Lecture 1: Introduction to Calculus PDF

EEE406: Mathematics for Engineering Week 1: Introduction to Calculus 1 What is Calculus? Calculus, a cornerstone of modern mathematics, emerged in the 17th century through the work of Isaac Newton and Gottfried Wilhelm Leibniz. This pow...

EEE406: Mathematics for Engineering Week 1: Introduction to Calculus 1 What is Calculus? Calculus, a cornerstone of modern mathematics, emerged in the 17th century through the work of Isaac Newton and Gottfried Wilhelm Leibniz. This powerful mathematical tool revolutionized our understanding of change and accumulation in the natural world. The word Calculus comes from Latin meaning "small stone", because it is like understanding something by looking at small pieces. 2 What is Calculus? Calculus is divided into two main branches: 1.Differential Calculus: cuts something into small pieces to find how it changes focuses on rates of change and slopes of curves. 2.Integral Calculus: joins (integrates) the small pieces together to find how much there is; deals with accumulation of quantities and areas under curves. 3 What is Calculus? The importance of calculus extends far beyond pure mathematics. It's essential in numerous fields, including: Physics: Describing motion, energy, and waves Engineering: Optimizing designs and analyzing systems Economics: Modeling growth and predicting market trends Biology: Studying population dynamics and reaction rates Computer Graphics: Creating realistic animations and simulations By providing tools to analyse continuous change, calculus enables us to model and predict complex phenomena, making it indispensable in modern science, technology, and decision-making processes. 