Calculus 1, Chapter 1 Fall 2021 Table des matières Chapter 1: Functions of one real variable: differentiability and related topics ...................................... 1 1. Inverse functions ............................................................................................................................. 1 1.1 Bijections (bijective functions, invertible functions) ............................................................. 1 1.2 Continuous bijection................................................................................................................. 3 1.3 Examples of inverse functions ................................................................................................ 3 2. Differentiability................................................................................................................................ 5 2.1. Differentiability at a point ........................................................................................................ 5 2.2 Digression: Variables vs functions ............................................................................................. 7 2.3. Applying derivatives to computation of limits ....................................................................... 10 Chapter 1: Functions of one real variable: differentiability and related topics 1. Inverse functions 1.1 Bijections (bijective functions, invertible functions) Definition: a bijection is a function between the elements of two sets, where each element of one set is paired with exactly one element of the other set, and each element of the other set is paired with exactly one element of the first set. There are no unpaired elements. Definition 1.1.1 (formal): a bijective function f: X → Y is a one-to-one (injective) and onto (surjective) mapping of a set X to a set Y: ∀y ∈ 𝑌, ∃! 𝑥 ∈ X such that 𝑓(𝑥) = 𝑦. 1 Calculus 1, Chapter 1 An injective non surjective mapping An injective surjective mapping (bijection) Fall 2021 A non-injective surjective mapping A non-injective nonsurjective mapping Detecting injectivity on a graph: Non-injective, because there are two pre-images for any strictly positive number: 𝑓(−2) = 𝑓(2) = 4 This graph does not pass the horizontal line test Theorem 1. 1. 1 and definition 1. 1. 2: Let f be a bijection from X to Y. Then there exists a unique function denoted f -1, called the inverse of f, defined on X with values in Y, such that ∀𝑥 ∈ X, 𝑓 ∘ 𝑓 −1 (𝑥) = 𝑥 𝑎𝑛𝑑 ∀𝑥 ∈ Y, 𝑓 −1 ∘ 𝑓(𝑥) = 𝑥. (We can rewrite that statement in terms of functions: 𝑓 ∘ 𝑓 −1 = 𝑖𝑑𝑋 , identity mapping 𝑖𝑑𝐸 is defined by ∀𝑥 ∈ X, 𝑥 ↦ 𝑥) 2 𝑓 −1 ∘ 𝑓 = 𝑖𝑑𝑌 , where the Calculus 1, Chapter 1 Fall 2021 1.2 Continuous bijection Theorem 1. 