1 Derivatives of Piecewise Defined Functions


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1 MATH 1010E University Matematics Lecture Notes (week 4) Martin Li 1 Derivatives of Piecewise Define Functions For piecewise efine functions, we often ave to be very careful in computing te erivatives. Te i erentiation rules (prouct, quotient, cain rules) can only be applie if te function is efine by ONE formula in a neigboroo of te point were we evaluate te erivative. If we want to calculate te erivative at a point were two i erent formulas meet, ten we must use te efinition of erivative as it of i erence quotient to correctly evaluate te erivative. Let us illustrate tis by te following example. Example 1.1 Fin te erivative f 0 (x) at every x 2 R for te piecewise efine function 5 2x wen x<0, f(x) = x 2 2x +5 wen x 0. Solution: We separate into 3 cases: x<0, x>0 an x = 0. For te first two cases, te function f(x) is efine by a single formula, so we coul just apply i erentiation rules to i erentiate te function. f 0 (x) =(5 2x) 0 = 2 for x<0, f 0 (x) =(x 2 2x + 5) 0 =2x 2 for x>0. At x = 0, we ave to use te efinition of erivative as it of i erence quotient. First of all, f(0) = 0 2 2(0) + 5 = 5. Ten we calculate te leftan an rigtan its: f() f(0) (5 2) 5 2= 2, f() f(0) ( ) ) = 2. +( 1
2 Since bot of tem exists an are equal, we ave f 0 f() f(0) (0) = 2. Terefore, putting all of tese togeter, we see tat f is i erentiable for every x 2 R an f 0 2 wen x apple 0, (x) = 2x 2 wen x>0. Remark 1.2 From te example above, we see tat te erivative f 0 (x) is still a continuous function (ceck tis!). Tis is not always true for any function! (Have you seen a counterexample? See Homework 2) Example 1.3 Consier te function efine by ax + b wen x apple 1, f(x) = ax 3 + x +2b wen x> 1, for wat value(s) of a, b 2 R is te function f i erentiable at every x 2 R? Solution: First, it is easy to see tat for ANY a, b 2 R, te function f is i erentiable at every x 6= 1sincef is efine by a polynomial on ( 1, +1) an ( 1, 1). Te only catc is at te point x = 1. If f is i erentiable at x = 1, it must also be continuous at x = 1. Terefore, we nee f(x) =f( x! 1 1). Now, f( 1) = a + b an te leftan an rigtan its are f(x) =a( x! 1 1)3 +( 1) + 2b = a +2b 1, f(x) =a( 1) + b = a + b. x! 1 + If f is continuous, ten bot of tese its must be te same an equal to f( 1). Hence, we ave a + b = a +2b 1 ) b =1. Now, we take b = 1. To fin te value of a wic make f i erentiable at x = 1, we require te it f( 1+) f( 1) 2
3 to exists, wic is equivalent to te statement tat te leftan an rigtan its exist an are equal. Te left an it is f( 1+) f( 1) Te rigt an it is f( 1+) f( 1) + [a( 1+) + 1] ( a + 1) a = a. [a( 1+) 3 +( 1+) + 2] ( a + 1) + a[( 1+) 3 + 1] + + a[( 1+) 2 ( 1+) + 1] + + = 3a +1. Terefore, if we set tem equal to eac oter, we obtain te conition a =3a +1 ) a = 1 2. In summary, we ave a = x 2 R. 1/2 an b =1iff is i erentiable at every 2 Di erentiation Rules II: Prouct an Quotient Rules Teorem 2.1 If f an g are i erentiable functions, ten bot teir prouct fg an quotient f/g are i erentiable an we ave (1) Prouct Rule: (2) Quotient Rule: provie tat g(x) 6= 0. [f(x)g(x)] 0 = f 0 (x)g(x)+f(x)g 0 (x), f(x) 0 = g(x)f 0 (x) f(x)g 0 (x) g(x) (g(x)) 2, Remark 2.2 Observe tat te i erentiation rule [kf(x)] 0 = kf 0 (x) were k is a constant is just a special case of prouct rule by taking g(x) k, wic as g 0 (x) =0. 3