4 Fundamental Concepts A function is a relation between a set of inputs (domain) and a set of possible outputs (range), where each input corresponds to exactly one output. For every value of x there is a corresponding value for y x is called the independent variable y is called the dependent variable. y is said to be a function of x and is written as 𝑦 = 𝑓(𝑥) The value of 𝑓(𝑥) when 𝑥 = 0 is denoted as 𝑓(0) The value of 𝑓(𝑥) when 𝑥 = 5 is denoted as 𝑓(5) etc. Types of Function: Linear, quadratic, exponential, logarithmic, trigonometric, etc. 5 Functions Expression: 𝑦 = 4𝑥 2 − 4𝑥 − 3 → 𝑓 𝑥 = 4𝑥 2 − 4𝑥 − 3 If 𝑥 = 2 2 𝑓 2 =4 2 −4 2 −3=4×4−4×2−3=5 So 𝑓 2 = 5 6 Functions If 𝑓 𝑥 = 4𝑥 2 − 4𝑥 − 3 2 𝑓 −2 = 4 −2 − 4 −2 − 3 = 16 − −8 − 3 = 21 Examples to try: Evaluate: 𝑓 0 𝑓 3 𝑓 −3 7 Functions If 𝑓 𝑥 = 4𝑥 2 − 4𝑥 − 3 2 𝑓 −2 = 4 −2 − 4 −2 − 3 = 16 − −8 − 3 = 21 Examples to try: Evaluate: 𝑓 0 = 4 0 2 − 4 0 − 3 = −3 𝑓 3 = 4 3 2 − 4 3 − 3 = 21 𝑓 −3 = 4 −3 2 − 4 −3 − 3 = 45 8 Limits A limit describes the value that a function approaches as the input (usually x) gets closer to a specific value. Key points: Notation:𝐥𝐢𝐦 𝒇(𝒙) = 𝑳 means 𝒇(𝒙) approaches 𝑳 as 𝒙 𝒙→𝒂 approaches 𝒂 The limit may exist even if the function is undefined at that point Limits are fundamental to defining derivatives and continuity 9  Limits lim 𝒇(𝒙) = 𝑳 𝒙→𝒂 means 𝒇(𝒙) approaches 𝑳 as 𝒙 approaches 𝒂 E.g. lim 𝐬𝐢𝐧 (𝒙) = 𝟎 𝒙→𝟎 lim 𝐜𝐨𝐬(𝒙) = 𝟏 𝒙→𝟐𝝅 lim𝝅 𝐭𝐚𝐧(𝒙) =? 𝒙→ 𝟐 10 Limits and Continuity lim𝝅 𝐭𝐚𝐧(𝒙) =? 𝒙→ 𝟐 𝜋 We can see graphically that if we approach (90𝑜 ) 2 from the left, tan(𝑥) approaches ∞. This can be written as lim𝝅− 𝐭𝐚𝐧(𝒙) = ∞ 𝒙→ 𝟐 𝜋 As we approach 2 (90𝑜 ) from the right, tan(𝑥) approaches −∞. lim+ 𝐭𝐚𝐧(𝒙) = −∞ 𝝅 𝒙→ 𝟐 As lim+ tan(x) ≠ lim π− tan(x) we say that this function π x→ 2 x→ 2 is discontinuous. 