2. 1: Any function continuous on an interval I, with values in an interval J and strictly increasing (respectively decreasing) on I is a bijection from I to J. Furthermore, its inverse is a continuous function, too, and it is strictly increasing (respectively decreasing) on J Theorem 1. 2. 2 : In a Cartesian coordinate system, the curves of a bijection and its inverse are symmetric to each other with respect to the diagonal line y = x. Here, 𝑓(𝑥) = − 1 𝑥−2 (blue line), 1 𝑔(𝑥) = 𝑓 −1 (𝑥) = 2 − 𝑥 (red line) 1.3 Examples of inverse functions I’ll give you a couple of classical examples of inverse functions. The second example leads to a «new» function, non-reduceable to ordinary algebraic or trigonometric functions or xponentials. We’ll study the properties of some of such «new» functions below. The demonstration of their existence (a consequence of theorem 1. 2. 1) is left as an exercise for the students. Example 1: n-th root Definition 1. 3. 1 : Let n be a natural nonzero number. The function defined on ℝ+ by 𝑥 ↦ 𝑥 𝑛 is a bijection from ℝ+ to ℝ+ , which results in its being invertible. Its inverse is called 𝑛 the n-th root, denoted 𝑥 ↦ √𝑥 . 𝑛 1 Remark: we also use the notation √𝑥 = 𝑥 𝑛 . 3 Calculus 1, Chapter 1 Example 2: arctan Definition 1. 3. 2 : The function tan is is a bijection from 𝜋 𝜋 (− 2 ; 2 ) to ℝ, so that it is invertible. Its inverse function is called arctangent, denoted arctan (in French: also arctg; in English: also atan and tan−1 ). Properties of arctan arctan(0) = 0 arctan(1) = 𝜋 4 arctan(−1) = − 𝜋 4 lim arctan(𝑥) = 𝑥→+∞ 𝜋 2 lim arctan(𝑥) = − 𝑥→−∞ 𝜋 2 4 Fall 2021 Calculus 1, Chapter 1 Fall 2021 2. Differentiability 2.1. Differentiability at a point Definition 2. 1. 1: Let f be a function defined in a neighborhood of a real number a. f is differentiable at a if there exists a real numberℓ called the derivative of f at a, denoted 𝑓 ′ (𝑎) (so that 𝑓 ′ (𝑎) = ℓ) such that: 𝑓(𝑥) − 𝑓(𝑎) =ℓ 𝑥→𝑎 𝑥−𝑎 lim This can also be written as lim ℎ→0 𝑓(𝑎+ℎ)−𝑓(𝑎) ℎ =ℓ Theorem 2. 1. 1 and definition 2. 1. 2: Let f be a function defined in a neighborhood of a. Then the following two statements are equivalent: 1. f is differentiable at a and its derivative at a is ℓ, 2. in a neighborhood of a, the following equality holds: 𝑓(𝑥) = 𝑓(𝑎) + ℓ(𝑥 − 𝑎) + (𝑥 − 𝑎)𝜀(𝑥), where lim 𝜀(𝑥) = 0 𝑥→𝑎 That equality is called a polynomial approximation of order 1 (a linear approximation) of 𝑓(𝑥) at a. Proof (obtaining an explicit expression for 𝜀 ): 𝑓 is differentiable at 𝑎 and its derivative at 𝑎 equals ℓ: lim 𝑥→𝑎 𝑓(𝑥)−𝑓(𝑎) 𝑥−𝑎 =ℓ ⇔ lim ( 𝑥→𝑎 𝑓(𝑥)−𝑓(𝑎) 𝑥−𝑎 − ℓ)=0 Consider a function 𝜀 defined at a neighborhood of a by 𝜀(𝑥)= 𝑓(𝑥)−𝑓(𝑎) 𝑥−𝑎 − ℓ. Then 𝑓(𝑥)=𝑓(𝑎)+(𝑥−𝑎) ℓ +(𝑥−𝑎)𝜀(𝑥) and lim 𝜀(𝑥)=0. QCD 𝑥→𝑎 Remark: One can define in a similar way the notion of differentiability on the right and on the left of a, using lim− 𝑓(𝑥) and lim+ 𝑓(𝑥) 𝑥→𝑎 𝑥→𝑎 Theorem 2. 