SMA 2101 Calculus I PDF

SMA 2101: CALCULUS I ⃝Francis c O. Ochieng [email protected] Department of Pure and Applied Mathematics Jomo Kenyatta Unive...

SMA 2101: CALCULUS I ⃝Francis c O. Ochieng [email protected] Department of Pure and Applied Mathematics Jomo Kenyatta University of Agriculture and Technology Course content Functions: definition, domain, range, codomain, composition (or composite), inverse. Limits, continuity and differentiability of a function. Differentiation by first principle and by rule for xn (integral and fractional n). Other techniques of differentiation, i.e., sums, products, quotients, chain rule; their applications to algebraic, trigonometric, logarithmic, exponential, and inverse trigonometric functions all of a single variable. Implicit and parametric differentiation. Applications of differentiation to: rates of change, small changes, stationary points, equations of tangents and normal lines, kinematics, and economics and financial models (cost, revenue and profit). Introduction to integration and its applications to area and volume. References Calculus: Early Transcendentals (8th Edition) by James Stewart Calculus with Analytic Geometry by Roland E. Larson, Robert P. Hostetler and Bruce H. Edwards; 5th edition Calculus and Analytical Geometry (9th edition) by George B. Thomas and Ross L. Finney Advanced Engineering Mathematics (10th ed.) by Erwin Kreyszig Calculus by Larson Hostellem Lecture 1 1 Functions To understand the word function, we consider the following scenario and definitions. For example, the growth of a sidling is an instance of a functional relation, since the growth may be affected by variations in temperature, moisture, sunlight, etc. If all these factors remain constant, then the growth is a function of time. Definition 1.1 (Variables). A variable is an object, event, time period, or any other type of category you are trying to measure. 1  ⃝Francis c Oketch Consider the formula used for calculating the volume of a sphere of radius r. 4 V = πr3 (1) 3 Then, i) V and r vary with different spheres. Hence, they are called variables. 4 ii) π and are constants, irrespective of the size of the sphere. 3 There are two types of variables, i.e., independent and dependent variables. Definition 1.2 (Independent and dependent variables). Independent variable refers to the input value while dependent variable refers to the output value. For example from formula (1), the volume, V , depends on the value of the radius, r, of the sphere. In this case, r is called the independent variable while V is called the dependent variable since it is affected by the variation of r. Similarly, for the function y = ax2 + bx + c, a, b and c are constants, x is the independent variable and y is the dependent variable. Definition 1.3 (Function). A function is a rule that assigns/associates each element in the independent set, say X, to a unique element in the dependent set, say Y. Examples of functions are i) Linear functions e.g., y = x + 5 ii) Quadratic functions e.g., y = x2 − 2x + 5 iii) Cubic functions e.g., y = x3 − 1 iv) Quartic functions e.g., y = 2x4 + x3 − 1 v) Trigonometric functions e.g., y = sin(2x + 5) vi) Logarithmic functions (log to base 10) e.g., y = log(3x + 1) vii) Natural logarithmic functions (log to base e ≈ 2.71828) e.g., y = ln(5x + 1) viii) Inverse of trigonometric functions e.g., y = tan−1 (2x + 1) ix) Exponential functions e.g., y = e2x+1 x) Absolute value functions e.g., y = |x|. This function is defined as { −x, if x < 0 y = |x| = x, if x ≥ 0 → Note: in the above examples the variable y depends on the variable x. Thus, we say that the dependent variable y is a function of the independent variable x. Using function notation, we write y = f (x), where f is a function. The function f (x) is read as f of x, meaning that f depends on x. 2 1.1 Domain, Range and Codomain ⃝Francis c Oketch 1.1 Domain, Range and Codomain Definition 1.4 (Domain). A domain consists of all the elements in the independent set (i.e., the set of inputs), X, for which the function is defined. Definition 1.5 (Range). A range refers to a set of all the images of the elements in the domain. Definition 1.6 (Codomain). A codomain consists of all the elements in the dependent set (i.e., the set of outputs), Y. For example, consider the diagram below Example(s): 1. Find the domain and range of the following functions. (a) f (x) = (x − 4)2 + 5 Solution  Since f (x) is defined (or is a real number) for any real number x, the domain of f is the interval (−∞, ∞). √  Let y = (x − 4)2 + 5. Making x the subject, we have x = 4 ± y − 5. This function is defined if y − 5 ≥ 0 or y ≥ 5. Therefore, the range is the interval [5, ∞). (b) f (x) = 2x2 − 5x + 1 Solution  Since f (x) is defined (or is a real number) for any real number x, the domain of f is the interval (−∞, ∞).  