11 Average vs Instantaneous Rate of Change Average rate of change Total Change = Time Elapse e. g. Overall average speed Overall Distance = Overall Time 12 Average vs Instantaneous Rate of Change Instantaneous rate of change Small Change = Time Elapsed dS dS e.g. Instananeous speed = dt dt 13 Average vs Instantaneous Rate of Change Instantaneous rate of change Small Change = Time Elapsed dS e.g. Instananeous speed = dt 14 Average vs Instantaneous Rate of Change Instantaneous rate of change Small Change = Time Elapsed dS e.g. Instananeous speed = dt 15 The Derivative of a Function To find the derivative of an arbitrary function 𝑦 = 𝑓(𝑥), we start with an arbitrary point 𝑃(𝑥, 𝑓(𝑥)) on the graph of 𝑓 and using the fact that the gradient of the secant line through the point P and nearby point 𝑄(𝑥 + ℎ, 𝑓 𝑥 + ℎ ) is ∆𝑦 𝑓 𝑥 + ℎ − 𝑓 𝑥 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 = = ∆𝑥 ℎ Q 𝑓(𝑥 + ℎ) 𝑓 𝑥 + ℎ − 𝑓(𝑥) f(𝑥) P ℎ 𝑥 𝑥+ℎ The Derivative of a Function The slope of the curve at P is then the limit of the gradient as Q approaches P along the curve: ∆𝑦 𝑓 𝑥+ℎ −𝑓 𝑥 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 = lim = lim 𝑄→𝑃 ∆𝑥 ℎ→0 ℎ Q P The Derivative of a Function The slope of the curve at P is then the limit of the gradient as Q approaches P along the curve: ∆𝑦 𝑓 𝑥+ℎ −𝑓 𝑥 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 = lim = lim 𝑄→𝑃 ∆𝑥 ℎ→0 ℎ Q P The Derivative of a Function The slope of the curve at P is then the limit of the gradient as Q approaches P along the curve: ∆𝑦 𝑓 𝑥+ℎ −𝑓 𝑥 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 = lim = lim 𝑄→𝑃 ∆𝑥 ℎ→0 ℎ P Q The Derivative of a Function Definition: The derivative of a function 𝑓 is the function 𝑓′ whose value 𝑥 is ′ 𝑓 𝑥+ℎ −𝑓 𝑥 𝑓 𝑥 = lim ℎ→0 ℎ The most common notations for the derivative of a function 𝑦 = 𝑓(𝑥) besides 𝑓′(𝑥) are: 𝒅𝒚 𝑑𝑓 𝑑 𝒚′ , , , 𝑓 𝑥 𝑎𝑛𝑑 𝐷𝑥 𝑓 𝒅𝒙 𝑑𝑥 𝑑𝑥 21 Rules of Differentiation The basic rule (also known as the Power Rule) of differentiation is If 𝑦 = 𝑐𝑥 𝑛 𝑑𝑦 then the derivative is given by 𝑑𝑥 𝑑𝑦 = 𝑐𝑛𝑥 𝑛−1 𝑑𝑥 This rule will also work when n is negative or a fraction. 