simpler computations, not requiring us to o the two one-sie limits separately, yiel the erivatives on the interiors of the branches on which f(x) is efine by the ifferent formulas. A similar but two-sie attempt at x = / woul have complications which cannot be overcome (impossible alreay just ue to the lack of continuity there recall the erivative cannot exist without continuity), an so ultimately that limit will not exist an so f ( /) oes not exist.

17 .. FIRST DIFFERENTIATION RULES; LEIBNIZ NOTATION 33 Exercises 1. Fin the following erivatives. Expan polynomials first where necessary. [ (a) x 199x + 7 ]. [ ] 1 (b) x + x. [ (c) t 7 19t ]. t () [(x + 9)(x 3)]. [ (e) 10 9y 8 ]. y (f) (x + 5).. Show that ( x x 3) ( ( x ))( ( x 3 )). Why oes this not violate Theorem.. (i.e., (.1))? 3. Give an alternate proof of the integer power rule, Theorem.10 by using a n b n = (a b) ( a n + a n b +a n 3 b + + ab n + b n). (Hint: a n will be your f(x+) term, an b n will be f(x).). Suppose s(t) = 3t t+19. Fin v(t). Also fin when the particle is moving to the right (v > 0), an when it is moving left (v < 0). 5. Graph the function f(x) = x x, showing all x-intercepts, all local maxima an minima. (See Example..5, page 31.) 6. Use cos(α+β) = cosαcosβ sinαsin β to prove (.19), page 317: cosx = sin x. It may be helpful to see the proof for the erivative of sinx. 7. Graph f(x) = sin x + cosx for x [, ], showing where this function is increasing an where it is ecreasing. 8. Graph f(x) = x+ cosx over a reasonable interval, showing where this function is increasing an where it is ecreasing. Also show its behavior as x ±. 9. 1(f) above must be expane ( multiplie out ) before using the power rule. The answer is 8x + 0. Compute 1(f) above using instea the technique in Footnote 15, page 309. (There u = x + 5). 10. The voltage V across a resistor in an electrical circuit is the prouct of the current I an the resistance R. (This is Ohm s Law, iscusse at length in Section.3.) If both the current an resistance vary with time t an are given by I = 3 + t + 0.1t, R = 0 0.t, fin the time rate of change of V (i.e., fin V t ) when t = 1.5. Here V is in Volts an t in secons. 11. Fin the general rate of change V/s of the volume of a cube with respect to the length s of one sie, given V = s 3. Then fin the specific rate when s = 10 cm. What are its units? 1. Fin the rate of change A/r of the area of a circle with respect to its raius, recalling that A = r. Does the formula for A/r look familiar? What is its value when r = 15 cm?