Let y = 2x2 − 5x + 1 or 2x√ 2 − 5x + (1 − y) = 0. Making x the subject (use quadratic 5 ± 25 − 8(1 − y) formula), we have x =. This function is defined if 25 − 8(1 − y) ≥ 0 4 [ ) 17 or y ≥ −. Therefore, the range is the interval − 178 ,∞. 8 4 (c) f (x) = 2 x − 5x + 6 Solution → Note: 4/0 = ∞ (infinity), vvvv large value, undefined, indeterminate.  The function f (x) is defined when the denominator is nonzero, i.e., if x2 − 5x + 6 ̸= 0. Solving yields x ̸= 2 and x ̸= 3. Therefore, the domain of f includes all the real numbers of x except x = 2 and x = 3, i.e., the set (−∞, ∞)\{2, 3} or (−∞, 2) ∪ (2, 3) ∪ (3, ∞). 4 ( )  Let y = 2 or x2 − 5x + 6 − y4 = 0. Making x the subject (use quadratic x − 5x + 6 formula), we have √ ( ) 5± 25 − 4 6 − 4 y x= 2 ( ) This function is defined if 25 − 4 6 − 4 y ≥ 0 or y ≥ −16. Therefore, the range is the interval [−16, ∞). 3 1.2 Evaluation of functions ⃝Francis c Oketch √ (d) f (x) = x−1 Solution  Since f (x) is defined (or is a real number) if x − 1 ≥ 0 or x ≥ 1, the domain of f is the interval [1, ∞). √  Let y = x − 1. Making x the subject, we have x = y 2 + 1. This function is defined for any real number y. Therefore, the range is the interval (−∞, ∞). (e) f (x) = 2|x − 3| + 4 Solution  Since f (x) is defined for all real numbers, the domain of f is the interval (−∞, ∞).  Since for all |x − 3| ≥ 0, the function f (x) = 2|x − 3| + 4 ≥ 4. Therefore, the range is all the values of y for which y ≥ 4 or the interval [4, ∞). Exercise: 1. Find the domain and range of the following functions. (a) f (x) = 6 − x2. [ans: domain (−∞, ∞), range (−∞, 6]] 6 + 3x (b) f (x) =. [ans: domain (−∞, 0.5) ∪ (0.5, ∞), range (−∞, 1.5) ∪ (1.5, ∞)] 1 − 2x x+5 (c) f (x) =. [ans: domain (−∞, 2) ∪ (2, ∞), range (−∞, 1) ∪ (1, ∞)] x−2 √ (d) f (x) = 4 − 2x + 5. [ans: domain (−∞, 2], range (−∞, ∞)] √ x2 − 16 [ ] (e) f (x) =.[ans: domain (−∞, −4) ∪ [4, 6) ∪ (6, ∞), range − √2 , √2 \{−1, 1}] x2 − 2x − 24 3 3 1.2 Evaluation of functions This involves replacing x in the function by the suggested value and retaining the rule of the function. Example(s): f (x + h) − f (x) 1. Given f (x) = 2x + 1. Find: (i) f (0), (ii) f (1), (iii) f (x + 2), and (iv) for h ̸= 0. h Solution i) f (0) = 2(0) + 1 = 0 + 1 = 1 ii) f (1) = 2(1) + 1 = 2 + 1 = 3 iii) f (x + 2) = 2(x + 2) + 1 = 2x + 4 + 1 = 2x + 5 f (x + h) − f (x) [2(x + h) + 1] − [2x + 1] 2x + 2h + 1 − 2x − 1 2h iv) = = = = 2. h h h h f (x + h) − f (x) 2. Given f (x) = 3x2 − 2x + 4. Find: (i) f (0), (ii) f (−1), (iii) f (x + 2), and (iv) h for h ̸= 0. Solution i) f (0) = 3(0)2 − 2(0) + 4 = 0 + 0 + 4 = 4 ii) f (−1) = 3(−1)2 − 2(−1) + 4 = 3 + 2 + 4 = 9 iii) f (x + 2) = 3(x + 2)2 − 2(x + 2) + 4 = 3(x2 + 4x + 4) − 2x − 4 + 4 = 3x2 + 10x + 12 4 1.3 Composite functions ⃝Francis c Oketch iv) [ ] [ ] f (x + h) − f (x) 3(x + h)2 − 2(x + h) + 4 − 3x2 − 2x + 4 = h ( 2 h ) ( ) 3x + 6hx + 3h2 − 2x − 2h + 4 − 3x2 − 2x + 4 6hx + 3h2 − 2h = = h h = 6x + 3h − 2 3. Given f (x) = x2 − 4x + 3. Find: (i) f (1), (ii) f (2), (iii) f (a), and (iv) f (a + h). Solution i) f (x) = x2 − 4x + 3 ⇒ f (1) = 12 − 4(1) + 3 = 0 ii) f (x) = x2 − 4x + 3 ⇒ f (2) = 22 − 4(2) + 3 = −1 iii) f (x) = x2 − 4x + 3 ⇒ f (a) = a2 − 4a + 3 iv) f (x) = x2 − 4x + 3 ⇒ f (a + h) = (a + h)2 − 4(a + h) + 3 4. Given ϕ(θ) = 2 sin θ. Find: (i) ϕ( π2 ), (ii) ϕ(0), and (iii) ϕ( π3 ). Solution (π) i) ϕ(θ) = 2 sin θ ⇒ ϕ( π2 ) = 2 sin 2 =2 ii) ϕ(θ) = 2 sin θ ⇒ ϕ(0) = 2 sin (0) = 0 √ (π) 3 √ iii) ϕ(θ) = 2 sin θ ⇒ ϕ( π3 ) = 2 sin 3 =2× = 3 2 Exercise: f (x + h) − f (x) (a) Given f (x) = x3 + 2x + 1, find: (i) f (0), (ii) f (−a), (iii) f (x + 2), and (iv) for h h ̸= 0. 1 g(x + h) − g(x) (b) Given g(x) = √ , find: (i) f (0), (ii) f (1), (iii) f (x + 2), and (iv) for h ̸= 0. x+1 h 6 − 2x p(x + h) − p(x) (c) Given p(x) = , find: (i) f (0), (ii) f (−1), (iii) f (2 − x), and (iv) for h ̸= 0. 1 + 3x h (d) If f (x) = 2x2 − 4x + 1, find (i) f (1), (ii) f (0), (iii) f (2), (iv) f (a), and f (x + h). (e) If f (x) = (x − 1)(x + 5), find (i) f (1), (ii) f (0), (iii) f (2), (iv) f (a + 1), and f ( a1 ). (f) If f (θ) = cos θ, find (i) f ( π2 ), (ii) f (0), (iii) f ( π3 ), (iv) f ( π6 ), and (v) f (π). f (3.001) − f (3) (g) If f (x) = x2 , find (i) f (3), (ii) f (3.1), (iii) f (3.01), (iv) f (3.001), and. 0.001 (h) If ϕ(x) = 2x , find (i) ϕ(0), (ii) ϕ(1), and (iii) ϕ(0.5). 