22 Note: Rules of Indices 𝑥 = 1 × 𝑥1 𝑥2 = 1 × 𝑥2 1 1 𝑥 −1 = 1× 1 = 𝑥 𝑥 𝑥0 = 1 1 𝑥 −2 = 𝑥2 𝑎 𝑎 2 2 𝑏 3 𝑥 = 𝑏 𝑥 [e.g. x = 3 x ] Exponent rules part 1 | Exponents, radicals, and scientific notation | Pre-Algebra | Khan Academy - YouTube23 Basic Differentiation (Power Rule) 𝑛 𝑑𝑦 𝑦 = 𝑐𝑥 → = 𝑐𝑛𝑥 𝑛−1 𝑑𝑥 (Note: in these examples c=1) 𝑑𝑦 If 𝑦 = 𝑥 5 → = 5𝑥 5−1 = 5𝑥 4 𝑑𝑥 𝑑𝑦 If 𝑦 = 𝑥 −10 → = −10 𝑥 −10−1 = −10𝑥 −11 𝑑𝑥 1 𝑑𝑦 1 If 𝑦 = = 𝑥 −1 → = −1 𝑥 −1−1 −2 = −1𝑥 = − 2 𝑥 𝑑𝑥 𝑥 𝑑𝑦 If 𝑦 = 𝑥 → = (1)𝑥 1−1 = 𝑥 0 = 1 𝑑𝑥 24 Basic Differentiation 𝑑𝑦 𝑦= 𝑐𝑥 𝑛 → = 𝑐𝑛𝑥 𝑛−1 𝑑𝑥 Examples to try: Find the derivative of 𝑦 = 𝑥 −2 𝑦 = 𝑥2 25 Basic Differentiation 𝑑𝑦 𝑦= 𝑐𝑥 𝑛 → = 𝑐𝑛𝑥 𝑛−1 𝑑𝑥 Examples to try: Find the derivative of −2 𝑑𝑦 𝑦=𝑥 → = −2𝑥 −3 𝑑𝑥 𝑑𝑦 𝑦= 𝑥2 → = 2𝑥 1 = 2𝑥 𝑑𝑥 26 Basic Differentiation 𝑑𝑦 𝑦= 𝑐𝑥 𝑛 → = 𝑐𝑛𝑥 𝑛−1 𝑑𝑥 1 𝑑𝑦 1 1−1 1 −1 If 𝑦 = 𝑥 2 → = 𝑥2 = 𝑥 2 𝑑𝑥 2 2 2 2 − 𝑑𝑦 2 − −1 2 −5 If 𝑦 = 𝑥 3 → = − 𝑥 3 =− 𝑥 3 𝑑𝑥 3 3 2 2 3 𝑑𝑦 2 −1 2 −1 2 1 2 1 2 If 𝑦 = 𝑥2 =𝑥 → 3 = 𝑥 3 = 𝑥 = × 1 3 = ×3 = 𝑑𝑥 3 3 3 3 𝑥 33 𝑥 𝑥3 27 Basic Differentiation 𝑑𝑦 𝑦= 𝑐𝑥 𝑛 → = 𝑐𝑛𝑥 𝑛−1 𝑑𝑥 Examples to try: Find the derivative of 4 𝑦= 𝑥7 1 −3 𝑦=𝑥 28 Basic Differentiation 𝑑𝑦 𝑦= 𝑐𝑥 𝑛 → = 𝑐𝑛𝑥 𝑛−1 𝑑𝑥 Examples to try: Find the derivative of 4 𝑑𝑦 4 −3 𝑦= 𝑥7 → = 𝑥 7 𝑑𝑥 7 − 1 𝑑𝑦 1 −4 𝑦=𝑥 3 → =− 𝑥 3 𝑑𝑥 3 29 Rules for Differentiation- 1: Differentiating a constant Many functions are combinations of simpler ones whose derivative is already known. For this situation rules have been developed which will help us to differentiate. Rule 1: If 𝑦 = 𝑐, where c is a constant, 𝑑𝑦 =0 𝑑𝑥 𝑑𝑦 e.g. 