1. Higher Maths 1 3 Differentiation 1 OUTCOME SLIDE UNIT

2. Higher Maths 1 3 Differentiation 2 OUTCOME SLIDE UNIT NOTE The History of Differentiation Differentiation is part of the science of Calculus, and was first developed in the 17th century by two different Mathematicians. Differentiation, or finding the instantaneous rate of change, is an essential part of: Gottfried Leibniz(1646-1716) Germany • Mathematics and Physics • Chemistry • Biology • Computer Science • Engineering • Navigation and Astronomy Sir Isaac Newton(1642-1727) England

3. D ÷ ÷ × S T 1 1 Higher Maths 1 3 Differentiation 3 OUTCOME SLIDE UNIT NOTE C Calculating Speed 10 8 B Example 6 Distance (m) Calculate the speed for each section of the journey opposite. 4 A 2 0 0 2 4 6 8 4 speed in A = 1.33 m/s ≈ Time (seconds) 3 Notice the following things: 5 speed in B = 5 m/s = 1 • the speed at each instant is not the same as the average 2 speed in C = 0.4 m/s = 5 • speed is the same as gradient average speed = 1.22 m/s ≈ y D 9 S = m = = x T

4. Higher Maths 1 3 Differentiation 4 OUTCOME SLIDE UNIT y D NOTE S = m Instantaneous Speed = = x T Distance (m) Distance (m) Time (seconds) Time (seconds) In reality speed does not often change instantly. The graph on the right is more realistic as it shows a gradually changing curve. The journey has the same average speed, but the instantaneous speed is different at each point because the gradient of the curve is constantly changing. How can we find the instantaneous speed?

5. Higher Maths 1 3 Differentiation 5 OUTCOME SLIDE UNIT NOTE Introduction to Differentiation Differentiate means ‘find out how fast something is changing in comparison with something else at any one instant’. ‘rate of change ofdistancewith respect totime’ D speed = T ‘rate of change ofspeed with respect totime’ S acceleration = T y ‘rate of change of -coordinate with respect to -coordinate’ gradient = y x x

6. x x x y y y Higher Maths 1 3 Differentiation 6 OUTCOME SLIDE UNIT NOTE Estimating the Instantaneous Rate of Change The diagrams below show attempts to estimate the instantaneous gradient (the rate of change of with respect to ) at the point A. x y y y y A A A x x x Notice that the accuracy improves as gets closer to zero. The instantaneous rate of change is written as: y dy x as approaches 0. = x dx

7. Higher Maths 1 3 Differentiation 7 OUTCOME SLIDE UNIT NOTE Basic Differentiation dy The instant rate of change ofywith respect toxis written as . dx By long experimentation, it is possible to prove the following: y xn = If How to Differentiate: n dy • multiply by the power• reduce the power by one –1 nx = then dx dy Note that describes both the rate of change and the gradient. dx

8. –1 –1 amxm+bnxn+… Higher Maths 1 3 Differentiation 8 OUTCOME SLIDE UNIT NOTE Differentiation of Expressions with Multiple Terms The basic process of differentiation can be applied to everyx-term in an algebraic expession. Important y axm+bxn+… = Expressions must be written as the sum of individual terms before differentiating. dy = dx How to Differentiate: • multiply every x-term by the power • reduce the power of every x-term by one

9. Higher Maths 1 3 Differentiation 9 OUTCOME SLIDE UNIT NOTE Examples of Basic Differentiation Example 1 dy 7 y 3x4– 5x3 ++9 Find for = x2 dx y 3x4– 5x3 +7x-2+9 = this disappears because dy 12x3– 15x2 –14x-3 = 9x0 9 = dx (multiply by zero) 14 12x3– 15x2 – = x3

10. Higher Maths 1 3 Differentiation 10 OUTCOME SLIDE UNIT NOTE Examples of Basic Differentiation Example 2 (x+3)(x–5) y Find the gradient of the curve = x2 at the point (5,0). dy 2x-2+ 30x-3 = x²–2x–15 dx y = x2 2 30 + = x2 x3 2 15 1 – – = x x2 dy 2 30 At x= 5, + = dx 25 125 1–2x-1–15x-2 = 8 = disappears (multiply by zero) 25

11. Higher Maths 1 3 Differentiation 11 OUTCOME SLIDE UNIT NOTE The Derived Function It is also possible to express differentiation using function notation. Newton Leibniz dy xn f(x) f′(x) = and If dx n f′(x) –1 nx mean exactly the same thing written in different ways. = then The word ‘derived’ means ‘produced from’, for example orange juice is derived from oranges. f′(x) f(x) The derived function is the rate of change of the function with respect to . x