4 Before we go into te proof of tese rules, let us first look at a few examples of ow to apply tese rules to elp us calculate erivatives. Example 2.3 x [(x + 2)(x2 + 1)] = (x + 2) 0 (x 2 + 1) + (x + 2)(x 2 + 1) 0 = (1)(x 2 + 1) + (x + 2)(2x) = 3x 2 +4x +1. x [sin x cos x] = (sinx)0 cos x +sinx(cos x) 0 = cos x cos x +sinx( sin x) = cos 2x. x x 2 +1 x +1 = (x + 1)(x2 + 1) 0 (x 2 + 1)(x + 1) 0 (x + 1) 2 = (x + 1)(2x) (x2 + 1)(1) (x + 1) 2 = x2 +2x 1 (x + 1) 2. x sin x x = x(sin x)0 sin x(x) 0 = x 2 x cos x sin x x 2. Note tat for te last two examples, te calculation is vali only for x 6= 1 an x 6= 0respectively. sin x Question: Doesteit x!0 x x exist? If we efine sin x f(x) = x wen x 6= 0, 1 wen x =0, ten oes f 0 (0) exists? If so, is it relate to te it at te beginning? Question: Calculation te erivatives of all te trigonometric an yperbolic functions! Now, we come back to te proof of te prouct rule an quotient rule. 4
5 Proof of Prouct Rule: Using te efinition of erivative, [f(x)g(x)] 0 f(x + )g(x + ) f(x)g(x) [f(x + )g(x + ) f(x)g(x + )] + [f(x)g(x + ) f(x)g(x)] f(x + ) g(x + ) f(x) g(x + ) + f(x) g(x) = g(x)f 0 (x)+f(x)g 0 (x). In te last equality, we ave also use tat g(x + ) =g(x) sinceg is continuous (even i erentiable) by assumption. Proof of Quotient Rule: Using te efinition of erivative, f(x) 0 f(x+) g(x+ g(x) f(x) g(x) f(x + )g(x) f(x)g(x + ) g(x)g(x + ) f(x+)g(x) f(x)g(x) + g(x)g(x + ) f(x) g(x)g(x + ) = g(x)f 0 (x) f(x)g 0 (x) (g(x)) 2. g(x) f(x+) f(x) f(x)g(x) f(x)g(x+) g(x+) g(x) Again we ave use te continuity of g in te last equality. 3 Composite Functions Apart from aition, subtraction, multiplication an ivision to get new functions, tere is anoter useful way to obtain new functions from ol calle composition. Definition 3.1 Given two functions f : D! E an g : E! F,wecan efine te composite function g f : D! F by g f(x) :=g(f(x)). 5
6 If we tink of functions as assignments, ten g f means assigning x to f(x) first an ten we furter assign f(x) tog(f(x)). Note tat in orer for te composition g f to be well efine, te omain of g must contain te image of f. However, we o not require eiter f or g to be injective or surjective. Question: Wic of te following statement is true? f an g are injective ) g f an g are injective ) g f is injective? f is injective? g is not injective ) g g is not surjective ) g f is not injective? f is not surjective? Let us look at one example. Consier two functions efine by f : R! R, f(x) := cos x, te composition g g : R! R, g(y) :=y 2, f : R! R is ence g f(x) =g(f(x)) = g(cos x) = cos 2 x. Note tat we can also o te composition in a i erent orer for tis example. We can form f g : R! R wic is efine by f g(y) =f(g(y)) = f(y 2 ) = cos y 2. Note tat even in tis case bot g f an f g are efine, tey are NOT equal to eac oter. Terefore, te orering is important wen we talk about composition: Remark 3.2 In general, we ave g f 6= f g even wen bot are wellefine. 6