1. 2: Any function differentiable at a point is continuous at that point. Remark: The converse of that theorem is false: Theorem 2. 1. 3: Let f be a function differentiable at a, with the derivative at a equal to ℓ. Let us denote Cf the curve of f in an X-Y Cartesian coordinate system. Then the slope of the tangent line to Cf drawn at the point whose X-coordinate is a, is ℓ. 5 Calculus 1, Chapter 1 Remark: In a similar way, one can define left and right tangent lines at a point where the left derivative does not equal the right derivative. Sometimes one calls such one-side tangent lines “halftangent lines”. Remark. If f is not differentiable at a but lim 𝑥→𝑎 𝑓(𝑥)−𝑓(𝑎) 𝑥−𝑎 Fall 2021 = ±∞, Cf has a vertical tangent line (or half- tangent line) at the point whose X-coordinate is a. Definition 2. 2. 1 : Let f be a function defined on an interval I and differentiable at all points of I. Then we say that f is differentiable on I and we use f’ to denote the function defined on I such that at any point x I, f’(x) equals the derivative of f at that point. 2.2: Differentiability on an interval; derivative as a function Definition 2. 2. 1 : Let f be a function defined on an interval I and differentiable at all points of I. Then we say that f is differentiable on I and we use f’ to denote the function defined on I such that at any point x I, f’(x) equals the derivative of f at that point. One often uses Leibniz’ notation: instead of 𝑓 ′ (𝑥) one writes 𝑑𝑓 writes 𝑑𝑥 (𝑎) or 𝑑𝑓 | 𝑑𝑥 𝑥=𝑎 𝑑𝑓 ; 𝑑𝑥 instead of 𝑓′(𝑎) one . In many regards, that notation is easier to use. In what follows it will be assumed that the formulas used to compute derivatives of basic functions are known. If this is not the case, refresh your knowledge in any way you wish, in particular (if you prefer) by reading the corresponding pages of the Wisconsin course available on your Moodle page. Here, we’ll explain just how to differentiate composite functions and inverse functions. Theorem 2. 2. 