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Bridge of Don Academy – Department of Mathematics Mr B. A. Willox CALCULUS – DIFFERENTIATION What do these pieces of notation mean: y dy and and f ( x) x dx and 2y d2y and and f ( x) x2 dx 2 What is derivative? What does it mean to differentiate? How do we differentiate? What is the derivative of: (a) y sin x derivative rate of change formula for gradient of a tangent to a curve velocity (where x is distance and y is time) the rate at which the rate of change is changing second derivative acceleration (where x is distance and y is time) formula for determining the nature of a stationary point and (b) y cos x ? Differentiation from First Principles Let us differentiate y x 2 from first principles. The derivative represents the instantaneous gradient at any given point. y To find the gradient at the point (x, x2) we can take a point nearby, (x + h, (x + h)2) and calculate the gradient between these points. The closer the points are the better our result i.e. h 0 our calculation improves. (x + h)2 x2 x f ( x) lim x 0 lim x 0 lim x 0 lim x 0 2x Page 1 x x+h ( x h) 2 x 2 ( x h) x x 2 2 xh h 2 x 2 h 2 2 xh h h 2x h Bridge of Don Academy – Department of Mathematics Mr B. A. Willox Chain Rule (differentiating a function of a function) If f(x) = g(h(x)) then or, in Leibniz notation, if f’(x) = g’(h(x)).h’(x) y = g(h(x)) then let u = h(x) dy dy du . dx du dx The unofficial Chain Rule is “Differentiate the outer function then multiply by the derivative of the inner function”. This can easily be expanded to more complex functions: if y = g(h(k(x))) then let u = k(x) y = g(h(u)) and let v = h(u) y = g(v) dy dy dv du dx dv du dx hence, Examples: 1. Differentiate y = (3x2 + 4x + 5)3. Let u = 3x2 + 4x + 5 y = u3 dy dy du dx du dx 3u 2 (6 x 4) 3(3x 2 4 x 5)(6 x 4) 6(3x 2 4 x 5)(3x 2) 2. Calculate dy when y = sin2(3x). dx Let u = 3x y = sin2u dy dy dv du dx dv du dx 2v cos u 3 2(sin u) cos u 3 6 sin 3x cos 3x 3(2sin 3x cos 3x) 3sin 6 x 3. Complete 15v 2 2v7 v 3v 4 Page 33, Exercise 3B – questions 1(b, a), 2(b, c), 4, 3c Page 2 and let v = sin u y = v2 Bridge of Don Academy – Department of Mathematics Mr B. A. Willox There can be some confusion with the notations for inverse trigonometric functions. a 1 1 . a We know that a2 a a We also know that sin 2 x (sin x)2 (sin x) (sin x) This implies that sin 1 x (sin x) 1 and 1 sin x so therefore the power for a trigonometric function comes after the ratio and before the algebraic term. but we know that this is not the case. [N.B. In many countries (and much of Britain is trying to implement this change) the inverse trigonometric functions are called arcsine, arccosine and arctangent.] To avoid this confusion, we denote functions as follows: inverse sine as sin 1 x , inverse cosine as cos 1 x and inverse tangent as tan 1 x , just as we have always done. 1 cosec x sin x 1 sec x cos x 1 cot x tan x However, we will denote the following functions as: Using the Chain Rule, find: dy 4. when y cosec x where x is defined on a suitable domain. dx y cosec x 5. Page 3 1 (sin x) 1 sin x dy dx dy when y sec x where x is defined on a suitable domain. dx i.e. cosecant i.e. secant i.e. cotangent. Bridge of Don Academy – Department of Mathematics Mr B. A. Willox The Product Rule If f(x) = g(x).h(x) then f’(x) = g’(x).h(x) + g(x).h’(x) or, using Leibniz notation, if y = uv then dy du v u dv dx dx dx [N.B. we do “derivative of 1st times 2nd untouched plus 1st untouched times derivative of 2nd”.] Proof f ( x) u ( x) v( x) f ( x h) f ( x) h u( x h) v( x h) u ( x) v( x) lim 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 v ( x h) u ( x h ) u ( x ) u ( x ) v ( x h) v ( x ) lim