(d by dx notation aka Leibniz notation)

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1 n Prerequisites: Differentiating, sin and cos ; sum/difference and chain rules; finding ma./min.; finding tangents to curves; finding stationary points and their nature; optimising a function. Maths Applications: Higher derivatives; integration. Real-World Applications: Particle motion; chemical reaction rates; generic rates of change. Differentiation from First Principles Recall from Higher that the derivative of a function f is ined by, f () lim h 0 f ( + h ) f ( ) h This frightening formula gives a recipe for calculating the derivative of any function. Using this recipe to calculate the derivative of a function is called differentiation from first principles. Although not required, it is useful to see just how painful it is to differentiate a function from first principles, as we can then be grateful for simpler differentiation techniques (which come from the inition). Eample 1 Differentiate f () 1 from first principles. f () lim h 0 ( + h) h 1 1 Multiplying the numerator and denominator by ( h) gives, f () lim h (( + h) )(( + h) + ) 1 1 (( + ) + ) h h Multiplying out the numerator and simplifying gives, M Patel (April 01) 1 St. Machar Academy

2 f () lim h 0 + h h h 1 1 (( + ) + ) Further simplification and rewriting the denominator gives, f () lim h h + Taking the limit as h 0 finally gives, f () 1 + f () 1 Notation for Derivatives The derivative of a function can be written in different ways. Below are listed some common types. Notation: The derivative of a function y f () is written variously as, y or f () (Dash notation aka prime notation or Lagrange notation) df d or d d f () or dy d (d by d notation aka Leibniz notation) Dy or Df (Big D notation aka Euler notation) The derivative of a function of time t has an alternative notation. Notation: The derivative of a function of time t is often written as, f (t) df dt (Dot notation aka Newton notation) M Patel (April 01) St. Machar Academy

3 Higher-Order Derivatives Sometimes, the derivative of a function can be differentiated. Given a function y f (), the higher-order derivative of order n (aka the n th derivative) is ined by, n d f d n d d d d d f d d d d d n times An important special case of this is the second derivative, d f d d df d d In the Euler and Lagrange notations, higher-order derivatives are written respectively as, D n f and f ( n ) () this Lagrange notation mainly being used for derivatives usually greater than or equal to 4 (as then the dashes become too numerous). To avoid ambiguity, sometimes the case when n 1 is called the first derivative. The derivatives we consider are technically called ordinary derivatives, in comparison to partial derivatives, which are derivatives of functions that are of more than 1 variable. Most interesting equations in science involve partial derivatives (as the description of most phenomena depends on more than 1 physical quantity). Eample Find the first and second derivatives of f () 4 + sin. f () 4 + cos f () 1 9 sin M Patel (April 01) St. Machar Academy

4 Eample Find the smallest value of n for which f ( n ) () 0 when f (). f () f () 6 f () 6 f (4 ) () 0 Clearly, all other higher derivatives will be 0 too, so n 4. Product and Quotient Rules The sum, difference and chain rules for differentiating functions are assumed in this course. In order to differentiate more types of functions, new rules will be introduced, the first to enable differentiation of a product of functions, the second to differentiate a quotient. Theorem (Product Rule): Given functions f and g, the derivative of their product is given by, (f g ) f g + f g Theorem (Quotient Rule): Given functions f and g, the derivative of their quotient (f / g ) is given by, 1 (f / g ) g (f g f g ) The Product Rule is easy to remember because of the + sign. The Quotient Rule takes more care because of the sign and the division by g. M Patel (April 01) 4 St. Machar Academy

5 Try writing both these rules in Leibniz notation and Euler notation to see which form is easier (or preferable) to remember. Eample 4 Differentiate h () cos. Let f () and g () cos. The following way of setting out the working is useful, f (), g () cos f (), g () sin Now substitute everything into the Product Rule, h () f () g () + f () g () h () ( ) cos + ( sin ) h () cos sin h () ( cos sin ) Eample 5 Differentiate h () ( 1) ( 1 ), simplifying as much as possible. Let f () and g () ( 1). Then, f (), g () ( 1) f (), g () ( 1) and g ( ) ( 1) 6 (remember rules of indices power of a power). Then apply the Quotient Rule to get, h () ( ) ( 1) ( ( 1) ) ( 1) 6 M Patel (April 01) 5 St. Machar Academy

6 h () ( 1) ( ( 1) ) 6 ( 1) h () ( 1) ( ) ( 1) ( + ) h () 4 ( 1) Reciprocal Trigonometric Functions 6 There are new trigonometric functions that will be studied in this course. They are ined using the sine, cosine and tangent functions. The secant function is ined by, sec 1 cos The domain and range of the secant function are, dom (sec ) R π R : + n π, n Z ran (sec ) R ( 1, 1) The cosecant function is ined by, cosec 1 sin The domain and range of the cosecant function are, dom (cosec ) R { R : n π, n Z } ran (cosec ) R ( 1, 1) M Patel (April 01) 6 St. Machar Academy