1 : The Chain Rule. More detail: https://en.wikipedia.org/wiki/Chain_rule Let f be a function differentiable on an interval I with values on an interval J, and let g be a function differentiable on J. Then the composition 𝑔 ∘ 𝑓 is differentiable on I and ∀𝑥 ∈ 𝐼, (𝑔 ∘ 𝑓)′ (𝑥) = (𝑔′ ∘ 𝑓)(𝑥) × 𝑓 ′ (𝑥). Why is this called “Chain Rule”? Because we can re-write this formula in Leibniz notation, where it becomes much easier to generalize the result on compositions of more than two functions. Let’s first prove the formula. 6 Calculus 1, Chapter 1 Fall 2021 Proof : Let 𝑎∈I, then (𝑔∘𝑓)(𝑥)−(𝑔∘𝑓)(𝑎) 𝑥−𝑎 = (𝑔∘𝑓)(𝑥)−(𝑔∘𝑓)(𝑎) 𝑓(𝑥)−𝑓(𝑎) × 𝑓(𝑥)−𝑓(𝑎) 𝑥−𝑎 = 𝑔[𝑓(𝑥)]−𝑔[𝑓(𝑎)] 𝑓(𝑥)−𝑓(𝑎) × 𝑥−𝑎 𝑓(𝑥)−𝑓(𝑎) Let 𝑋 = 𝑓(𝑥). When 𝑥 → 𝑎, 𝑋 → 𝑓(𝑎) because 𝑓 is continuous at 𝑎. 𝑔[𝑓(𝑥)]−𝑔[𝑓(𝑎)] 𝑓(𝑥)−𝑓(𝑎) 𝑥→𝑎 lim But lim 𝑥→𝑎 𝑓(𝑥)−𝑓(𝑎) 𝑥−𝑎 Therefore, lim 𝑔(𝑋)−𝑔[𝑓(𝑎)] 𝑋−𝑓(𝑎) 𝑥→𝑎 = lim = 𝑔’[𝑓(𝑎)] = (𝑔′ ∘ 𝑓)(𝑎) because 𝑔 is differentiable at 𝑓(𝑎). = 𝑓′(𝑎) because 𝑓 is differentiable at 𝑎. (𝑔∘𝑓)(𝑥)−(𝑔∘𝑓)(𝑎) 𝑥−𝑎 𝑥→𝑎 = (𝑔’ ∘ 𝑓)(𝑎) × 𝑓′(𝑎) QED. In Leibniz notation, this formula can be written as 𝑑𝑔 𝑑𝑥 𝑑𝑔 = 𝑑𝑓 𝑑𝑓 𝑑𝑥 (to be compared with (𝑔 ∘ 𝑓)′ (𝑥) = (𝑔′ ∘ 𝑓)(𝑥) × 𝑓 ′ (𝑥) ). which looks like as if we were dealing with fractions and then simplified by df. When we write 𝑑𝑧 𝑑𝑥 𝑑𝑧 𝑑𝑦 = 𝑑𝑦 ⋅ 𝑑𝑥 instead of (𝑔 ∘ 𝑓)′ (𝑥) = 𝑓 ′ (𝑥)(𝑔′ ∘ 𝑓)(𝑥), it becomes easy to generalize it to compositions of more than two functions: if 𝑧 is a function of 𝑦 which is a function of 𝑡 which is a function of 𝑥, then 𝑑𝑧 𝑑𝑧 𝑑𝑦 𝑑𝑡 = ⋅ ⋅ 𝑑𝑥 𝑑𝑦 𝑑𝑡 𝑑𝑥 and so on. You can see why the rule is called the Chain Rule. Remark. We have a peculiar situation here where the mathematical notation we introduced is not very helpful in itself. The formula using the composition notation is difficult to memorize. There exists a much easier way of writing the same thing, but it uses a different notation. d𝑦 When we re-write the chain rule in Leibniz's notation, using d𝑥 instead of y ′ (𝑥), we consider z to be a function of the variable y which is itself a function of x (y and z are, therefore, dependent variables), and so, z becomes a function of x as well. In that approach, the notation z denotes a variable, not a function! z as a function of y is a different function (mapping) than z as a function of x! 2.2 Digression: Variables vs functions Suppose you study kinetic energy. The kinetic energy of a body of mass 𝑚 moving at a speed 𝑣 is 𝐸= 𝑚𝑣 2 . 2 We can express this same energy in terms of the momentum 𝑝 = 𝑚𝑣: 𝑝2 𝐸 = 2𝑚. The energy 𝐸 is the same variable in the two formulas. But evidently, 𝐸 is not the same function of 𝑣 as it is a function of 𝑝 ! In Lebnitz notation, a letter denotes a variable rather than a function. 