h 0 h v ( x h) u ( x h ) u ( x ) u ( x ) v ( x h) v ( x ) lim lim h 0 h 0 h h u( x h) u( x) lim u( x) v( x h) v( x) lim v( x h) h 0 h0 h h v( x) u '( x) u ( x) v '( x) f '( x) lim h 0 “central” terms introduced for convenience u '( x) v( x) u( x) v '( x) Examples: 6. Differentiate y = (2x – 3)(3x + 5). dy (2)(3x 5) (2 x 3)(3) dx 6 x 10 6 x 9 12 x 1 7. Calculate dy when y = sin2x . cos2x where x is defined on a suitable domain. dx dy d sin 2 x cos 2 x sin 2 x d cos 2 x dx dx dx 2 2sin x cos x cos 2 x sin x 2sin 2 x 2sin x cos x cos 2 x 2sin 2 x sin 2 x sin 2 x cos 2 x 2sin x sin 2 x 2 sin 2 x cos 2 x 2sin 2 x Page 4 [ N .B. 2sin x cos x sin 2 x] Bridge of Don Academy – Department of Mathematics 8. Mr B. A. Willox Calculate f’(x) when f(x) = (x2 + 2)(x – 1)2. f '( x) 2 x ( x 1) 2 ( x 2 2) 2( x 1) 2( x 1) x( x 1) ( x 2 2) 2( x 1)(2 x 2 x 2) 9. Calculate 3x 2 3 dy when y sin 2 3x where x is defined on a suitable domain. dx x Page 35, Exercise 4A – questions 1(a, c, e), 3(b, h, j) Page 36, Exercise 4B – questions 1(b, c), 5, 6 Page 5 Bridge of Don Academy – Department of Mathematics Mr B. A. Willox The Quotient Rule If f ( x) g ( x) then h( x) f '( x) g '( x) h( x) g ( x) h '( x) (h( x)) 2 dy or, using Leibniz notation, if y u then dx v dudx v u dvdx v2 [N.B. derivative of top bottom – top derivative of bottom, all over bottom squared.] f ( x) Proof 1 g ( x) g ( x) h( x) h( x) f '( x) g '( x) h( x) g ( x) h( x) h '( x) 1 2 using the Product Rule and the Chain Rule h '( x) g '( x) 1 g ( x) 2 h( x) h( x) g '( x) h( x) h '( x) 2 g ( x) 2 h( x) h( x) g '( x) h( x) g ( x) h '( x) 2 h( x) Examples: 10. 2 Differentiate y x 3x 5 . x2 dy (2 x 3)( x 2) ( x 2 3x 5)(1) dx ( x 2) 2 2 2 2 x 7 x 6 x2 3x 5 ( x 2) 2 x 4 x 2 1 ( x 2) 11. 2 Calculate d 5x . dx 3 2 x d 5x 2 d 5x 2 dx 3 2 x dx 3 2 x 1 2 1 1 10 x (3 2 x) 2 5 x 2 1 (3 2 x) 2 (2) 2 2 1 3 2x 2 10 x 3 2 x 5 x 2 Page 6 1 3 2x 3 2x 1 10 x 3 2 x 5 x 2 3 2x 3 2x 3 2x 10 x(3 2 x) 5 x 2 3 2 x (3 2 x) 15 x(2 x) 3 3 2x 2 Bridge of Don Academy – Department of Mathematics Mr B. A. Willox 12. IMPORTANT!! Calculate the derivative of y = tan x. y tan x sin x cos x dy cos x cos x sin x ( sin x) dx (cos x) 2 2 2 cos x 2 sin x cos x 12 cos x sec 2 x 13. IMPORTANT!! Calculate the derivative of y = cot x. 1 cos x tan x sin x dy ( sin x) sin x cos x cos x dx sin 2 x 2 2 sin x 2cos x sin x cosec2 x y cot x 14. IMPORTANT!! y sec x Calculate the derivative of y = sec x. 1 (cos x) 1 cos x dy 1(cos x) 2 ( sin x) dx sin2x cos x sin x 1 cos x cos x tan x sec x 15. IMPORTANT!! Calculate the derivative of y = cosec x. y cosec x 1 (sin x) 1 sin x dy 1(sin x) 2 (cos x) dx cos2 x sin x cos x 1 sin x sin x cot x cosec x Page 37, Exercise 5A – questions 1b, 2c, 3, 5, 7 Page 38, Exercise 6 – questions 1c, 1d, 4, 6 Page 40, Exercise 7 – questions 23 Page 7 using the Chain Rule Bridge of Don Academy – Department of Mathematics Mr B. A. Willox The Exponential and Logarithmic Functions Why is e 2 718281828 ? What is e? Imagine there is a very generous (or stupid) banker that is offering a savings interest rate of 100% per annum. If I put £1 into the bank, after 1 year I will have £2. Due to their intellect, they believe that offering 50% interest every 6 months will work out the same. However, now if I put £1 into the bank, after 1 year I will have £1 (1 5)2 £2 25 . Their brilliance continues and they decide to offer an interest rate of Now, if I put £1 into the bank, after 1 year I will now have £1 (1 100 % payable every day. 365 1 365 ) £2 714567482 . 