7 The cotangent function is ined by, cot cos sin The domain and range of the cotangent function are, dom (cot ) R { R : n π, n Z } ran (cot ) R Derivatives of Trigonometric Functions The derivatives of tan, sec, cosec and tan can be obtained by using the quotient rule and the derivatives of sin and cos. d tan sec d d sec sec tan d d cosec cosec cot d d cot cosec d Derivatives of ep and ln The derivatives of ep ( e ) and ln ( to obtain. The results will be stated here. d d e e d d ln 1 log ) require more effort e M Patel (April 01) 7 St. Machar Academy

8 Eample 6 Differentiate p () e sin. p () (e ) sin + e (cos ) p () e ( sin + cos ) Eample 7 Differentiate m () ln 4. m () 4 ( ) ln 4 4 (ln 4 ) Rectilinear Motion ln 4 m () (ln 4 ) ( ln 4 1) m () (ln 4 ) The derivative of a function gives a measure of the rate of change of a function with respect to a certain variable. Historically, Calculus was invented to satisfy the needs of describing physical phenomena. In particular, the rate of change of functions with respect to time played a very important role in understanding dynamics through mathematics. Motion in a straight line is called rectilinear motion. Unless otherwise stated, time will be measured in seconds and distance in metres. M Patel (April 01) 8 St. Machar Academy

9 The displacement of a particle from the origin O is the location of the particle from O after time t. Displacement is denoted by s (t). The velocity of a particle is ined by, v (t) ds dt s (t) The acceleration of a particle is ined by, a (t) dv dt d s dt ṡ (t) Eample 8 The displacement of a particle from the origin after time t is described by the function s (t) 4t t. Find the velocity and acceleration of the particle when t. s (t) 8t 9t s () 8() 9() s () 57 m/s ṡ (t) 8 18t ṡ () 8 18() ṡ () 46 m/s Differentiability of a Function Recall that a function is differentiable at a point if a unique tangent line can be drawn through that point. This can be understood in terms of tangent lines approaching from the left and right to the point in question. As tangent lines approach from the left and right, the gradients of these tangent lines should approach the same value. The function should be smooth and curvy at the point in question. M Patel (April 01) 9 St. Machar Academy

10 The left-hand derivative of f at a is ined by, f (a) lim h 0 f (a + h) f (a) h The right-hand derivative of f at a is ined by, f + (a) lim + h 0 f (a + h) f (a) h Theorem: A function is differentiable at a point if and only if both the left-hand and right-hand derivatives eist at that point and are equal to each other. Thus, to determine if a function is not differentiable at a point, it suffices to show one of the following (i) the left-hand derivative does not eist (ii) the right-hand derivative does not eist (iii) if they both eist, they are not equal. In terms of the graph of a function, this usually means identifying a sharp corner (aka kink ) or an endpoint (at an endpoint, obviously one of the derivatives won t eist). Eample 9 Determine where the function f ined by, f () (0 ) ( > ) is differentiable and where it is not differentiable. First sketch the graph of f. M Patel (April 01) 10 St. Machar Academy

11 y For 0 < <, f 1, so f is differentiable for 0 < <. For >, f 0, so f is differentiable for >. The only unclear points are where 0 and. For 0, the lefthand derivative does not eist, so f is not differentiable at 0. For, f () 1 and f + () 0, so f is not differentiable at. Critical Points A function may have points where the derivative is 0 or where the derivative does not eist. Such points have a special name. A function f has a critical point at a if either f (a) 0 or f (a) does not eist. a is called the critical number and f (a) is called the critical value. Note that all stationary points are critical points, but not all critical points are stationary points. Referring to the previous eample, f has critical points at (0, 0), (, ) and (, ) ( > ). At (0, 0) and (, ), the derivative doesn t eist, whereas for (, ) ( > ) the derivative is 0 (i.e. all the points (, ) ( > ) are stationary points). Etrema of a Function It is important to know the maimum and minimum values of a function. Maima and minima of a function are known as etrema. M Patel (April 01) 11 St. Machar Academy