𝑦 = 5 = 5𝑥 0 → = 5 × 0 × 𝑥 0−1 = 0 𝑑𝑥 30 Rules for Differentiation 2: Differentiating a function containing a constant 𝑑𝑦 If 𝑦 = 𝑐 ∙ 𝑓 𝑥 then = 𝑐 ∙ 𝑓′ 𝑥 𝑑𝑥 Examples: 𝑦 = 3𝑥 5 = 3 ∙ 𝑥 5 𝑑𝑦 → = 3 × 5𝑥 5−1 = 3 5𝑥 4 = 15𝑥 4 𝑑𝑥 𝑦 = 3 cos 𝑥 𝑑𝑦 → = 3 × (− sin 𝑥) = −3 sin 𝑥 𝑑𝑥 7 𝑦 = = 7𝑥 −1 𝑥 𝑑𝑦 −1−1 −2 7 → = 7 −1 𝑥 = −7𝑥 = − 2 𝑑𝑥 𝑥 31 2: Functions containing a constant 𝑑𝑦 If 𝑦 = 𝑐 ∙ 𝑓 𝑥 then ′ =𝑐∙𝑓 𝑥 𝑑𝑥 Examples to try: Find the derivative of 𝑦 = 5𝑥 𝑦 = 4𝑥 2 𝑦 = 6𝑥 4 𝑦 = 3𝑥 −2 𝑦 = −2𝑥 −4 32 2: Functions containing a constant 𝑑𝑦 If 𝑦 = 𝑐 ∙ 𝑓 𝑥 then = 𝑐 ∙ 𝑓′ 𝑥 𝑑𝑥 Find the derivative of: 𝑑𝑦 𝑦 = 5𝑥 → =5 𝑑𝑥 2 𝑑𝑦 𝑦 = 4𝑥 → = 8𝑥 𝑑𝑥 4 𝑑𝑦 𝑦 = 6𝑥 → = 24𝑥 3 𝑑𝑥 − 2 𝑑𝑦 𝑦 = 3𝑥 → = −6𝑥 −3 𝑑𝑥 −4 𝑑𝑦 𝑦 = −2𝑥 → = 8x −5 𝑑𝑥 33 2: Functions containing a constant 𝑑𝑦 If 𝑦 = 𝑐 ∙ 𝑓 𝑥 then = 𝑐 ∙ 𝑓′ 𝑥 𝑑𝑥 Examples to try: (Using supplied tables) Find the derivative of: 𝑦 = 4 sin 𝑥 𝑦 = 6 tan 𝑥 𝑦 = 2 sec 𝑥 34 2: Functions containing a constant 𝑑𝑦 If 𝑦 = 𝑐 ∙ 𝑓 𝑥 then = 𝑐 ∙ 𝑓′ 𝑥 𝑑𝑥 Examples to try: Find the derivative of: 𝑑𝑦 𝑦 = 4 sin 𝑥 → = 4 cos 𝑥 𝑑𝑥 𝑑𝑦 𝑦 = 6 tan 𝑥 → = 6sec 2 x 𝑑𝑥 𝑑𝑦 𝑦 = 2 sec 𝑥 → = 2 sec 𝑥 tan 𝑥 𝑑𝑥 35 2: Functions containing a constant 𝑑𝑦 If 𝑦 = 𝑐 ∙ 𝑓 𝑥 then = 𝑐 ∙ 𝑓′ 𝑥 𝑑𝑥 Examples to try: (Using supplied tables) Find the derivative of: 𝑦 = 4 sin 𝑥 𝑦 = 6 tan 𝑥 𝑦 = 2 sec 𝑥 36 2: Functions containing a constant 𝑑𝑦 If 𝑦 = 𝑐 ∙ 𝑓 𝑥 then = 𝑐 ∙ 𝑓′ 𝑥 𝑑𝑥 Examples to try: Find the derivative of: 𝑑𝑦 𝑦 = 4 sin 𝑥 → = 4 cos 𝑥 𝑑𝑥 𝑑𝑦 𝑦 = 6 tan 𝑥 → = 6sec 2 x 𝑑𝑥 𝑑𝑦 𝑦 = 2 sec 𝑥 → = 2 sec 𝑥 tan 𝑥 𝑑𝑥 37 Rules for Differentiation 3: Differentiating the Sum of functions 𝑑𝑦 If 𝑦 is the sum of two or more functions of 𝑥, then is the sum of 𝑑𝑥 their derivatives. 𝑑𝑦 𝑑𝑢 𝑑𝑣 If 𝑦 = 𝑢 𝑥 + 𝑣(𝑥) then = +. 