Teaching Differentiation: A Rare Case for the Problem of the Slope of the Tangent Line

Transcription

1 Teacing Differentiation: A Rare Case for te Problem of te Slope of te Tangent Line arxiv: v1 [mat.ho] 29 Apr 2018 Roman Kvasov Department of Matematics University of Puerto Rico at Aguadilla Aguadilla, Puerto Rico 00604, USA May 2, 2018 Abstract In tis article we discuss an important students misconception about derivatives, tat te epression of te derivative of te function contains te information as to weter te function is differentiable or not were te epression is undefined. As a working eample we consider a typical Calculus problem of finding te orizontal tangent lines of a function. Following te standard procedure, we derive te epression for te derivative using Product Rule. Te searc for te values of te independent variable, tat make te derivative equal zero, leads to missing te unique solution of te problem. We sow tat in tis case, even toug te epression of te derivative is undefined, te function indeed possesses te derivative at te point. We also provide te metodological treatment of suc functions, wic can be effectively used in te classroom. Key words: matematics education, calculus, derivative. 1 Introduction Derivate is one of te most important topics not only in matematics, but also in pysics, cemistry, economics and engineering. Every standard Calculus course provides a variety of eercises for te students to learn ow to apply te concept of derivative. Te types of problems range from finding an equation of te tangent line to te application of differentials and advanced curve sketcing. Usually, tese 1

2 eercises eavily rely on suc differentiation tecniques as Product, Quotient and Cain Rules, Implicit and Logaritmic Differentiation [1]. Te definition of te derivative is ardly ever applied after te first few classes and its use is not muc motivated. Like many oter topics in undergraduate matematics, derivative gave rise to many misconceptions [2], [3], [4]. Just wen te students seem to learn ow to use te differentiation rules for most essential functions, te application of te derivative brings new issues. A common students error of determining te domain of te derivative from its formula is discussed in [5] and some interesting eamples of te derivatives, defined at te points were te functions temselves are undefined, are provided. However, te unt for misconceptions takes anoter twist for te derivatives undefined at te points were te functions are in fact defined. Te epression of te derivative of te function obtained using differentiation tecniques does not necessarily contain te information about te eistence or te value of te derivative at te points, were te epression for te derivative is undefined. In tis article we discuss a type of continuous functions tat ave te epression for te derivative undefined at a certain point, wile te derivative itself at tat point eists. We sow, ow relying on te formula for te derivative for finding te orizontal tangent line of a function, leads to a false conclusion and consequently to missing a solution. We also provide a simple metodological treatment of similar functions suitable for te classroom. 2 Calculating te Derivative In order to illustrate ow deceitful te epression of te derivative can be to a students eye, let us consider te following problem. Problem Differentiate te function f () = sin ( 2 ). For wic values of from te interval [ 1, 1] does te grap of f () ave a orizontal tangent? Problems wit similar formulations can be found in many Calculus books [1], [6], [7]. Following te common procedure, let us find te epression for te derivative of 2

3 te function f () applying te Product Rule: f () = ( 3 ) ( ) sin 2 + ( sin ( 2)) 3 = 1 3 sin ( 2) + 2 cos ( 2) 3 2 = 62 cos 2 + sin (1) Similar to [1], we find te values of were te derivative f () is equal to zero: 6 2 cos 2 + sin 2 = 0 (2) Since te epression for te derivative (1) is not defined at = 0, it is not ard to see tat for all values of from [ 1, 1] distinct from zero, te left-and side of (2) is always positive. Hence, we conclude tat te function f () does not ave orizontal tangent lines on te interval [ 1, 1]. However, a closer look at te grap of te function f () seems to point at a different result: tere is a orizontal tangent at = 0 (see Figure??). First, note tat te function f () is defined in = 0. In order to verify if it as a orizontal tangent at tis point, let us find te derivative of te function f () using definition: f f (0 + ) f (0) (0) sin ( 2 ) ( ) sin ( 2 ) 2 3 sin ( 2 ) 2 = = 0 since eac of te its above eists. We see tat, indeed, te function f () possesses a orizontal tangent line at te point = 0. 3 Closer Look at te Epression for te Derivative Wat is te problem wit te standard procedure proposed by many tetbooks and repeated in every Calculus class? Te eplanation lies in te following premise: te 3

4 epression of te derivative of te function does not contain te information as to weter te function is differentiable or not at te points were it is undefined. As it is pointed out in [5], te domain of te derivative is determined a priori and terefore sould not be obtained from te formula of te derivative itself. In te eample above te Product Law for derivatives requires te eistence of te derivatives of bot functions at te point of interest. Since te function is not differentiable in zero, te Product Rule cannot be applied. In order to see wat eactly appens wen we apply te Product Rule, let us find te epression for te derivative using definition of te derivative: f f ( + ) f () () + sin ( + ) 2 sin ( 2 ) ( ) = sin ( 2) + ( sin ( + ) 2 sin ( 2 ) ) sin ( 2) + sin ( + ) 2 sin ( 2 ) sin ( 2) + 2 cos ( 2) wic seems to be identical to te epression (1). Students are epected to develop a skill of deriving similar results and know ow to find te derivative of te function using definition of te derivative only. But ow legal are te performed operations? 4