7 Te notation g f may be confusing sometimes since we are writing g first an ten f but te composition means oing f first an ten g. One goo way to memorize tis is tat wen g f acts on x, wewriteg f(x). It is f wic its x first, an ten followe by g. Tis is te reason wy we use tis convention. Also, in te above example, we write g f as a function of x an f g as a function of y. Wen comparing tem as functions, te name of te variable (x an y) are irrelevant. It is te rule of assignment tat etermines te function. For example, we consier f(x) =x an g(y) =y as te same function. It will become clear later tat it is useful to keep using x for elements in te omain of f an y as elements in te coomain of f. Composition is a rater nice operation wic preserves many of te analytic properties of f an g. Teorem 3.3 If (i) f is continuous at x 0,an (ii) g is continuous at y 0 := f(x 0 ), ten g f is continuous at x 0. In oter wors, composition of continuous functions is continuous. Te proof is rater intuitive. Wen x is close to x 0,teny := f(x) is also close to y 0 := f(x 0 ) by te continuity of f. On te oter an, since y is close to y 0,wemustaveg(y) close to g(y 0 ). Tis is te same as saying tat g(f(x)) is close to g(f(x 0 )). A rigorous proof can given using te efinition of continuity. Intereste stuents can try to write own a complete proof as an exercise. We ave alreay use tis teorem implicitly many times wen we evaluate its. For example, wen we compute x!0 cos2 x = (cos 0) 2 =1, we ave use te continuity of cos an square function wic justifies te irect substitution wit x = 0 to compute te it. 4 Di erentiation Rules III: Cain Rule Composition of i erentiable functions is also i erentiable. Teorem 4.1 (Cain Rule) If 7
8 (i) f is i erentiable at x 0,an (ii) g is i erentiable at y 0 := f(x 0 ), ten g f is i erentiable at x 0 an (g f) 0 (x 0 )=g 0 (f(x 0 )) f 0 (x 0 ). Remark 4.2 Te cain rule simply says tat te erivative of composition is just te prouct of erivatives. Yet tis is one of te most confusing rules among all te i erentiation rules we ave seen so far. Te reason is tat we ave to be careful at wic points we are evaluating te erivatives! We are NOT evaluating g 0 at x 0, but at f(x 0 ) instea. If you recall te stepbystep assignment picture of composite functions, you see tat it is inee te only way for te formula to make sense since g 0 (x 0 ) is not even efine as x 0 is not in te omain of g! Let us turn to some examples. Example 4.3 Calculate te erivatives of te following functions: (i) cos x 2, (ii) (x 2 + 1) 7, (iii) e sin x, (iv) e sin x2. Solution: (i) Let f(x) =x 2 an g(y) = cos y, teng te function we want to i erentiate. Note tat f(x) = cos x 2 is f 0 (x) =2x an g 0 (y) = sin y. Terefore, applying cain rule we get (g f) 0 (x) =g 0 (f(x))f 0 (x) =( sin(x 2 )) (2x) = 2x sin x 2. (ii) Let f(x) =x 2 +1 an g(y) =y 7.Weavef 0 (x) =2x an g 0 (y) =7y 6, Terefore, x (x2 + 1) 7 = 7(x 2 + 1) 6 (2x) = 14x(x 2 + 1) 6. 8
9 Terefore, te cain rule basically says tat we can i erentiate layerbylayer, starting from te outermost layer. In tis example, we first i erentiation te function of raising to power 7 an ten i erentiate into te bracket. (iii) Using te concept of i erentiating layerbylayer, we o not nee to efine f an g every time. Since we know te erivative of e y is just e y an te erivative of sin x is cos x, we get x esin x = e sin x cos x. (iv) For functions involving more tan two layers, we just i erentiate tem one by one. Here, te outer layer is exponential function, te mile layer is sin an te inner later is x 2,so x esin x2 = e sin x2 (cos x 2 ) (2x). Note tat wen i erentiating te outer layers, you keep te inner layers uncange. Tis makes sure tat you are evaluating at te correct point as