𝑑𝑥 𝑑𝑥 𝑑𝑥 Examples: 𝑑𝑦 𝑑 𝑑 𝑦= 2𝑥 2 + 4𝑥 3 → = 2𝑥 2 + 4𝑥 3 = 4𝑥 + 12𝑥 2 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑦 𝑑 𝑑 𝑑 𝑦= 7𝑥 5 − 6𝑥 2 + sin 𝑥 → = 7𝑥 5 + −6𝑥 2 + sin 𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 = 35𝑥 4 − 12𝑥 + cos 𝑥 38 3: Sum of functions 𝑑𝑦 𝑑𝑢 𝑑𝑣 If 𝑦 = 𝑢 𝑥 + 𝑣(𝑥) then = + 𝑑𝑥 𝑑𝑥 𝑑𝑥 Example: 1 − 1 𝑦 = 3 + 𝑒 𝑥 + cos 𝑥 = 𝑥 3 + 𝑒 𝑥 + cos 𝑥 𝑥 𝑑𝑦 1 −4 𝑥 1 → = − 𝑥 3 + 𝑒 − sin 𝑥 = − 3 + 𝑒 𝑥 − sin 𝑥 𝑑𝑥 3 3 𝑥4 39 3: Sum of functions 𝑑𝑦 𝑑𝑢 𝑑𝑣 If 𝑦 = 𝑢 𝑥 + 𝑣(𝑥) then = + 𝑑𝑥 𝑑𝑥 𝑑𝑥 Example: 3 5 1 −2 1 −3 𝑦= + 3 = 3𝑥 + 5𝑥 𝑥 𝑥 𝑑𝑦 3 −3 5 − 4 3 5 → =− 𝑥 2− 𝑥 3=− − 3 𝑑𝑥 2 3 2 𝑥3 3 𝑥4 40 3: Sum of functions 𝑑𝑦 𝑑𝑢 𝑑𝑣 If 𝑦 = 𝑢 𝑥 + 𝑣(𝑥) then = + 𝑑𝑥 𝑑𝑥 𝑑𝑥 Examples to try: Find the derivative of 𝑦 = 𝑥2 + 𝑥 𝑦 = 4𝑥 2 + 3𝑥 −3 41 3: Sum of functions 𝑑𝑦 𝑑𝑢 𝑑𝑣 If 𝑦 = 𝑢 𝑥 + 𝑣(𝑥) then = +. 𝑑𝑥 𝑑𝑥 𝑑𝑥 Examples to try: Find the derivative of 𝑑𝑦 𝑦= 𝑥2 +𝑥 → = 2𝑥 + 1 𝑑𝑥 2 −3 𝑑𝑦 𝑦 = 4𝑥 + 3𝑥 → = 8𝑥 − 9𝑥 −4 𝑑𝑥 42 Rules for Differentiation 4: The Product Rule If 𝑦 = 𝑓 𝑥 𝑔(𝑥) where 𝑓(𝑥) and 𝑔(𝑥) are both functions of 𝑥, and y is the product (multiple of these functions) then 𝑑𝑦 = 𝑓′ 𝑥 𝑔 𝑥 + 𝑓(𝑥)𝑔′(𝑥) 𝑑𝑥 43 4: The Product Rule 𝑑𝑦 If 𝑦 = 𝑓 𝑥 𝑔(𝑥) then = 𝑓 ′ 𝑥 𝑔(𝑥) + 𝑓 𝑥 𝑔′(𝑥) 𝑑𝑥 Examples: 𝑦 = 4𝑥 3 sin 𝑥 𝑓 𝑥 = 4𝑥 3 → 𝑓 ′ 𝑥 = 12𝑥 2 𝑔 𝑥 = sin 𝑥 → 𝑔′ 𝑥 = cos 𝑥 𝑑𝑦 So = 12𝑥 2 sin 𝑥 + 4𝑥 3 cos 𝑥 𝑑𝑥 44 4: The Product Rule 𝑑𝑦 If 𝑦 = 𝑓 𝑥 𝑔(𝑥) then = 𝑓 ′ 𝑥 𝑔(𝑥) + 𝑓 𝑥 𝑔′(𝑥) 𝑑𝑥 Example: 𝑦 = 𝑥 2 cos 𝑥 𝑓 𝑥 = 𝑥 2 → 𝑓 ′ 𝑥 = 2𝑥 𝑔 𝑥 = cos 𝑥 → 𝑔′ 𝑥 = −sin 𝑥 𝑑𝑦 So = 2𝑥 cos 𝑥 − 𝑥 2 sin 𝑥 𝑑𝑥 45 4: The Product Rule 𝑑𝑦 If 𝑦 = 𝑓 𝑥 𝑔(𝑥) then = 𝑓 ′ 𝑥 𝑔(𝑥) + 𝑓 𝑥 𝑔′(𝑥) 𝑑𝑥 Examples to try: Find the derivative of 𝑦 = 𝑥 sin 𝑥 𝑦 = 𝑥 ln 𝑥 𝑦 = 2𝑥 2 tan 𝑥 𝑦 = 𝑥 3 cot 𝑥 𝑦 = 𝑥 2 sin 𝑥 𝑦 = 15𝑥 3 cos 𝑥 46 4: The Product Rule 𝑑𝑦 If 𝑦 = 𝑓 𝑥 𝑔(𝑥) then = 𝑓 ′ 𝑥 𝑔(𝑥) + 𝑓 𝑥 𝑔′(𝑥) 𝑑𝑥 Example: 𝑑𝑦 𝑦 = 𝑥 sin 𝑥 → = sin 𝑥 + 𝑥 cos 𝑥 𝑑𝑥 𝑑𝑦 𝑦 = 𝑥 ln 𝑥 → = ln 𝑥 + 1 𝑑𝑥 2 𝑑𝑦 𝑦 = 2𝑥 tan 𝑥 → = 4𝑥 tan 𝑥 + 2𝑥 2 sec 2 𝑥 𝑑𝑥 3 𝑑𝑦 𝑦 = 𝑥 cot 𝑥 → = 3𝑥 2 cot 𝑥 − 𝑥 3 cosec 2 