1.3 Composite functions The composition of functions is a function of another function. Consider the function f with domain A and range B, and the function g with domain D and range E. If B is a subset of D, then the composite function (gof )(x) is the function with domain A and range E such that (gof )(x) = g(f (x)) For example, given f (x) = 2x + 1 and g(x) = 5x − 3. Then, (gof )(x) = g(f (x)) = g(2x + 1) = 5(2x + 1) − 3 = 10x + 2 5  ⃝Francis c Oketch Similarly, (f og)(x) = f (g(x)) = f (5x − 3) = 2(5x − 3) + 1 = 10x − 5 → Note: (f og)(x) ̸= (gof )(x). Exercise: √ 1. Given f (x) = x2 − 1, g(x) = x − 1 and h(x) = x. Find: (a) (f og)(x) (b) (hog)(x) (c) (gog)(x) (d) (gohof )(x) 2. Consider the functions f (x) = x2 + 1 and g(x) = 1/x. Evaluate (a) (f og)(4) (b) (gof )(−1/2) √ 3. If f (x) = x and g(x) = 4x + 2, find the domain of (f og)(x). [ans: x ≥ −0.5 or (−∞, −0.5]] Lecture 2 2 Limits of functions Definition 2.1 (Basic limit definition). Let f (x) be a function and let a and L be real numbers. If f (x) approaches L as x approaches a from either RHS or LHS of a (but is not equal to a), then we say that f (x) has limit L as x approaches a, and is mathematically written as:. = L. lim f (x) x→a Diagrammatically, we have → Note: lim f (x) is the value that f (x) x→a approaches as x approaches a, and a does not have to be in the domain of f. 2.1 Properties of limits Theorem 2.1. Suppose lim f (x) = L1 and lim g(x) = L2. Then, x→a x→a [ ] [ ] 1. [Addition/subtraction rule] lim [f (x) ± g(x)] = lim f (x) ± lim g(x) = L1 ± L2 x→a x→a x→a [ ] 2. [Scalar multiple] lim [λf (x)] = λ lim f (x) = λL1 , where λ is a constant. x→a x→a [ ] [ ] 3. [Product rule] lim [f (x) · g(x)] = lim f (x) · lim g(x) = L1 · L2 x→a x→a x→a [ ] lim f (x) f (x) L1 4. [Quotient rule] lim = x→a = , provided g(a) ̸= 0. x→a g(x) lim g(x) L2 x→a √ ( )1/n √ 5. lim n f (x) = lim f (x) = (L1 )1/n = n L1 x→a x→a → Note: if f (x) = c (where c is a constant), then lim [f (x)] = lim [c] = c x→a x→a 6 2.2 Techniques of evaluating limits of functions ⃝Francis c Oketch 2.2 Techniques of evaluating limits of functions  Direct substitution (DS) The required limit is obtained by just plugging in the value of input, say x, into the given function, say f (x). Example(s): (a) Evaluate lim 3x3 − x2 + 2x + 5. x→2 Solution lim (3x3 − x2 + 2x + 5) = 3 lim x3 − lim x2 + 2 lim x + lim 5) x→2 x→2 x→2 x→2 x→2 = 3(2 ) − (2 ) + 2(2) + 5 3 2 = 29 x2 − 1 (b) Evaluate lim. x→1 x + 1 Solution x2 − 1 D.S 12 − 1 0 lim = = =0 x→1 x + 1 1+1 2  Factorization If on direct substitution we get the indeterminate form 0/0, then it means that there is a common factor in both the numerator and denominator. In this case, we perform factorization first so as to simplify the given function. → Note: if the polynomial in the numerator is of degree greater than the degree of the polynomial in the denominator, we first need to perform long division. Example(s): x2 + x − 6 (a) Evaluate lim x→2 x−2 Solution x2 + x − 6 (x − 2)(x + 3) lim = lim x→2 x−2 x→2 x−2 D.S = lim (x + 3) = 2 + 3 x→2 = 5 x2 + 3x + 2 (b) Evaluate lim x→−2 2x2 − 8 Solution x2 + 3x + 2 (x + 2)(x + 1) lim = lim x→−2 2x2 − 8 x→−2 2(x + 2)(x − 2) x + 1 D.S −2 + 1 −1 = lim = = x→−2 2(x − 2) 2(−2 − 2) −8 1 = 8 x3 − 1 (c) Evaluate lim. x→1 x2 − 1 7 2.2 Techniques of evaluating limits of functions ⃝Francis c Oketch Solution ( ) x3 − 1 x−1 lim 2 = lim x + 2 (long division) x→1 x − 1 x→1 x −1 [ ] x−1 = lim x + (factorization) x→1 (x − 1)(x + 1) ( ) 1 D.S 1 1 = lim x + = 1+ =1+ x→1 x+1 1+1 2 3 = 2  Limits at infinity In this case, we first divide the numerator and denominator by the highest power of x in the denominator. Example(s): 5x3 − 1 (a) Evaluate lim. x→∞ 4x3 − 2x − 7 Solution 1 1 5x3 − 1 5− 5− 5−0 lim = lim x3 D.S = ∞ = x→∞ 4x3 − 2x − 7 x→∞ 2 7 2 7 4−0−0 4− 2 − 3 4− − x x ∞ ∞ 5 = 4  Rationalization Suppose there exists a surd in either the numerator or denominator or both. Then, we first need to multiply both the numerator and denominator by the conjugate of the factor containing the surd (in either the numerator or denominator) and then simplify the resulting function. After rationalization, we perform a direct substitution. → Note: in case the surds appear in both the numerator and denominator, then we rationalize the denominator. Example(s): √ (a) Evaluate lim x2 − 4x − x. x→∞ Solution (√ ) √ (√ ) x2 − 4x + x lim x2 − 4x − x = lim x2 − 4x − x √ x→∞ x→∞ x2 − 4x + x x2 − 4x − x2 −4x = lim √ = lim √ x→∞ x − 4x + x 2 x→∞ x − 4x + x 2 1 −4x · −4 lim (√ x lim √ = ) 1 = x→∞ x→∞ x2 − 4x + x · 4 1− +1 x x D.S −4 −4 −4 = √ =√ = 4 1−0+1 1+1 1− +1 ∞ = −2 8 2.2 Techniques of evaluating limits of functions ⃝Francis c Oketch √ x−3 (b) Evaluate lim. x→9 x − 9 Solution √ √ √ x−3 ( x − 3)( x + 3) lim = lim √ x→9 x − 9 x→9 (x − 9)( x + 3) (x − 9) 1 = lim √ = lim √ x→9 (x − 9)( x + 3) x→9 x+3 D.S 1 1 = √ = 9+3 3 + 3 1 = 6 Exercise: √ √ (a) lim x2 − 2 − x2 + x. x→∞ Solution (√ ) √ √ √ √ √ x2 − 2 + x2 + x lim x2 − 2 − x2 + x = lim x2 − 2 − x2 + x · √ √ x→∞ x→∞ x2 − 2 + x2 + x (x2 − 2) − (x2 + x) −2 − x = lim √ √ = lim √ √ x→∞ x −2+ x +x 2 2 x→∞ x − 2 + x2 + x 2 2 2 − −1 − −1 = lim √ x √ D.S = √ ∞ √ x→∞ 2 1 2 1 1− 2 + 1+ 1− + 1+ x x ∞ ∞ 1 = − 2 x3 − 1 (b) Evaluate lim. [ans: 3] x→1 x − 1 √ 1− x (c) Evaluate lim. [ans: 1/2] x→1 1 − x 5x2 − 3x + 2 (d) Evaluate lim. [ans: 1/2] x→∞ 10x2 − x + 100 √ √ 1+x− 1−x (e) Evaluate lim. [ans: 1] x→0 x → Note: A function which grows arbitrarily large as x goes to positive or negative infinity is said to have an infinite limit. Infinity is not a real number, so if a function has infinite limit, we say that the limit does not exist. Lecture 3 Theorem 2.2 (Squeeze law (sandwich theorem)). Suppose that f (x) ≤ g(x) ≤ h(x) holds for all x around a, except possibly at x = a. If lim f (x) = lim h(x) = L, then lim g(x) = L. x→a x→a x→a Example(s): ( ) 1 1. Find lim x sin. x→0 x2 + x Solution 9 2.2 Techniques of evaluating limits of functions ⃝Francis c Oketch We know that sin θ is sandwiched between −1 and 1 i.e., −1 ≤ sin(θ) ≤ 1. Therefore, ( ) 1 As −1 ≤ sin ≤1 x + x2 ( ) 1 ⇒ −x ≤ x sin ≤x x + x2 ( ) 1 ⇒ − lim (x) ≤ lim x sin ≤ lim (x) x→0 x→0 x + x2 x→0 ( ) 1 ⇒ 0 ≤ lim x sin ≤0 x→0 x + x2 ( ) 1 ⇒ lim x sin =0 x→0 x + x2 Exercise: sin h (1 − cos h) 1. [Assignment 1 ] Prove that lim = 1 and lim = 0. h→0 h h→0 h Proof. Consider the following unit circle. Let the length of line OA be a units, AB is b units, CD be c units and angle AOB be h. Here, OB = OC = 1 unit. Now, cos h = a, sin h = b and tan h = c. From the figure, the area of triangle OAB is less than that of the sector OCB which is also less than that of 1 h 1 triangle OCD i.e., ab ≤ · π(1)2 ≤ (1)c. 2 2π 2 Thus, 1 1 1 cos h sin h ≤ h ≤ tan h 2 2 2 sin h Multiply through by 2 and using the identity tan h = , we have cos h sin h cos h sin h ≤ h ≤ cos h Taking reciprocals, we have 1 1 cos h ≥ ≥ cos h sin h h sin h 1 sin h Multiplying though by sin h yields ≥ ≥ cos h, which can be rewritten as cos h h sin h 1 cos h ≤ ≤ h cos h sin h 1 sin h Taking limit as h → 0, we have lim cos h ≤ lim ≤ lim. That is, 1 ≤ lim ≤ 1. h→0 h→0 h h→0 cos h h→0 h Hence, by the squeeze law we get sin. h lim =1 h→0 h 10 2.3 One-Sided Limit ⃝Francis c Oketch Also, [ ] [ ] (1 − cos h) (1 − cos h) (1 + cos h) 1 − cos2 h 1 lim = lim · = lim · h→0 h h→0 h (1 + cos h) h→0 h 1 + cos h [ ] [ ] sin2 h 1 sin h sin h = lim · = lim · h→0 h 1 + cos h h→0 h 1 + cos h [ ][ ] [ ] sin h sin h sin h D.S 0 = lim lim = (1) lim = h→0 h h→0 (1 + cos h) h→0 (1 + cos h) (1 + 1) = 0 Therefore, (1 − cos. h) = 0 lim h→0 h 2.3 One-Sided Limit Definition 2.2 (Left-Hand Limit). If a function f (x) approaches the number L as x approaches the real number a from the LHS of a, then we say that L is the left-hand limit of f at x = a and is written as: lim f (x). = L. − x→a Definition 2.3 (Right-Hand Limit). If a function f (x) approaches the number L as x approaches the real number a from the RHS of a, then we say that L is the right-hand limit of f at x = a and is written as: lim f (x). = L. + x→a → Note: the limit of f (x) as x approaches a exists if both left-hand limit and right-hand limit exist and are equal at x = a. In that case, we have lim f (x) = lim f.(x) = lim f (x) = L x→a− x→a+ x→a Example(s):    x 3 if x < 1 (a) Consider the function defined by f (x) = 1 if x = 1 Evaluate lim f (x).   2−x x→1 if x > 1. Solution (i) LHL: lim f (x) = lim (x3 ) = 13 = 1 x→1− x→1− (ii) RHL: lim f (x) = lim (2 − x) = 2 − 1 = 1 x→1+ x→1+ (iii) Since the result (i) = (ii), we get lim f (x) = 1 x→1 Exercise:    x − 2x 2 if x < 1 (a) Consider the function defined by f (x) = 2 if x = 1. Evaluate lim f (x) and   3x − 4 x→1− if x > 1. lim f (x). x→1+ { 2 − 3x if x ≤ 1 (b) Consider the function defined by f (x) =. Does lim f (x) exist? 2x3 if x > 1 x→1 { x if x ̸= 0 (c) Find the value of lim f (x) where f (x) = x→0 1 if x = 0. 11 2.3 One-Sided Limit ⃝Francis c Oketch  Meaning of absolute value functions To separate (or split) the function contained in the absolute value function, do the following: i) First identify the reference point by equating the interior term to zero. ii) Investigate the signs of the interior expression to the left and the right of the reference point. For example,  If f (x) = |x − 3|. The reference point is x − 3 = 0 ⇒ x = 3. Thus, { −(x − 3) if x < 3 f (x) = +(x − 3) if x ≥ 3  If f (x) = 5 + |x + 5|. The reference point is x + 5 = 0 ⇒ x = −5. Thus, { 5 − (x + 5) if x < −5 f (x) = 5 + (x + 5) if x ≥ −5   1   if x < 0  2 + x 1  If f (x) =. The reference point is x = 0. Thus, f (x) =. 