1. Higher Maths 1 3 Differentiation UNIT OUTCOME SLIDE

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

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

4. Instantaneous Speed NOTE Higher Maths 1 3 Differentiation UNIT OUTCOME Time (seconds) Distance (m) Time (seconds) Distance (m) 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? D T S = y x = m = SLIDE

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

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

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

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

9. Examples of Basic Differentiation NOTE Higher Maths 1 3 Differentiation UNIT OUTCOME Find for dy dx y = 3 x 4 – 5 x 3 + + 9 7 x 2 y = 3 x 4 – 5 x 3 + 7 x -2 + 9 dy dx = 12 x 3 – 15 x 2 – 14 x -3 9 = 9 x 0 this disappears because = 12 x 3 – 15 x 2 – 14 x 3 SLIDE (multiply by zero) Example 1

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

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

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

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

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

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

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

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

18. Positive and Negative Infinity NOTE Higher Maths 1 3 Differentiation UNIT OUTCOME SLIDE Example ∞ The symbol is used for infinity. + ∞ – ∞ ‘ positive infinity’ ‘ negative infinity’ The symbol means approaches’. y = 5 x 3 + 7 x 2 For very large , the value of becomes insignificant compared with the value of . 5 x 3 ± x as x , ∞ y 5 x 3 5× ( ) 3 + ∞ = + ∞ 5× ( ) 3 – ∞ = – ∞ REMEMBER as x + ∞ ∞ +1 = ∞ y + ∞ and as x – ∞ y – ∞ 7 x 2

19. Higher Maths 1 3 Differentiation UNIT OUTCOME SLIDE x 7 – 7 + 7 slope – 0 + dy dx Curve Sketching NOTE To sketch the graph of any function, the following basic information is required: • the stationary points and their nature • the x -intercept ( s ) and y -intercept • the value of y as x approaches positive and negative infinity solve for y = 0 and x = 0 solve for = 0 and use nature table dy dx y = x 3 – 2 x 4 – ∞ y + ∞ as x ∞ y -2 x 4 + ∞ y – ∞ as x and as x Example

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