𝑥 𝑑𝑥 2 𝑑𝑦 𝑦 = 𝑥 sin 𝑥 → = 2𝑥 sin 𝑥 + 𝑥 2 cos 𝑥 𝑑𝑥 𝑑𝑦 𝑦 = 15𝑥 3 cos 𝑥 → = 45𝑥 2 cos 𝑥 − 15𝑥 3 sin 𝑥 𝑑𝑥 47 Rules for Differentiation 5: The Quotient Rule 𝑓 𝑥 If 𝑦 = where 𝑓(𝑥) and 𝑔(𝑥) are both functions of x, and y is the 𝑔 𝑥 quotient (one function divided by the other) of them then 𝑑𝑦 𝑓 ′ 𝑥 𝑔 𝑥 − 𝑓 𝑥 𝑔′ 𝑥 = 2 𝑑𝑥 𝑔 𝑥 48 5: The Quotient Rule 𝑓 𝑥 𝑑𝑦 𝑓′ 𝑥 𝑔 𝑥 −𝑓 𝑥 𝑔′ 𝑥 If 𝑦 = then = 2 𝑔 𝑥 𝑑𝑥 𝑔 𝑥 3 cos 𝑥 Example: 𝑦 = 5𝑥 3 𝑓 𝑥 = 3 cos 𝑥 → 𝑓 ′ 𝑥 = −3 sin 𝑥 𝑔 𝑥 = 5𝑥 3 → 𝑔′ 𝑥 = 15𝑥 2 𝑑𝑦 −3 sin 𝑥 5𝑥 3 − 3 cos 𝑥 15𝑥 2 −15𝑥 3 sin 𝑥 − 45𝑥 2 cos 𝑥 = 3 2 = 𝑑𝑥 5𝑥 25𝑥 6 49 5: The Quotient Rule 𝑓 𝑥 𝑑𝑦 𝑓′ 𝑥 𝑔 𝑥 −𝑓 𝑥 𝑔′ 𝑥 If 𝑦 = then = 2 𝑔 𝑥 𝑑𝑥 𝑔 𝑥 𝑒 𝑥 +𝑥 Example: 𝑦 = sin 𝑥 𝑓 𝑥 = 𝑒𝑥 + 𝑥 → 𝑓′ 𝑥 = 𝑒𝑥 + 1 𝑔 𝑥 = sin 𝑥 → 𝑔′ 𝑥 = cos 𝑥 𝑑𝑦 𝑒𝑥 + 1 sin 𝑥 − 𝑒 𝑥 + 𝑥 cos 𝑥 𝑒 𝑥 sin 𝑥 + sin 𝑥 − 𝑒 𝑥 cos 𝑥 − 𝑥 cos 𝑥 = 2 = 𝑑𝑥 sin 𝑥 sin2 𝑥 50 5: The Quotient Rule 𝑓 𝑥 𝑑𝑦 𝑓′ 𝑥 𝑔 𝑥 −𝑓 𝑥 𝑔′ 𝑥 If 𝑦 = then = 2 𝑔 𝑥 𝑑𝑥 𝑔 𝑥 Examples to try: Find the derivative of 𝑥+7 𝑦= 𝑥−9 cos 𝑥 𝑦= 1 + sin 𝑥 𝑥 𝑦= 𝑥+1 51 5: The Quotient Rule 𝑓 𝑥 𝑑𝑦 𝑓′ 𝑥 𝑔 𝑥 −𝑓 𝑥 𝑔′ 𝑥 If 𝑦 = then = 2 𝑔 𝑥 𝑑𝑥 𝑔 𝑥 Examples to try: Find the derivative of 𝑥 + 7 𝑑𝑦 𝑥−9 − 𝑥−7 2 𝑦= → = = − 𝑥 − 9 𝑑𝑥 𝑥−9 2 𝑥−9 2 cos 𝑥 𝑑𝑦 − sin 𝑥 1 + sin 𝑥 − cos 𝑥 cos 𝑥 − sin 𝑥 1 + sin 𝑥 − cos 2 𝑥 𝑦= → = = 1 + sin 𝑥 𝑑𝑥 1 + sin 𝑥 2 1 + sin 𝑥 2 𝑥 𝑑𝑦 1+𝑥 −𝑥 1 𝑦= → = = 𝑥 + 1 𝑑𝑥 1+𝑥 2 1+𝑥 2 52 Rules for Differentiation 6: The Chain Rule(or Function of a Function Rule) Let 𝑢(𝑥) be a function of x and let 𝑓(𝑢) be a function of u, then we can form the composite function 𝑦 ∘ 𝑢 𝑥 = 𝑦(𝑢 𝑥 ) 𝑑𝑦 𝑑𝑦 𝑑𝑢 Then = ∙ 𝑑𝑥 𝑑𝑢 𝑑𝑥 53 6: The Chain Rule 𝑑𝑦 𝑑𝑦 𝑑𝑢 Let 𝑦 ∘ 𝑢 𝑥 = 𝑦(𝑢 𝑥 ) Then = ∙ 𝑑𝑥 𝑑𝑢 𝑑𝑥 Example: 𝑦 = sin(6𝑥 + 1) 𝑑𝑢 𝑢 = 6𝑥 + 1 → =6 𝑑𝑥 𝑑𝑦 𝑦 = sin 𝑢 → = cos 𝑢 𝑑𝑢 So then 𝑑𝑦 𝑑𝑦 𝑑𝑢 = ∙ = cos u ∙ 6 = 6 cos(6𝑥 + 1) 𝑑𝑥 𝑑𝑢 𝑑𝑥 54 6: The Chain Rule 𝑑𝑦 𝑑𝑦 𝑑𝑢 Let 𝑦 ∘ 𝑢 𝑥 = 𝑦(𝑢 𝑥 ) Then = ∙ 𝑑𝑥 𝑑𝑢 