2 − |x|     1  if x ≥ 0 2−x Example(s): |5x| (a) Evaluate lim. x→0 x Solution The reference point is 5x = 0 ⇒ x = 0. Thus, we have { −(5x) |5x| x if x < 0 f (x) = = +(5x) x x if x > 0 Now, −5x (i) LHL: lim f (x) = lim = −5 x→0− x→0− x 5x (ii) RHL: lim f (x) = lim =5 x→0 + x→0 x + |5x| (iii) Since (i) ̸= (ii), therefore, lim does not exist. The above problem possesses a one-sided x→0 x limits. ( ) |x − 7| (b) Evaluate lim. x→7 (x − 7) Solution The reference point is x − 7 = 0 ⇒ x = 7. Thus, we have { −(x−7) |x − 7| x−7 = −1 if x < 7 f (x) = = x−7 +(x−7) x−7 = +1 if x > 7 Now, (i) LHL: lim f (x) = lim (−1) = −1 x→7− x→0− (ii) RHL: lim f (x) = lim (+1) = 1 x→7+ x→0+ 12  ⃝Francis c Oketch (iii) Since (i) ̸= (ii), therefore, lim f (x) does not exist. The above problem possesses a one-sided x→7 limits. Exercise: ( ) x+6 (a) Evaluate lim. x→6 |x + 6| ( ) 1 (b) Evaluate lim 2 +. x→1 |x − 1| 3 Continuity of a function A function f (x) is said to be continuous at a point x = a if the following three conditions are satisfied: i) f (a) is finite, i.e., f (x) must be defined at x = a. ii) lim f (x) exists (i.e., LHL=RHL at x = a) x→a iii) lim f (x) = f (a), i.e., (ii)=(i) x→a → Note: if at least one of these conditions is not satisfied, then f (x) is discontinuous at x = a. In this case, we say that the point a is a discontinuity of f (i.e., f (x) has some gaps or jumps at x = a). Example(s):  2   x −1  if x < −1 (a) Discuss the continuity of the function f (x) = x+1 at x = −1    x2 − 3 if x ≥ −1 Solution We need to test the three conditions for continuity: (i) f (−1) = (−1)2 − 3 = −2 (defined). (ii) lim f (x): x→−1 ( ) x2 − 1 (x  − 1) +1)(x  LHL: lim f (x) = lim = lim  = lim (x − 1) = −2 x→−1− x→−1− x+1 x→−1− (x  +1) x→−1− RHL: lim f (x) = lim (x2 − 3) = 1 − 3 = −2 x→−1+ x→−1+ Since LHL=RHL=-2, therefore, lim f (x) = −2 x→−1 (iii) So, as lim f (x) = f (−1) therefore, f (x) is continuous on (−4, 4). x→−1 2x4 − 6x3 + x2 + 3 (b) Discuss the continuity of the function f (x) = at x = 1. x−1 Solution 2x4 − 6x3 + x2 + 3 Clearly, the function f (x) = is discontinuous at x = 1. However, the point x−1 of discontinuity can be removed by first simplifying the given function. Thus, by long division 13  ⃝Francis c Oketch we have 2x3 − 4x2 − 3x − 3 ) x−1 2x4 − 6x3 + x2 +3 − 2x4 + 2x3 − 4x3 + x2 4x3 − 4x2 − 3x2 3x2 − 3x − 3x + 3 3x − 3 0 Hence, the function can be rewritten in the simplest form f (x) = 2x3 − 4x2 − 3x − 3, which is now continuous at x = 1 [student to verify this]. Therefore, the original function is said to have a removable point of discontinuity. (c) Find the value of the constants in the give problems if f (x) is continuous everywhere in the real number line { 4+c if x < 1 i) Given f (x) =. Find c [ans: c = 2] 4x + 2 if x ≥ 1    if x < −1 15x ii) Given f (x) = ax + b if − 1 ≤ x < 2. Find a and b [ans: a = 3, b = 18]   12x if x > 2 Solution In these questions, we make use of the second condition of continuity in particular, i.e., LHL=RHL at any point x = a. i) The reference point is x = 1. Thus, LHL: lim f (x) = lim (4 + c) = 4 + c x→1− x→1− RHL: lim f (x) = lim (4x + 2) = 4 + 2 = 6 x→1+ x→1+ Since f (x) is continuous at x = 1, we have 4 + c = 6. Therefore, c = 2. ii)  Case 1: Taking the reference point as x = −1. Thus, LHL: lim f (x) = lim (15) = 15 x→−1− x→−1− RHL: lim f (x) = lim (ax + b) = −a + b x→−1+ x→−1+ Since f (x) is continuous at x = −1, we have −a + b = 15 (∗).  Case 2: Taking the reference point as x = 2. Thus, LHL: lim f (x) = lim (ax + b) = 2a + b x→2− x→2− RHL: lim f (x) = lim (12x) = 24 x→2+ x→2+ Since f (x) is continuous at x = 2, we have 2a + b = 24 (∗∗). Solving equations (∗) and (∗∗) simultaneously, we obtain a = 3 and b = 18. Exercise: 14  ⃝Francis c Oketch  3   x + 27  if x ̸= −3 (a) Discuss the continuity of the function f (x) = x+3.    27 if x = −3 (b) Find the value of A and B so that the following function is continuous for all x.  ( )   1 − cos x   A if x < 0   sin2 x     f (x) = 2x2 − x + B if 0 ≤ x ≤ 1         x2 + 2x − 3   if x > 1 2 x −1 Solution (( ( A(1 − cos(x)) A( −( (1( cos(x)) A lim f (x) = x→0− x→0− lim sin2 (x) = lim −( (1( x→0− ( ((( + cos(x)) = 2 cos(x))(1 lim f (x) = lim (2x2 − x + B) = B x→0+ x→0+ A Since f (x) to be continuous at x = 0, we have = B − − − (∗). 2 Also, lim f (x) = lim (2x2 − x + B) = 1 + B x→1− x→1− x2 + 2x − 3 (x  + 3) −1)(x  4 lim f (x) = lim = lim  = =2 x→1+ x→1+ x2 − 1 x→1  + (x− 1)(x + 1) 2 Since f (x) to be continuous at x = 1, we have 1 + B = 2 − − − (∗∗). Solving equations (∗) and (∗∗), we get A = 2, B = 1. (c) Find a and b so that the following functions are continuous ∀x ∈ R: i)    2, if x < 1     f (x) = ax + b, if 1 ≤ x < 2       6, if x ≥ 2 [ans: a = 4, b = −2] ii)    −2x, if x < 1     f (x) = b − ax2 , if 1 ≤ x < 4       −16x, if x ≥ 4 [ans: a =, b =] Lecture 4 15  ⃝Francis c Oketch 4 Derivative of functions df Definition 4.1 (First principle). The derivative of a function f (x) denoted by f ′ (x) or is the rate dx of change of f with respect to x, and is given by [ ] f ′ (x) = lim. + h) − f (x) , f (x h→0 h for all x for which this limit exists. The