12 There are types of etrema. Local. Endpoint. Global. A function f has a local (relative) maimum at a if an open interval I about a for which f (a) f () I. A function f has a local (relative) minimum at a if an open interval I about a for which f (a) f () I. Theorem: All local etrema occur at critical points. Stated differently, if f has a local etremum, then that local etremum point is also a critical point. Note that not all critical points are local etrema. Referring back to Eample 9, the point (0, 0) is not a local minimum, but it is a critical point. If a is an endpoint in dom f, f has an endpoint maimum at a if p dom f for which f (a) f () ( [a, p) or (p, a]). If a is an endpoint in dom f, f has an endpoint minimum at a if p dom f for which f (a) f () ( [a, p) or (p, a]). A function f function has a global (absolute) maimum at a if f (a) f () ( dom f ). A function f function has a global (absolute) minimum at a if f (a) f () ( dom f ). M Patel (April 01) 1 St. Machar Academy

13 Theorem: Every global etremum is either a local etremum or endpoint etremum. Eample 10 Classify the etrema for the function f with the graph shown. y D F C A G B E The small circles indicate the etent of where f is ined. The white circle means that the point is not in dom f, whereas the black circle means that the point is in dom f. Let s analyse the labelled points in turn. Remember, all etrema occur at critical points. A is not an endpoint ma., as A dom f (if A were in dom f, then A would be an endpoint ma.). A is not a critical point. B is a critical point, as f (B) < 0 and f + (B) > 0, so f (B). In particular, a local min. occurs at B, as all y values in the immediate vicinity of f (B) are bigger than f (B); the local min. is also a global min., as is clear from the graph. So, B is a global minimum. C is a critical point (even though f (C) > 0 and f + (C) > 0, the actual values are different and there is clearly no smooth transition between these values; the function is kinky at C, so f (C)). C is not a local M Patel (April 01) 1 St. Machar Academy

14 etremum, as there y values close to f (C) which are both bigger and smaller than f (C). C is a critical point. D is a critical point, as f (D) > 0 and f + (D) < 0, so f (D). D is obviously a local ma. and from the graph it clearly has the largest y value. So, D is a global maimum. E is a critical point, as f (E) 0. Thus, E is a stationary point. It is a local as well as a global min. So, E is a global minimum. F is clearly a critical point (for a similar reason cited for D). It is also a local ma., but not a global ma. So, F is a local maimum. G is a critical point, as f + (G). There are values to the left of G for which f () > f (G) and G is an endpoint. So, G is an endpoint minimum. Second Derivatives, Concavity and Points of Infleion The second derivative provides a useful measure of how curvy a function is and how the curviness changes. A function f is concave up on an interval if f () 0 on that interval. A function f is concave down on an interval if f () 0 on that interval. A function f has a point of infleion (inflection) at a if there is a change of concavity as the graph passes through a. The above initions lead to an alternative way of spotting infleion points. Theorem: If a function f has a point of infleion at a, then either f () or f () 0. M Patel (April 01) 14 St. Machar Academy

15 Note that if a function has f (a) 0, this does not necessarily mean 4 that a is a P of I. For eample, f () has f () 1 and thus f (0) 0. However, (0, 0) is clearly a minimum (sketch the graph). Infleion points may be further classified depending on the value of f. A function f has a horizontal (stationary) point of infleion at a if f (a) 0. A function has a non-horizontal (non-stationary) point of infleion at a if f (a) 0. For stationary points, there is a useful test for finding local maima and minima. Theorem (Second Derivative Test): If f (a) 0 and f (a) > 0, then f has a local minimum at a. If f (a) 0 and f (a) < 0, then f has a local maimum at a. If f (a) 0 and f (a) 0, then f may have a local maimum, a local minimum or a (horizontal) point of infleion. In the third case, the P of I must be horizontal, as f 0. A nature table is normally used to find maima, minima and infleion points. The Second Derivative Test gives a quick way of deciding whether a function has maima or minima. In the third case, a nature table must be drawn to determine the stationary point. Eample 11 Find all maima, minima and infleion points of f () sin in the interval (0, π). f () cos. The stationary points occur at (π/, 1) and (π/, 1). M Patel (April 01) 15 St. Machar Academy

16 f () sin. f (π/) 1 < 0 and f (π/) 1 > 0. So, (π/, 1) is a global maimum and (π/, 1) is a global minimum. As neither the ma. nor min. have f 0, any infleion points will be found by solving this equation. So, solving sin 0 gives π. When π, f (π) 0 and f (π) 1 0. Thus, (π, 0) is a non-horizontal P of I. If a function f has a P of I (and f eists at the point in question), then f 0. Hence, if f 0, there do not eist any infleion points. Eample 1 Show that f () points. + 5 has no stationary points and no infleion Rewriting f using long division gives, f () f () ( + ) For stationary points, f () 0. However, the equation, 8 ( + ) 0 obviously has no solutions. Hence, there are no maima or minima. For any infleion points, we need to analyse the second derivative, 16 f () ( + ) Similarly, the equation 16 ( + ) 0 has no solutions. Thus, there are no infleion points. M Patel (April 01) 16 St. Machar Academy