𝑑𝑥 3 Example: 𝑦 = 𝑥2 +2 2 𝑑𝑢 𝑢 =𝑥 +2→ = 2𝑥 𝑑𝑥 3 𝑑𝑦 𝑦=𝑢 → = 3𝑢2 𝑑𝑢 So then 𝑑𝑦 𝑑𝑦 𝑑𝑢 2 = ∙ = 3u2 ∙ 2𝑥 = 6xu2 = 6x x 2 + 2 𝑑𝑥 𝑑𝑢 𝑑𝑥 55 6: The Chain Rule 𝑑𝑦 𝑑𝑦 𝑑𝑢 Let 𝑦 ∘ 𝑢 𝑥 = 𝑦(𝑢 𝑥 ) Then = ∙ 𝑑𝑥 𝑑𝑢 𝑑𝑥 Example: 𝑦 = cot(3𝑥 2 − 14) 2 𝑑𝑢 𝑢 = 3𝑥 − 14 → = 6𝑥 𝑑𝑥 𝑑𝑦 𝑦 = cot 𝑢 → = − cosec 2 𝑢 𝑑𝑢 So then 𝑑𝑦 𝑑𝑦 𝑑𝑢 = ∙ = − cosec 2 𝑢 ∙ 6𝑥 = −6𝑥 cosec 2 𝑢 = −6x cosec 2 (3𝑥 2 − 14) 𝑑𝑥 𝑑𝑢 𝑑𝑥 56 6: The Chain Rule 𝑑𝑦 𝑑𝑦 𝑑𝑢 Let 𝑦 ∘ 𝑢 𝑥 = 𝑦(𝑢 𝑥 ) Then = ∙ 𝑑𝑥 𝑑𝑢 𝑑𝑥 Examples to try: Find the derivative of 𝑦 = cos(𝑥 2 + 2𝑥) 𝑦 = sin(𝑥 2 ) 𝑦 = ln(𝑥 3 + 𝑥) 𝑦 = ln(sin 𝑥) 57 6: The Chain Rule 𝑑𝑦 𝑑𝑦 𝑑𝑢 Let 𝑦 ∘ 𝑢 𝑥 = 𝑦(𝑢 𝑥 ) Then = ∙ 𝑑𝑥 𝑑𝑢 𝑑𝑥 Examples to try: Find the derivative of 2 𝑑𝑦 𝑦 = cos(𝑥 + 2𝑥) → = − 2𝑥 + 2 sin 𝑥 2 + 2𝑥 𝑑𝑥 2 𝑑𝑦 𝑦 = sin(𝑥 ) → = 2𝑥 cos 𝑥 2 𝑑𝑥 𝑑𝑦 2𝑥 2+1 𝑦 = ln(𝑥 3 + 𝑥) → = 3 𝑑𝑥 𝑥 +𝑥 𝑑𝑦 cos 𝑥 𝑦 = ln(sin 𝑥) → = 𝑑𝑥 sin 𝑥 58 Summary of Rules 𝑑𝑦 Basic rule: 𝑦 = 𝑐𝑥 𝑛 → = 𝑐𝑛𝑥 𝑛−1 𝑎𝑥 𝑑𝑦 Differentiating a constant: 𝑦=𝑐→ =0 𝑑𝑥 Differentiating a function containing a constant: 𝑦 = 𝑐 𝑓 𝑥 → 𝑦 = 𝑐𝑓′(𝑥) 𝑑𝑦 Differentiating a sum of functions: 𝑦=𝑢 𝑥 +𝑣 𝑥 → = 𝑢′ 𝑥 + 𝑣′(𝑥) 𝑑𝑥 𝑑𝑦 Product Rule: 𝑦=𝑓 𝑥 𝑔 𝑥 → = 𝑓 ′ 𝑥 𝑔(𝑥) + 𝑓 𝑥 𝑔′(𝑥) 𝑑𝑥 𝑓 𝑥 𝑑𝑦 𝑓′ 𝑥 𝑔 𝑥 −𝑓 𝑥 𝑔′ 𝑥 Quotient Rule: 𝑦= → = 2 𝑔 𝑥 𝑑𝑥 𝑔 𝑥 𝑑𝑦 𝑑𝑦 𝑑𝑓 𝑑𝑔 Chain Rule: 𝑦=𝑓 𝑔 𝑥 → = = ∙ 𝑑𝑥 𝑑𝑥 𝑑𝑔 𝑑𝑥 59 Next Week Continue Differentiation Rules Stationary points Applications TODAY Practical Session 1: Tutorial MF124/125 Differentiation Questions Practical Session 2: Lab MF124/125 Introduction to MATLAB 60 Further Resources Combining rules for differentiation https://www.khanacademy.org/math/ap-calculus-ab/ab-differentiation- 2-new/ab-3-5b/v/differentiating-using-multiple-rules-strategy Common misunderstandings using the Chain Rule https://www.khanacademy.org/math/ap-calculus-ab/ab-differentiation- 2-new/ab-3-1a/v/common-chain-rule-misunderstandings