7 Calculus 1, Chapter 1 Fall 2021 The Chain Rule, example 1 Consider the function 2 𝑦 = 𝑒 sin(𝑥 ) . It can be decomposed as a composition of three functions: 𝑦 = 𝑓(𝑢) = 𝑒 𝑢 𝑢 = 𝑔(𝑣) = sin 𝑣 𝑣 = ℎ(𝑥) = 𝑥 2 𝑑𝑦 𝑑𝑢 Their derivatives are: = 𝑓 ′ (𝑢) = 𝑒 𝑢 𝑑𝑢 = 𝑔′ (𝑣) = cos 𝑣 𝑑𝑣 𝑑𝑣 = ℎ′ (𝑥) = 2𝑥 𝑑𝑥 In Leibniz notation, the result can be written: 𝑑𝑦 𝑑𝑦 𝑑𝑢 𝑑𝑣 | | | = ⋅ ⋅ 𝑑𝑥 𝑑𝑢 𝑢=𝑔(ℎ(𝑎)) 𝑑𝑣 𝑣=ℎ(𝑎) 𝑑𝑥 𝑥=𝑎 2 = 𝑒 sin(𝑥 ) ⋅ cos(𝑥 2 ) ⋅ 2𝑥 The Chain Rule, example 2 3 Consider the function 𝑧 = 𝑓(𝑥) = √1 − 𝑥 2 , defined on [−1, 1 ] 2 1 We can write 𝑓 = 𝑣 ∘ 𝑢 where 𝑢(𝑥) = 1 − 𝑥² and 𝑣(𝑥) = √𝑥 0 Or we can write 𝑧 = 𝑣(𝑢(𝑥)), which is the same. -2 -1 -1 𝑢 is differentiable on [-1, 1] as a polynomial, and it has values in [0, +∞[. -2 𝑑𝑢 -3 Moreover, ∀𝑥 ∈ [−1, 1], 𝑢′ (𝑥) = 𝑑𝑥 = −2𝑥. 𝑣 is differentiable on (0; +∞) (this is assumed to be known). 𝑑𝑣 f(x) 0 1 f'(x) 1 Moreover, ∀𝑢 ∈ (0; +∞), 𝑣 ′ (𝑢) = 𝑑𝑢 = 2 𝑢. √ 𝑑𝑓 𝑑𝑣 𝑑𝑢 𝑑𝑥 Therefore, 𝑑𝑥 = 𝑑𝑢 = 1 2√1−𝑥 2 (−2𝑥) = − 𝑥 √1−𝑥 2 If we consider that 𝑧 is the same variable whether it’s represented as a function of 𝑥 (as 𝑓(𝑥)) or a function of 𝑢 (as 𝑣(𝑢)), we can write 𝑑𝑧 𝑑𝑥 𝑑𝑧 𝑑𝑢 𝑑𝑥 = 𝑑𝑢 which has the form stated in the Chain Rule. A problem: 𝒗 is not differentiable on [𝟎; +∞). Therefore, 𝑓(𝑥) is only differentiable on (-1, 1): ∀𝑥 ∈ (−1, 1), 𝑢(𝑥) > 0 Therefore, 𝑓 is differentiable on (-1;1) as a composition of differentiable functions, and 8 2 Calculus 1, Chapter 1 ∀𝒙 ∈ (−𝟏, 𝟏)𝒇′ (𝒙) = 𝒖′ (𝒙) × 𝒗′ ∘ 𝒖(𝒙) = −𝟐𝒙 × Fall 2021 𝟏 𝟐√𝟏 − 𝒙𝟐 =− 𝒙 √𝟏 − 𝒙𝟐 One cannot say anything offhand concerning the differentiability of f at 𝑥 = ±1. It has to be studied separately. Differentiating an inverse function Theorem 2. 2. 2: Let f be a bijection between an interval I and an interval J, differentiable in I and whose derivative does not become zero on J. Then f -1 is differentiable on J and its derivative is given by 1 ∀𝑥 ∈ 𝐽, (𝑓 −1 )′ (𝑥) = . 𝑓 ′ (𝑓 −1 (𝑥)) In Leibniz notation, this can be expressed in an easy-to-memorize form: If 𝑦 = 𝑓(𝑥) and there exists an inverse function 𝑓 −1 , then 𝑥 = 𝑓 −1 (𝑦), and 𝑑𝑥 𝑑𝑦 = 1 𝑑𝑦 𝑑𝑥 . Example 1: The exponential function is differentiable on ℝ and is a bijection between ℝ and (0, + ∞). Moreover, ∀𝑥 ∈ ℝ, exp′ (𝑥) = exp (𝑥) ≠ 0 The log function is the inverse of exp, therefore it is differentiable on (0, + ∞) and ∀𝑥 ∈ (0; +∞), ln′ (𝑥) = 1 1 = exp (ln(𝑥)) 𝑥 Proof: 𝑥 = 𝑒𝑦 𝑦 = ln 𝑥 𝑑𝑥 𝑑𝑦 = 𝑒𝑦 𝑑𝑦 1 1 1 = = 𝑦= 𝑑𝑥 𝑑𝑥 𝑒 𝑥 𝑑𝑦 𝜋 𝜋 𝜋 𝜋 Example 2: The function tan is differentiable on(− 2 ; 2 ) and is a bijection between (− 2 ; 2 ) and ℝ. Moreover, ∀𝑥 ∈ ℝ, tan′ (𝑥) = 1 + tan²(𝑥) ≠ 0 The arctan is the inverse of function tan, therefore it is differentiable on ℝ and ∀𝑥 ∈ ℝ, arctan′ (𝑥) = 1 tan′ (arctan(𝑥)) 𝑦 = arctan 𝑥 = 1 1 = 1 + tan²(arctan(𝑥)) 1 + 𝑥² 𝑥 = tan 𝑦 𝑑𝑥 𝑑𝑦 9 = 1 + tan2 𝑦 Calculus 1, Chapter 1 𝑑𝑦 1 1 1 = = = 2 𝑑𝑥 𝑑𝑥 1 + tan 𝑦 1 + 𝑥 2 𝑑𝑦 Fall 2021 2.3. Applying derivatives to computation of limits Theorem 2. 