process of finding the derivative f ′ (x) is called differentiation of f (x). The above relation is called first principle of differentiation or differentiation by the definition or differentiation of first kind. Geometrically, consider the curve y = f (x) and let ∆x = h. Here, ∆y = f (x + h) − f (x). So, gradient of the secant ∆y f (x + h) − f (x) line through points A and B is =. ∆x h Taking limit as h → 0 yields dy f (x + h) − f (x) = lim = f ′ (x) dx h→0 h Therefore, dy. ′ = f (x) dx Example(s): (a) Use first principle of differentiation to find the derivative of the function f (x) = x2. Solution Given f (x) = x2 , we have f (x + h) = (x + h)2. By the first principle of differentiation, we have f (x + h) − f (x) (x + h)2 − x2 f ′ (x) = lim = lim h→0 h h→0 h x + 2hx + h − x 2 2 2 2hx + h2 D.S = lim = lim = lim (2x + h) = (2x + 0) h→0 h h→0 h h→0 = 2x 1 (b) Use first principle of differentiation to find the derivative of the following functions: (i) f (x) = √ x and (ii) f (x) = x. Solution 1 1 i) Given f (x) = , we have f (x + h) =. By the first principle of differentiation, we x x+h have ( ) f (x + h) − f (x) 1 x+h − 1 x f ′ (x) = lim = lim h→0 h h→0 h x − (x + h) −h −1 D.S −1 = lim = lim = lim = h→0 hx(x + h) h→0 hx(x + h) h→0 x(x + h) x(x + 0) 1 = − 2 x 16 4.1 Basic differentiation rules ⃝Francis c Oketch √ √ ii) Given f (x) = x, we have f (x + h) = x + h. By the first principle of differentiation, we have √ √ ′ f (x + h) − f (x) x+h− x f (x) = lim = lim h→0 h h→0 h √ √ √ √ ( x + h − x)( x + h + x) x+h−x = lim √ √ = lim √ √ h→0 h( x + h + x) h→0 h( x + h + x) h 1 D.S 1 = lim √ √ = lim √ √ = √ √ h→0 h( x + h + x) h→0 x+h+ x x+0+ x 1 = √ 2 x Exercise: (a) Use first principle of differentiation to find the derivative of the following functions. i) f (x) = −x3 + 3x2 + 4 [ans: f ′ (x) = −3x2 + 6x] 3x 3 ii) f (x) = [ans: f ′ (x) = ] 1 − 5x (1 − 5x)2 −7 + 5x −29 iii) f (x) = [ans: f ′ (x) = ] −3 − 2x (−3 − 2x)2 √ 3 iv) f (x) = 6x + 2 − 5 [ans: f ′ (x) = √ ] 6x + 2 1 −1 v) f (x) = √ [ans: f ′ (x) = √ √ ] x+2 2 x( x + 2)2 4.1 Basic differentiation rules  The derivative of a constant dc If f (x) = c (a constant) for all x, then f ′ (x) = 0 for all x. That is, = f ′.(x) = 0. dx Proof. Given f (x) = c ⇒ f (x + h) = c. Thus, from the first principle, we have f (x + h) − f (x) c−c 0 f ′ (x) = lim = lim = lim = 0 h→0 h h→0 h h→0 h  The power rule ′ If f (x) = xn for n ∈ R, then f (x) =. nx n−1. That is, bring down the power and reduce the power by one. n(n − 1) n−2 2 Proof. Given f (x) = xn ⇒ f (x + h) = (x + h)n = xn + nxn−1 h + x h + · · · + hn. 2! Thus, from the first principle, we have ( ) n(n − 1) n−2 2 xn nxn−1 h + + x h + · · · + hn − xn f (x + h) − f (x) 2! f ′ (x) = lim = lim h→0 h h→0 h ( ) n(n − 1) nx n−1 h+ xn−2 h + ··· + h 2 n 2! = lim h→0 h ( ) n(n − 1) = lim nxn−1 + xn−2 h + · · · + hn−1 h→0 2! ( ) D.S n(n − 1) n−2 = nxn−1 + x (0) + · · · + (0)n−1 = nxn−1 2! 17 4.1 Basic differentiation rules ⃝Francis c Oketch For example, i) If f (x) = 6x5 , then f ′ (x) = 30x4. ii) If f (x) = x10 , then f ′ (x) = 10x9.  The derivative of a linear combination If f (x) and g(x) are differentiable functions of x and a and b are constants, then d [af (x) + bg(x)]. = af ′ (x) + bg ′ (x) dx Proof. Let y(x) = af (x) + bg(x). Thus, from the first principle, we have dy y(x + h) − y(x) [af (x + h) + bg(x + h)] − [af (x) + bg(x)] = lim = lim dx h→0 h h→0 h a[f (x + h) − f (x)] + b[g(x + h) − g(x)] = lim h→0 h f (x + h) − f (x) g(x + h) − g(x) = a lim + b lim h→0 h h→0 h ′ ′ = af (x) + bg (x) For example, dy i) If y = 24x + 8x5 , then = 24 + 40x4. dx dy ii) If y = 7x3 − 9x2 + 4x + 2, then = 21x2 − 18x + 4. dx  The product rule If u(x) and v(x) are differentiable functions of x, then the product u(x)v(x) is also a differentiable function of x, and d [u(x)v(x)] = u′ (x)v(x). + u(x)v′ (x) dx Proof. Let y(x) = u(x)v(x). Thus, from the first principle, we have dy y(x + h) − y(x) u(x + h)v(x + h) − u(x)v(x) = lim = lim dx h→0 h h→0 h u(x + h)v(x + h) − u(x)v(x + h) + u(x)v(x + h) − u(x)v(x) = lim h→0 h [u(x + h) − u(x)] v(x + h) + u(x) [v(x + h) − v(x)] = lim h→0 h [u(x + h) − u(x)] v(x + h) u(x) [v(x + h) − v(x)] = lim + lim h→0 h h→0 h [ ][ ] u(x + h) − u(x) v(x + h) − v(x) = lim lim v(x + h) + u(x) lim h→0 h h→0 h→0 h = u′ (x)v(x) + u(x)v′ (x) → Note: the product rule says that the derivative of the product of two functions is formed by multiplying the derivative of each function by the other function and then adding the results. In general, suppose y = u1 (x)u2 (x) · · · un (x), then dy = u′1 (x)u2 (x) · · · un (x) + u1 (x)u′2 (x). · · · un (x) + · · · + u1 (x)u2 (x) · · · u′ (x) n dx Example(s): 18 4.1 Basic differentiation rules ⃝Francis c Oketch (a) Find the derivative of f (x) = (1 − 5x2 )(6x2 − 4x + 1). Solution Let f (x) = uv, where u = 1 − 5x2 and v = 6x2 − 4x + 1. Differentiating yields u′ = −10x and v′ = 12x − 4. Therefore, f ′ (x) = u′ v + uv′ = (−10x)(6x2 − 4x + 1) + (1 − 5x2 )(12x − 4) = −60x3 + 40x2 − 10x + 12x − 4 − 60x3 + 20x2 = −120x3 + 60x2 + 2x − 4 (b) Find the derivative of y = (x − 2)(x2 + 6)(x4 + 1). Solution Let y = uvw, where u = x − 2, v = x2 + 6 and w = x4 + 1. Differentiating yields u′ = 1, v′ = 2x and w′ = 4x3. Therefore, dy = u′ vw + uv′ w + uvw′ dx = (1)(x2 + 6)(x4 + 1) + (x − 2)(2x)(x4 + 1) + (x − 2)(x2 + 6)(4x3 ) ( ) ( ) ( ) = x6 + x2 + 6x4 + 6 + 2x x5 + x − 2x4 + 2 + 4x3 x3 + 6x − 2x2 − 12 = x6 + x2 + 6x4 + 6 + 2x6 + 2x2 − 4x5 + 4x + 4x6 + 24x4 − 8x5 − 48x3 = 7x6 − 12x5 + 30x4 − 48x3 + 3x2 + 4x + 6  The quotient rule u(x) If u(x) and v(x) are differentiable functions of x, then the quotient (where v(x) ̸= 0) is also v(x) a differentiable function of x, and [ ] d u(x) u′ (x)v(x). − u(x)v′ (x) = dx v(x) [v(x)]2 u(x) Proof. Let y(x) =. Thus, from the first principle, we have v(x) u(x + h) u(x) − dy y(x + h) − y(x) v(x + h) v(x) u(x + h)v(x) − u(x)v(x + h) = lim = lim = lim dx h→0 h h→0 h h→0 hv(x)v(x + h) u(x + h)v(x) − u(x)v(x) + u(x)v(x) − u(x)v(x + h) = lim h→0 hv(x)v(x + h) [u(x + h) − u(x)] v(x) − u(x) [v(x + h) − v(x)] = lim h→0 hv(x)v(x + h) [u(x + h) − u(x)] [v(x + h) − v(x)] lim v(x) − u(x) lim = h→0 h h→0 h lim v(x)v(x + h) h→0 u′ (x)v(x) − u(x)v′ (x) = [v(x)]2 Example(s): 19 4.1 Basic differentiation rules ⃝Francis c Oketch 2x2 + 1 (a) Differentiate y =. x2 − 1 Solution u Let y = , where u = 2x2 + 1 and v = x2 − 1. Differentiating yields u′ = 4x and v ′ = 2x. v Therefore, dy vu′ − uv ′ (x2 − 1)(4x) − (2x2 + 1)(2x) 4x3 − 4x − 4x3 − 2x −6x = = = = 2 dx v2 (x2 − 1)2 (x2 − 1)2 (x − 1)2 x3 (b) Differentiate y =. x−1 Solution u Let y = , where u = x3 and v = x − 1. Differentiating yields u′ = 3x2 and v ′ = 1. v Therefore, dy vu′ − uv ′ (x − 1)(3x2 ) − (x3 )(1) 3x3 − 3x2 − x3 2x3 − 3x2 = = = = dx v 2 (x − 1) 2 (x − 1)2 (x − 1)2  The chain rule Suppose that y is a differentiable function of u and u is a differentiable function of x (i.e., y = y(u) and u = u(x)), then y is a (differentiable) function of x by extension (i.e., y = y(u(x))) and dy dy du =. · dx du dx → Note: chain rule is used when we want to differentiate a function of another function. Example(s): (a) Differentiate with respect to x the function y = (3x + 4)4. Solution dy du Let y = u4 , where u = 3x + 4. Differentiating yields = 4u3 and = 3. Therefore, du dx chain rule yields dy dy du = · = (4u3 )(3) = 12u3 = 12(3x + 4)3 dx du dx (b) Differentiate with respect to x the function y = (x2 + 3x)7. Solution dy du Let y = u7 , where u = x2 +3x. Differentiating yields = 7u6 and = 2x+3. Therefore, du dx chain rule yields dy dy du = · = (7u6 )(2x + 3) = 7(x2 + 3x)6 (2x + 3) dx du dx dy (c) Find if y = (1 − 3x2 )5. dx Solution dy du Let y = u5 , where u = 1 − 3x2. Differentiating yields = 5u4 and = −6x. Therefore, du dx chain rule yields dy dy du = · = (5u4 )(−6x) = −30x(u4 ) = −30x(1 − 3x2 )4 dx du dx 20 4.1 Basic differentiation rules ⃝Francis c Oketch ( )2 dy 1 + 2x (d) Find if y =. dx 1+x Solution 1 + 2x dy Let y = u2 , where u =. Differentiating yields = 2u and 1+x du du (1 + x)(2) − (1 + 2x)(1) 1 = 2 =. Therefore, chain rule yields dx (1 + x) (1 + x)2 [ ] [ ][ ] dy dy du 1 1 + 2x 1 2(1 + 2x) = · = (2u) 2 =2 2 = dx du dx (1 + x) 1+x (1 + x) (1 + x)3 √ √ √ (e) Differentiate with respect to x the function y = 1+ 1+ 1 + x. Solution √ √ √ dy Let u = 1 + x, v = 1 + u, and w = 1 + v. Then, y = w. Differentiating yields = 1, dw du 1 dv 1 dw 1 = 1 , = 1 , and = 1. Therefore, chain rule yields dx 2(1 + x) 2 du 2(1 + u) 2 dv 2(1 + v) 2 [ ][ ][ ] dy dy dw dv du 1 1 1 = · · · = (1) 1 1 1 dx dw dv du dx 2(1 + v) 2 2(1 + u) 2 2(1 + x) 2 1 1 = 1 1 1 = (√ ) (√ ) (√ ) 8(1 + v) 2 (1 + u) 2 (1 + x) 2 8 1 + v 1 +u 1+x 1 = (√ √ ) (√ ) √ √ (√ ) 8 1+ 1+ 1+x 1+ 1+x 1+x → Note: (Direct chain rule) Consider the function y = [f (x)]n. Then, direct chain rule yields dy. n−1 · f ′ (x) = n [f (x)] dx dy For example, if y = (1 − 3x2 )5 then DCR yields = 5(1 − 3x2 )4 (0 − 6x) = −30x(1 − 3x2 )4. dx Exercise: (a) Use chain rule to differentiate the following functions i) y = (3x2 + 5)3 ii) y = (3x3 + 5x)2 1 iii) y = (7x2 − 4) 3 iv) y = (6x2 − 4x)−2 v) y = (3x2 − 5)− 3 2 vi) y = (1 + x4 − 2x3 )4 (1 − 4x2 )3 √ dy 1+x dy 1 (b) Find when y =. [ans: = 3 ] dx 1−x dx 1 (1 + x) 2 (1 − x) 2 √ √ √ (c) Differentiate with respect to x the function y = x+ x+ x. Lecture 4 21 4.2 Derivative of trigonometric functions ⃝Francis c Oketch 4.2 Derivative of trigonometric functions  Derivative of sin x and cos x d d [sin x] = cos x and. [cos x] = − sin x dx dx Proof. i) Sine Let f (x) = sin x. Thus, from the first principle of differentiation