365 If we continue this pattern to pay interest every second then we would have 1 £1 1 31536000 31536000 £2 718281615 after 1 year. As you can see, we are getting closer and closer to the value for e. e is the natural growth function as it allows for ‘doubling’ with the manmade constraints of how we measure time. The exponential function is defined as follows, 2 3 4 e x 1 x x x x ..... 2! 3! 4! Note, that if x = 1 then Page 8 e 1 1 1 1 1 1 1 ..... 2 6 24 120 720 2 718 (to 4 sig. fig.) Bridge of Don Academy – Department of Mathematics Mr B. A. Willox Derivative of ex If then 2 3 4 y = e x 1 x x x x ..... 2! 3! 4! dy 0 1 x 1 x 2 1 x 3 ..... dx 2 6 2 3 0 1 x x x ..... 2! 3! x e i.e. d ex ex dx Examples: 16. Differentiate y = e6x. y = eu Let u = 6x dy dy du dx du dx eu 6 6e6 x 17. Calculate the derivative of y = xex. Using the Product Rule, Page 9 dy 1 ex x ex dx e x (1 x) Bridge of Don Academy – Department of Mathematics Mr B. A. Willox Derivative of logex (i.e. lnx) y = logex = lnx x = ey dx e y x dy dy 1 dx x [turn both sides upside-down] i.e. d ln x 1 x dx Examples: 18. Find dy when y = ln(5x2 – 2). dx y = ln u Let u = 5x2 – 2 dy dy du dx du dx 1 10 x u x 10 2 5x 2 19. Calculate the derivative of y = ln(cos x). y = ln w Let w = cos x dy dy dw dx dw dx 1 ( sin x) w sin x cos x tan x 20. Differentiate y = eln3x. Let a = 3x y = elna and let b = ln a dy dy db da dx db da dx eb 1 3 a 3 eln a a 3 eln 3 x 3x 1 eln 3 x x Page 10 y = eb Bridge of Don Academy – Department of Mathematics Mr B. A. Willox Some cases can be simplified using the laws of logs (from Higher). Differentiate y ln 1 x2 . 1 x 2 21. y ln 1 x 2 ln 1 x 2 ln 1 x 2 1 x dy (2 x) 2x 2 dx 1 x 1 x2 2 x 1 x 2 2 x 1 x 2 1 x 2 1 x 2 2 4x 4 1 x 22. Calculate d ln x 2 1 . dx Let y ln x 2 1 and u = x2 + 1 or this can simplified as: dy dy dv du dx dv du dx 1 1 1 u 2 2x v 2 1 1 2x u 2 u x u 2x x 1 y = ln v v= u and Note that logs must be to the base e, otherwise, we must change the base. For example, y = log10x x = 10y ln x = ln 10y ln x = y ln 10 y = ln x 1 ln x ln10 ln10 Take log, to the base e, of both sides, dy dx Page 11 d ln x 2 1 d 1 ln( x 2 1) dx dx 2 1 d ln( x 2 1) 2 dx 2x 2 ( x 2 1) 1 1 ln10 x 1 x ln10 Bridge of Don Academy – Department of Mathematics Mr B. A. Willox Similarly, for y = 2x Take logs of both sides to the base e, ln y = ln 2x ln y = x ln 2 ln y x= ln 2 dx 1 dy y ln 2 dy y ln 2 2 x ln 2 dx Page 43, Exercise 8A – questions 1, 3(a, c, d), 5(c, e), 6 Page 44, Exercise 8B – questions 1 Page 12 Bridge of Don Academy – Department of Mathematics Mr B. A. Willox List of Standard Derivatives f’(x) xn nxn-1 sin x cos x cos x - sin x tan x sec2 x cot x - cosec2 x sec x sec x tan x cosec x - cosec x cot x ex ex ln x 1 x ax ax ln a loga x 1 x ln a We don’t need to memorise this last two Page 13 f(x) Bridge of Don Academy – Department of Mathematics Mr B. A. Willox Higher Order Derivatives – Part 1 We already know that for a function where x represents displacement and t represents time that dx dt represents the change in distance with respect to the time taken for that change i.e. velocity. d 2x Also, 2 represents the change in velocity with respect to the time taken for that change i.e. dt acceleration. Displacement, Velocity and Acceleration using Calculus On a displacement-time graph the gradient of a chord represents average velocity whereas the gradient of a tangentat a point represents instantaneous velocity (i.e. velocity at a certain time). So, v ds dt distance velocity On a velocity-time graph the gradient of a chord represents average acceleration whereas the gradient of a tangent at a point represents instantaneous acceleration. So, 2 a dv d s2 . dt dt Examples: 23. The displacement, s metres, of a point moving in a straight line after t seconds is given by s = 5t2 + t3. Find the velocity and acceleration after 4 seconds. s = 5t2 + t3 v ds 10t 3t 2 when t = 4, v = 40 + 48 = 88 dt the point has a velocity of 88 m/s after 4 seconds a dv 10 6t when t = 4, a = 10 + 24 = 34 dt the point has an acceleration of 34 m/s2 after 4 seconds. Page 14 Bridge of Don Academy – Department of Mathematics 24. A particle is moving in a straight line and its acceleration after t seconds is (20 – 6t) m/s2. (a) Find its velocity in terms of t given that its initial velocity is 5 m/s. (b) Find an expression for the particles displacement, s m, in terms of t given that t = 2 when s = 50. (a) a dv 20 6t dt v 20t 3t 2 c1 when t = 0, v = 5 (b) Remember: c1 = 5 v ds dt s 5t 10t 2 t 3 c2 when t = 2, s = 50 c2 = 8 v = 5 + 20t – 3t2 s = 8 + 5t +10t2 – t3 The area under the curve in a velocity-time graph gives the distance travelled. s = area under curve = Page 15 Mr B. A. Willox v dt dt dt ds s c where c is a constant ds Bridge of Don Academy – Department of Mathematics Mr B. A. Willox Higher Order Derivatives – Part 2 We also use derivatives for finding stationary points or for examining the behaviour of a curve. Here we see the graph of y f ( x) sloping up so ‘gradient’ is positive We can see that there are stationary points at the points marked with a . Points of inflection are marked with a . sloping up so ‘gradient’ is positive sloping down so ‘gradient’ is negative ‘horizontal’ so ‘gradient’ is zero sloping up so ‘gradient’ is positive By looking at the graph of the derivative we can see why these points must be stationary points. ‘horizontal’ so ‘gradient’ is zero Here we see the graph of y f ( x) Notice that the point of inflection is not a stationary point – it is an “increasing point of inflection”. The gradient was decreasing as in approached the point (i.e. concave down) then after the point the gradient was increasing (i.e. concave up). Therefore the point is a point of inflection. Here we see the graph of y f ( x ) . ☼ ☼ Page 16 Notice that the maximum stationary point from f ( x) gives us a negative value for f ( x) and That the minimum stationary point from f ( x) gives us a positive value for f ( x) . Bridge of Don Academy – Department of Mathematics Mr B. A. Willox To summarize: If (i) f (a) 0 then there is a maximum stationary point at the point a, f (a) (ii) f (a) 0 then there is a minimum stationary point at the point a, f (a) (iii) f (a) 0 then there is a point of inflection at the point a, f (a) . We can use these results to determine the nature of stationary points (as an alternative to the ‘nature table’). However, it is often more efficient to use a ‘nature table’. For example, if we found a point of inflection at point a, f (a) then we would need to use a ‘nature table’ to determine whether it was increasing or decreasing. Example: 25. Sketch the curve of the function y x 2 (2 x 3)( x 1) for 2 x 2 , indicating all roots, stationary points and points of inflection. 2 Roots occur when y = 0 x (2 x 3)( x 1) 0 3 x 0, x , x 1 2 y 2 x 4 x3 3x 2 dy dx Stationary points occur when Page 46, Exercise 9A – questions 1, 2, 4a, 4(c, d), 6, 8 Page 46, Exercise 9B – questions 1, 3 Page 48, Review – questions 2 to 10 Page 17 dy 0 dx Bridge of Don Academy – Department of Mathematics Mr B. A. Willox Examples from SQA Exams: 26. Given that f ( x) x e x , x 0 obtain and simplify f ( x ) . 27. Given f ( x) cos 2 x e tan x , and evaluate f . 4 Page 18 2 x 2 [2002: 4 marks] , obtain f ( x ) [2004: 3,1 marks] Bridge of Don Academy – Department of Mathematics 28. 29. Page 19 Mr B. A. Willox Obtain the derivative of each of the following functions: (a) f ( x) exp (sin 2 x) ; (b) y 4( x 2 1) . Write down the derivative of tan x. Show that 1 tan 2 x sec 2 x . Hence obtain tan 2 x dx . [2007: 3, marks] [2008: 1,1,2 marks]