3. 1. L'Hôpital’s Rule 0 0 When dealing with indeterminate forms of « » or « ∞ ∞ » types, it is often convenient to use L'Hôpital’s Rule. It says that if we have a quotient of two functions: 𝑓(𝑥) and 𝑔(𝑥), and • lim 𝑓(𝑥) = lim 𝑔(𝑥) = 0 𝑥→𝑐 𝑥→𝑐 lim 𝑓(𝑥) = lim 𝑔(𝑥) = ±∞ or 𝑥→𝑐 𝑥→𝑐 and: (i) 𝑓 and 𝑔 are differentiable in a neighborhood of c, (ii) 𝑔′ (𝑥) ≠ 0 in a neighborhood of c (iii) Then lim 𝑓′(𝑥) lim 𝑔′(𝑥) 𝑥→𝑐 𝑓(𝑥) 𝑥→𝑐 𝑔(𝑥) Important!!! exists, = lim 𝑓′(𝑥) 𝑥→c 𝑔′(𝑥) . The point c can be an end of an open interval, and in particular, c can be ± ∞. L’Hôpital’s Rule, examples Example 1: 𝑥 2 − 2𝑥 𝑥→0 tan(𝑥) Find lim Set 𝑓(𝑥) = 𝑥 2 − 2𝑥 and 𝑔(𝑥) = tan(𝑥). 𝑓(0) = 0 and 𝑔(0) = 0 𝑓 is differentiable on ℝ and ∀𝑥 ∈ ℝ, 𝑓 ′ (𝑥) = 2𝑥 − 2 𝜋 𝜋 𝜋 𝜋 𝑔 is differentiable on (− 2 ; 2 ) and ∀𝑥 ∈ (− 2 ; 2 ) , 𝑔′ (𝑥) = 1 + tan²(𝑥) Therefore, l'Hôpital’s rule yields 𝒙𝟐 − 𝟐𝒙 𝒇(𝒙) 𝒇′(𝒙) −𝟐 = 𝐥𝐢𝐦 = 𝐥𝐢𝐦 = = −𝟐 𝒙→𝟎 𝐭𝐚𝐧(𝒙) 𝒙→𝟎 𝒈(𝒙) 𝒙→𝟎 𝒈′(𝒙) 𝟏 𝐥𝐢𝐦 Example 2: 𝑒 𝑥² −1 Find lim 𝑥 sin(𝑥) 𝑥→0 0 (« 0 ») Set 𝑓(𝑥) = 𝑒 𝑥² − 1 and 𝑔(𝑥) = 𝑥 sin(𝑥). 𝑓(0) = 𝑔(0) = 0. 2 𝑓 ′ (𝑥) = 2𝑥𝑒 𝑥 . 𝑔′ (𝑥) = sin(𝑥) + 𝑥 cos(𝑥) 10 Calculus 1, Chapter 1 𝑒 𝑥² −1 lim 𝑥 sin(𝑥) 𝑥→0 = 𝑓′(𝑥) lim 𝑔′(𝑥) 𝑥→0 Fall 2021 2 = 2𝑥𝑒 𝑥 lim sin(𝑥)+𝑥 cos(𝑥) 𝑥→0 0 0 But this is still an indeterminate form of « » type! Let’s use l’Hôpital on that expression: 2 𝑡(𝑥) = 2𝑥𝑒 𝑥 , 𝑠(𝑥) = sin 𝑥 + 𝑥 cos 𝑥 2 𝑡 ′ (𝑥) = 𝑓 ′′ (𝑥) = 2𝑒 𝑥 + 4𝑥 2 𝑒 𝑥² 𝑠 ′ (𝑥) = 𝑔′′ (𝑥) = cos(𝑥) cos(𝑥) − 𝑥 sin(𝑥) = 2 cos 𝑥 − 𝑥 sin 𝑥 2 2 2𝑥𝑒 𝑥 2𝑒 𝑥 +4𝑥 2 𝑒 𝑥² 2 lim sin(𝑥)+𝑥 cos(𝑥) = lim 2 cos 𝑥 −𝑥 sin 𝑥 = 2=1 𝑥→0 𝑥→0 𝑒 𝑥² −1 Result: lim 𝑥 sin(𝑥)=1 𝑥→0 You can see that in this case, to obtain a result we had to use l'Hôpital’s rule twice. Example 3: Find lim 1 cos (𝑥 ) − 1 𝑥→+∞ 1 sin (𝑥 ) Here we are dealing with a limit at +∞, which means a one-side limit at the end of an open interval. This does not change the calculations. 1 𝑥 ′ 1 𝑥 1 𝑥 1 𝑥 𝑓′(𝑥) = (cos − 1) = − ( 2 ) (− sin ) = ( 2 ) sin 1 ′ 1 1 𝑥 1 𝑔′ (𝑥) = (sin 𝑥) = − (𝑥 2 ) cos 𝑥 1 sin 𝑥 𝑓′(𝑥) =− → 0 1 𝑥→+∞ 𝑔′(𝑥) cos 𝑥 Therefore, the limit we seek equals 0. (It is also possible to introduce a new variable 1 𝑥 𝑦 = and calculate the limit of cos 𝑦−1 sin 𝑦 at 𝑦 → 0+ ) L’Hôpital’s Rule, warnings I did not put here any example where l’Hôpital’s rule won’t work. You can find several such examples in the Wikipedia article https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule . All of them are related to violating one of the requirements (i) – (iii) listed when formulating the theorem. 11