5 y Figure 1: Grap of te function g () = cos ( 2 ) Let us consider eac of te following its: + 3 sin ( 2) sin ( + ) 2 sin ( 2 ) +. Te last tree its eist for all real values of te variable. However, te first it does not eist wen = 0. Indeed = + 2 5

6 Tis implies tat te Product and Sum Laws for its cannot be applied and terefore tis step is not justifiable in te case of = 0. Wen te derivation is performed, we automatically assume te conditions, under wic te Product Law for its can be applied, i.e. tat bot its tat are multiplied eist. It is not ard to see tat in our case tese conditions are actually equivalent to 0. Tis is precisely wy, wen we wrote out te epression for te derivative (1), it already contained te assumption tat it is only true for te values of tat are different from zero. Note, tat in te case of = 0 te application of te Product and Sum Laws for its is not necessary, since te term ( 3 + ) sin ( 2 ) vanises. Te correct epression for te derivative of te function f () sould be te following: { 6 2 cos( 2 )+sin( 2 ) f () = 3, if 0 2 0, if = 0 Te epression for te derivative of te function provides te correct value of te derivative only for tose values of te independent variable, for wic te epression is defined; it does not tell anyting about te eistence or te value of te derivative, were te epression for te derivative is undefined. Indeed, let us consider te function g () = cos ( 2) and its derivative g () g () = cos (2 ) 6 2 sin ( 2 ) 3 2 Similar to te previous eample, te epression for te derivative is undefined at = 0. Noneteless, it can be sown tat g () is not differentiable at = 0 (see Figure 2). Terefore, we provided two visually similar functions: bot ave te epressions for teir derivatives undefined in zero, owever, one of tese functions possesses a derivative, but te oter one does not. 4 Metodological Remarks Unfortunately, tere eist many functions similar to te ones discussed above and tey can arise in a variety of typical Calculus problems: finding te points were te tangent line is orizontal, finding an equation of te tangent and normal lines to te curve at te given point, te use of differentials and grap sketcing. Relying 6

7 only on te epression of te derivative for determining its value at te undefined points may lead to missing a solution (as in te eample discussed above) or to some completely false interpretations (as in te case of curve sketcing). As it was discussed above, te epression for te derivative does not provide any information on te eistence or te value of te derivative, were te epression itself is undefined. Here we present a metodology for te analysis of tis type of functions. Let f () be te function of interest and f () be te epression of its derivative undefined at some point 0. In order to find out if f () is differentiable at 0, we suggest to follow a list of steps: 1. Ceck if te function f () itself is defined at te point 0. If f () is undefined at 0, ten it is not differentiable at 0. If f () is defined at 0, ten proceed to net step. 2. Identify te basic functions tat are used in te formula of te function f (), tat are temselves defined at te point 0, but teir derivative is not (suc as, for eample, te root functions). 3. Find te derivative of te function f () at te point 0 using definition. Te importance of te first step comes from te fact tat most students tend to pay little attention to te functions domain analysis wen asked to investigate its derivative. Formally, te second step can be skipped, owever it will give te students te insigt into wic part of te function presents a problem and teac tem to identify similar cases in te future. te difficulty of accomplising te tird step depends on te form of te function and sometimes can be tedious. Neverteless, it allows te students to apply te previously obtained skills and encourages te review of te material. 7

8 y Figure 2: Grap of te function g () = cos ( 2 ) 5 Conclusion We discussed te misconception, tat te epression of te derivative of te function contains te information as to weter te function is differentiable or not at te points, were te epression is undefined. We considered a typical Calculus problem of looking for te orizontal tangent line of a function as an eample. We sowed ow te searc for te values tat make te epression of te derivative equal zero leads to missing a solution: even toug te epression of te derivative is undefined, te function still possesses te derivative at te point. We provided an eample of te function tat similarly as te epression for te derivative undefined, owever te function is not differentiable at te point. We also presented te metodological treatment of suc functions by applying te definition of te derivative, wic can be used in te classroom. 8

9 References [1] Stewart J. Single Variable Calculus: Early Transcendentals. Brooks/Cole; p [2] Muzangwa J., Cifamba P. Analysis of errors and misconceptions in te learning of calculus by undergraduate students. Acta Didactica Napocensia, Vol. 4:2; [3] Gur H., Barak B. Te Erroneous Derivative Eamples of Elevent Grade Students. Educational Sciences: Teory and Practice; p [4] Li, X. Cognitive Analysis of students errors and misconceptions in variables, equations and functions. Doctoral dissertation, Teas A&M University; [5] Rivera-Figueroa A., Ponce-Campuzano J. Derivative, maima and minima in a grapical contet. International Journal of Matematical Education in Science and Tecnology, Vol. 44:2; p [6] Larson R., Edwards B. Calculus (9t Edition). Brooks/Cole; p [7] Tomas G., Weir M., Hass J. Tomas Calculus: Early Transcendentals (12t Edition). Pearson; p