iscusse in Remark 4.2. Question: Use layer by layer i erentiation to evaluate te following erivatives: q x + p x x an x x p 1+x 2 Proof of Cain Rule: By efinition of erivative, (g f) 0 g(f(x)) g(f(x 0 )) (x 0 ) x!x 0 x x apple 0 g(f(x)) g(f(x0 )) x!x 0 f(x) f(x 0 ). f(x) f(x 0 ) x x 0 g(f(x)) g(f(x 0 )) f(x) f(x 0 ). x!x 0 f(x) f(x 0 ) x!x 0 x x 0 Let y = f(x) an y = f(x 0 ), since f is continuous at x 0,weavey! y 0 as x! x 0, terefore, we can rewrite te first it in te last line above in terms of y: g(y) g(y 0 ) f(x) f(x 0 ) = g 0 (y 0 )f 0 (x 0 )=g 0 (f(x 0 ))f 0 (x 0 ). y!y 0 y y 0 x!x 0 x x 0 Tis proves te cain rule. Question: Tere is a loopole in te above proof. Can you fin it an fix it? 9
10 5 Inverse Di erentiation We can use te cain rule to fin te erivative of te inverse f 1 of a function, if it exists. Teorem 5.1 Let f :(a, b)! (c, ) be a bijective i erentiable function wose erivative f 0 is continuous. Ten, te inverse function g :(c, )! (a, b) is i erentiable at every y suc tat f 0 (g(y)) 6= 0an g 0 (y) = 1 f 0 (g(y)). Remark 5.2 Note tat te inverse may not be i erentiable at were f 0 = 0. For example, consier f : R! R efine by f(x) =x 3, ten its inverse g(y) =y 1/3 exists but is not i erentiable at y =0. Proof of Teorem 5.1 Te proof tat g is i erentiable is muc more involve so we skip it ere. By te efinition of inverse, f(g(y)) = y for all y. Since te above equation ols for ALL y, we can i erentiate te wole equation on bot sie wit respect to y, applying cain rule on te left an sie, we obtain f 0 (g(y))g 0 (y) =1. If f 0 (g(y)) 6= 0, we can ivie it to te oter sie to obtain te formula of g 0 (y). Example 5.3 Sow tat x (ln y) = 1 y. Solution: Recall tat te exponential function exp(x) : R! (0, 1) is 11 an onto wose inverse is ln y :(0, 1)! R. We ave alreay seen tat xexp(x) =exp(x). Terefore, using inverse i erentiation, keeping in min tat y =exp(x), we ave y ln(y) = 1 exp(x) = 1 y. Example 5.4 Sow tat x sin 1 y = p 1. 1 y 2 10
11 Solution: Recall tat sin(x) :R! R is neiter 11 nor onto so te inverse oes not exists. However, if we restrict its omain an coomain to sin(x) : ( 2, 2 )! ( 1, 1), ten it becomes 11 an onto an ence tere is an inverse sin 1 y : ( 1, 1)! ( 2, 2 ). Using inverse i erentiation, since y =sinx, weave y sin 1 1 y = x sin x = 1 cos x = 1 p 1 sin 2 x = 1 p. 1 y 2 Note tat we coul ave restrict sin(x) to a i erent omain e.g. ( 2, 3 2 ). Te inverse function sin 1 y woul be i erent, but te erivative is te same! Te coice of sin 1 is a penomenon calle brancing. Question: Wat appens to y sin 1 y wen y!±1? Wat oes it correspon to in terms of te slope of te graps? Question: Discuss te omains an coomains of te inverses of oter trigonometric an yperbolic functions. Wat are teir erivatives? 6 More Examples We give a few more examples illustrating te use of all te i erentiation rules we iscusse. x (ln(x + p 1+x 2 )) = = = 1 x + p 1+x 2 x (x + p 1+x 2 ) 1 x + p 1+x [1 + x(1 + 2 x2 ) 1 2 ] 1 p. 1+x 2 We can use exp an ln to efine an arbitrary exponential function a b for ANY a>0 an b 2 R: a b := exp(b ln a). Hence, we can calculate te erivatives of tese general exponential functions: x 3x = x (exp(x ln 3)) = exp(x ln 3) ln 3 = 3x ln 3. x xx = x (exp(x ln x)) = exp(x ln x) [(1)(ln x)+(x)( 1 x )] = (1+ ln x)xx. Question: Calculate x (xxx ). 11
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