Higher Maths 1 3 Differentiation1

The History of Differentiation Differentiation is part of the science of Calculus, and was first developed in the 17 th century by two different Mathematicians. Gottfried Leibniz ( ) Germany Sir Isaac Newton ( ) England 2 Differentiation, or finding the instantaneous rate of change, is an essential part of: Mathematics and Physics Chemistry Biology Computer Science Engineering Navigation and Astronomy

Higher Maths 1 3 Differentiation Calculating Speed Time (seconds) Distance (m) D ST × ÷÷ Example Calculate the speed for each section of the journey opposite. A B C speed in A = 4 3 speed in B = m/s = speed in C = m/s = average speed = m/s ≈ 11 ≈ 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 = 3

Instantaneous Speed Higher Maths 1 3 Differentiation 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 = 4

Introduction to Differentiation Higher Maths 1 3 Differentiation 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’ yx 5

Estimating the Instantaneous Rate of Change Higher Maths 1 3 Differentiation The diagrams below show attempts to estimate the instantaneous gradient (the rate of change of with respect to ) at the point A. A y A x x y y A x 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 6

Basic Differentiation The instant rate of change of y with respect to x is written as. Higher Maths 1 3 Differentiation dy dx By long experimentation, it is possible to prove the following: dy dx = y = x nx n nxnx 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 7

Differentiation of Expressions with Multiple Terms Higher Maths 1 3 Differentiation y = ax m + bx n + … multiply every x -term by the power How to Differentiate: amx m + bnx n + … –1 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. 8

Examples of Basic Differentiation Higher Maths 1 3 Differentiation Find for dy dx y = 3 x 4 – 5 x x 2x 2 y = 3 x 4 – 5 x x dy dx = 12 x 3 – 15 x 2 – 14 x -3 9 = 9 x 09 x 0 this disappears because = 12 x 3 – 15 x 2 – 14 x 3x 3 9 (multiply by zero) Example 1

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

The Derived Function The word ‘derived’ means ‘produced from’, for example orange juice is derived from oranges. Higher Maths 1 3 Differentiation It is also possible to express differentiation using function notation. = f (x)f (x) nx n –1 If then f ′(x)f ′(x) = x nx n f ′(x)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) f (x)f (x) x 11

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 Higher Maths 1 3 Differentiation12 The gradient of any tangent to the function can be found by substituting the x -coordinate of intersection into. f (x)f (x) A B f ′(x)f ′(x) f (x)f (x) m AB = f ′(x)f ′(x)

f ′(x)f ′(x) Equations of Tangents Higher Maths 1 3 Differentiation13 y – b = m ( x – a ) To find the equation of a tangent: substitute gradient and point of intersection into y – b = m ( x – a )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)f (x) = x = x f ′(2)f ′(2) = × ( 2 ) = 6 m = y – 4 = 6 ( x – 2 ) 6 x – y – 8 = 0 substitute:

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

If a function is neither increasing or decreasing, the gradient is zero and the function can be described as stationary. Stationary Points Higher Maths 1 3 Differentiation15 There are two main types of stationary point. Minimum Maximum Turning PointsPoints of Inflection At any stationary point, or alternatively dy dx = 0 f ′(x)f ′(x) Rising Falling

Investigating Stationary Points Higher Maths 1 3 Differentiation16 Example Find the stationary point of f (x)f (x) = x 2 – 8 x + 3 x 2 – 8 x + 3 Stationary point given by f ′(x)f ′(x) = 2 x – 8 2 x – 8 2 x – 8 = 0 and determine its nature. x = 4 f ′(x)f ′(x) = 0 x 4 –4 – slope The stationary point at is a minimum turning point. – 0 + f ′(x)f ′(x) x = 4 Use a nature table to reduce the amount of working. ‘ slightly less than four ’ ‘ slightly more ’ gradient is positive

Investigating Stationary Points Higher Maths 1 3 Differentiation17 Example 2 Investigate the stationary points of y = 4 x 3 – x 44 x 3 – x 4 dy dx = 12 x 2 – 4 x 3 = 0 4 x 2 (3 – x )4 x 2 (3 – x ) = 0 4 x 24 x 2 = 0 x = 0 3 – x3 – x = 0 x = 3 or y = 0 y = 27 x 0 –0 – slope dy dx rising point of inflection stationary point at ( 0,0 ) : x 3 –3 – slope dy dx maximum turning point + 0 – stationary point at ( 3, 27 ) :

Positive and Negative Infinity Higher Maths 1 3 Differentiation18 Example ∞ The symbol is used for infinity. + ∞ – ∞ ‘positive infinity’ ‘negative infinity’ The symbol means ‘approaches’. y = 5 x x 2 For very large, the value of becomes insignificant compared with the value of. 5 x 35 x 3 ± x± x as x, ∞ y 5 x 35 x 3 5× ( ) 3 + ∞ = + ∞ – ∞ = – ∞ as x + ∞ ∞ +1 = ∞ y + ∞ and as x – ∞ y – ∞ 7 x 27 x 2

x 7 –7 – slope – 0 + dy dx Curve Sketching Higher Maths 1 3 Differentiation19 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

Graph of the Derived Function Higher Maths 1 3 Differentiation20 f ′(x)f ′(x) f (x)f (x) The graph of can be thought of as the graph of the gradient of. Example y = f ( x ) y = f ′ ( x ) f ′ ( x ) = 0 Positive Negative Gradient f ′ ( x ) > 0 f ′ ( x ) < 0 The roots of are f (x)f (x) f ′(x)f ′(x) given by the stationary points of.