and using the trigonometric identity sin(A + B) = sin A cos B + sin B cos A, we have f (x + h) − f (x) sin(x + h) − sin x f ′ (x) = lim = lim h→0 h h→0 h sin x cos h + sin h cos x − sin x − sin x(1 − cos h) + sin h cos x = lim = lim h→0 h h→0 h [ ] [ ] (1 − cos h) sin h = − sin x lim + cos x lim = (− sin x)(0) + (cos x)(1) h→0 h h→0 h = cos x ii) Cosine Similarly, let f (x) = cos x. Thus, from the first principle of differentiation and using the trigonometric identity cos(A + B) = cos A cos B − sin A sin B, we have f (x + h) − f (x) cos(x + h) − cos x f ′ (x) = lim = lim h→0 h h→0 h cos x cos h − sin x sin h − cos x − cos x(1 − cos h) − sin x sin h = lim = lim h→0 h h→0 h [ ] [ ] (1 − cos h) sin h = − cos x lim − sin x lim = (− cos x)(0) − (sin x)(1) h→0 h h→0 h = − sin x Example(s): (a) Differentiate the following functions wrt x: (i) y = sin(3x + 2), (ii) y = cos3 x, (iii) y = sin x sin(x2 ), (iv) y = x sin(x), (v) y = , and (vi) y = cos2 (3x). x Solution i) Given that y = sin(3x + 2). Let y = sin(u), where u = 3x + 2. Differentiating yields dy du = cos u and = 3. Therefore, chain rule yields du dx dy dy du = · = (cos u)(3) = 3 cos(3x + 2) dx du dx dy ii) Given that y = cos3 x. Let y = u3 , where u = cos x. Differentiating yields = 3u2 du du and = − sin x. Therefore, chain rule yields dx dy dy du = · = (3u2 )(− sin x) = −3 sin x cos2 x dx du dx 22 4.2 Derivative of trigonometric functions ⃝Francis c Oketch dy iii) Given that y = sin(x2 ). Let y = sin(u), where u = x2. Differentiating yields = cos u du du and = 2x. Therefore, chain rule yields dx dy dy du = · = (cos u)(2x) = 2x cos(x2 ) dx du dx iv) Given that y = x sin(x). Let y = uv, where u = x and v = sin(x). Differentiating yields u′ = 1 and v ′ = cos x. Therefore, product rule yields dy = uv ′ + vu′ = (x)(cos x) + (sin x)(1) = x cos x + sin x dx sin x u v) Given that y =. Let y = , where u = sin x and v = x. Differentiating yields x v u′ = cos x and v ′ = 1. Therefore, quotient rule yields dy vu′ − uv ′ (x)(cos x) − (sin x)(1) x cos x − sin x = = = dx v2 x2 x2 vi) Given that y = cos2 (3x). Let y = u2 , where u = cos(3x). Differentiating yields dy du = 2u and = −3 sin(3x). Therefore, chain rule yields du dx dy dy du = · = (2u)[−3 sin(3x)] = −6 cos(3x) sin(3x) dx du dx √ dy 1√ (b) If y = 1 + sin x, show that = 1 − sin x. dx 2 Solution dy 1 1 du = u− 2 and 1 Let y = u 2 , where u = 1 + sin x. Differentiating yields = cos x. du 2 dx Therefore, chain rule yields ( ) dy dy du 1 −1 cos x cos x = · = u 2 (cos x) = √ = √ dx du dx 2 2 u 2 1 + sin x √ √ √ 1 cos x 1 − sin x 1 cos x 1 − sin x 1 cos x 1 − sin x = √ √ = √ = √ 2 ( 1 + sin x)( 1 − sin x) 2 1 − sin2 x 2 cos2 x 1√ = 1 − sin x 2 dy (c) Find if y = sin(cos x). dx Solution dy du Let y = sin u, where u = cos x. Differentiating yields = cos u and = − sin x. du dx Therefore, chain rule yields dy dy du = · = (cos u)(− sin x) = − cos(cos x) sin x dx du dx In general, d d [sin (f (x))] = f ′ (x) cos (f (x)) and. [cos (f (x))] = −f ′ (x) sin (f (x)) dx dx Exercise: dy √ (a) Find if y = sin( x). dx 23 4.2 Derivative of trigonometric functions ⃝Francis c Oketch √ 1 + sin x dy 1 (b) If y = , show that =. 1 − sin x dx 1 − sin x (c) If m is a positive integer, find the differential coefficients with respect to x of (i) sinm x and (ii) sin(xm ). (d) Differentiate the following functions with respect to x: (i) y = sin 3x, (ii) y = cos(x2 ), √ cos 2x (iii) y = sin 2x, (iv) y = 4 sin2 ( x2 ), (v) y = sin x cos 2x, (vi) y = , and (vii) sin 3x y = 2 cos x + 2x sin x − x cos x. 2  Derivative of tan x, cot x, sec x and cosec x I Tangent sin x Let y = tan x =. Differentiating using quotient rule yields cos x dy (cos x)(cos x) − (sin x)(− sin x) cos2 x + sin2 x 1 = 2 = 2 = = sec2 x dx cos x cos x cos2 x d Therefore, [tan x]. = sec2 x. dx II Cotangent cos x Let y = cot x =. Differentiating using quotient rule yields sin x [ ] dy (sin x)(− sin x) − (cos x)(cos x) − sin2 x + cos2 x −1 = 2 = 2 = = −cosec2 x dx sin x sin x sin2 x d Therefore, [cot x] =. −cosec2 x. dx III Secant 1 Let y = sec x =. Differentiating using quotient rule yields cos x dy (cos x)(0) − (1)(− sin x) sin x 1 sin x = 2 = 2 = · = sec x tan x dx cos x cos x cos x cos x d Therefore, [sec x] =. sec x tan x. dx IV Cosecant 1 Let y = cosec x =. Differentiating using quotient rule yields sin x dy (sin x)(0) − (1)(cos x) − cos x −1 cos x = = = · = −cosec x cot x dx sin2 x sin2 x sin x sin x d Therefore, [cosec x] =.−cosec x cot x. dx In summary, we have f (x) f ′ (x) sin x cos x cos x − sin x tan x sec2 x cosec x −cosec x cot x sec x sec x tan x cot x −cosec2 x 24 4.2 Derivative of trigonometric functions ⃝Francis c Oketch Example(s): (a) Differentiate the following functions with respect to x: (i) y = tan 2x, (ii) y = cot 3x, (iii)