12. Higher Maths 1 3 Differentiation 12 OUTCOME SLIDE UNIT NOTE Tangents to Functions A tangent to a function is a straight line which intersects the function in only one place, with the same gradient as the function. f(x) B mAB = f′(x) A f(x) The gradient of any tangent to the function can be found by substituting thex-coordinate of intersection into . f′(x)

13. Higher Maths 1 3 Differentiation 13 OUTCOME SLIDE UNIT NOTE REMEMBER Equations of Tangents y–b=m(x–a) To find the equation of a tangent: Straight Line Equation • differentiate x • substitute -coordinate to find gradient at point of intersection y–b=m(x–a) • substitute gradient and point of intersection into 3 Example = x2 f′(x) 2 Find the equation of the tangent to the function at the point (2,4). 3 m = f′(2) = ×(2)2 = 6 2 1 y– 4= 6(x– 2) = x3 f(x) substitute: 2 6x–y– 8= 0

14. REMEMBER Higher Maths 1 3 Differentiation 14 OUTCOME SLIDE UNIT NOTE Increasing and Decreasing Curves Gradient The gradient at any point on a curve can be found by differentiating. Positive uphill slope dy Negative theny is increasing. >0 If dx downhill slope dy theny is decreasing. <0 If dx Alternatively, dy dy >0 f′(x) f(x) If then >0 <0 dx y dx is increasing. x f′(x) f(x) If then <0 dy <0 is decreasing. dx

15. ! Higher Maths 1 3 Differentiation 15 OUTCOME SLIDE UNIT NOTE Stationary Points If a function is neither increasing or decreasing, the gradient is zero and the function can be described as stationary. There are two main types of stationary point. Turning Points Points of Inflection Maximum Falling Rising Minimum dy f′(x) At any stationary point, or alternatively = 0 = 0 dx

16. ! Higher Maths 1 3 Differentiation 16 OUTCOME SLIDE UNIT NOTE Investigating Stationary Points Use a nature table to reduce the amount of working. Example Find the stationary point of x2–8x+3 f(x) = ‘slightly less than four’ ‘slightly more’ and determine its nature. x 4– 4+ Stationary point given by 4 f′(x) = 0 f′(x) – + 0 gradientis positive = 2x–8 f′(x) slope 2x–8 = 0 x The stationary point atis a minimum turning point. = 4 x = 4

17. Higher Maths 1 3 Differentiation 17 OUTCOME SLIDE UNIT stationary point at (0,0): NOTE Investigating Stationary Points x 0– 0+ 0 Example 2 dy + + Investigate the stationary points of 0 dx y 4x3–x4 = slope dy rising point of inflection 12x2–4x3 0 = = dx stationary point at (3, 27): 4x2(3–x) 0 = x 3– 3+ 3 4x2 3–x dy 0 0 = = or – + 0 dx x x 0 3 = = slope y y 0 27 = = maximum turning point

18. REMEMBER ! Higher Maths 1 3 Differentiation 18 OUTCOME SLIDE UNIT NOTE ∞ Positive and Negative Infinity The symbol is used for infinity. Example ∞ ‘positive infinity’ + y 5x3+ 7x2 = ∞ ‘negative infinity’ – ±x 7x2 For very large , the value ofbecomes insignificant compared with the value of . ∞ ∞ +1 = 5x3 The symbol means ‘approaches’. as x , y ∞ 5x3 ∞ ∞ 5×()3 + + = as x and as x ∞ ∞ – + ∞ ∞ 5×()3 – – = y y ∞ ∞ – +

19. x 7– 7+ 7 dy – + 0 dx slope Higher Maths 1 3 Differentiation 19 OUTCOME SLIDE UNIT NOTE Curve Sketching Example To sketch the graph of any function, the following basic information is required: y x3– 2x4 = as x ∞ • the stationary points and their nature dy y -2x4 solve for =0 and use nature table dx ∞ as x • the x-intercept(s) and y-intercept + y ∞ solve for y=0 and x=0 – ∞ and as x – • the value of y as x approachespositive and negative infinity y ∞ +

20. REMEMBER Higher Maths 1 3 Differentiation 20 OUTCOME SLIDE UNIT NOTE Graph of the Derived Function f′(x) The graph of can be thought of as the graph of the gradient of . Gradient f(x) Positive f′(x)> 0 Example y = f′(x) Negative f′(x)= 0 y f′(x)< 0 x The roots of are f′(x) given by the stationary y = f(x) f(x) points of . f′(x)= 0