Presentation on theme: "Higher Mathematics: Unit 1.3 Introduction to Differentiation"— Presentation transcript:
1Higher Mathematics: Unit 1.3 Introduction to Differentiation
2Required skills Before we start ………. You will need to remember work with Indices, as well as what you have learned about Straight Lines from Unit 1.1.Let’s recall the rules on indices …..
3Rules of indices a0 = 1 12.3140 = a-m = x -5 = = am × an = am+n Examplesa0 = 1=a-m =x -5 ==am × an = am+n2a3/2 × 3a1/2 =am ¸ an = am-nx2 ¸ x -3 =(am)n = amn(q2)3 =1
4What is Differentiation? Differentiation is the process of derivingf ′(x) from f(x). We will look at this process in a second.f ′ (x) is called the derived function or derivative of f(x).The derived function represents:the rate of change of the functionthe gradient of the tangent to the graph of the function.
5Tangents to curves y2 – y1 x2 – x1 B(x2,y2) A(x1,y1) C The derivative function is a measure of the gradient or slope of a function at any given point. This requires us to consider the gradient of a line.We can do this if we think about how we measure the gradient from Unit 1.1A(x1,y1)B(x2,y2)CThe gradient of AB = mABy2 – y1x2 – x1
6Tangents to curvesWe will investigate the tangent to a curve at a single point.The points A and B lie on a function. The line joining A and B is not a tangent to the curve at the point A. If we think about moving the point B towards A, the slope of the line will change.We can observe the results by watching this graphically. We will take the simple function f(x) = x2Click the graphic to investigate this function.
7Tangents to curves x f ′ (x) 2 4 1 -1 -2 -4 Observe the table of results and suggest a relationship between the value of the slope of the tangent and the value of x on the functionf(x) = x2It appears that the relationship may bef ′(x) = 2xCan we find this algebraically?xf ′ (x)241-1-2-4
8Tangents to curves x+h f(x+h) x f(x) We will look at a function and think about the gradient of the function at any given point. The function itself is not important, the process we go through to get the gradient is. We want to find the gradient at a point on a curve.f(x)A is the point (x, f(x)) and Bis a point on the function a short distance h from A.This gives B the coordinates(x+h, f(x+h))x+hf(x+h)xf(x)The line AB is shown on the diagram. We want to find the gradient of the curve at A. If we find the gradient of the line AB and move B towards A we should get the gradient at A.x
9Tangents to curves x+h f(x+h) x f(x) The gradient of the line AB would therefore bex+hf(x+h)xf(x)
10Tangents to curvesLook at what happens as we move B towards A, i.e. h is getting smaller.As the size of h gets smaller and smaller the line AB becomes the tangent to the curve at the point A. We refer to this as the limit as h → 0. Note that h cannot become 0 or we would not get a line AB!The gradient of the line AB becomes the same as the gradient of the curve at AThis is the value we were looking for.
11Putting it together…. Differentiate the function f(x) = x2 from first principles.The derivative is the same as the gradient of the tangent to the curve so we can go straight to the gradient formula we saw in the previous slides.The limit as h → 0 iswritten ash gets so small its effectively zero.
12What does the answer mean? For each and every point on the curve f(x)=x2 , the gradient of the tangent to the curve at the point x is given by the formulaf ′(x)=2xValue of xGradient of the tangent-3-2-11234-6-4-2246A table of results might make this clearer8Is there an easier way to do this?We can agree the observation graphically and algebraically.
14Rules for differentiation There are four rules for differentiating – remember these and you can differentiate anything …RuleExamplesf(x) = xn Þ f ′(x) = nxn-1f(x) = x6 Þ f ′ (x) =f(x) = cxn Þ f ′ (x) = cnxn-1f(x)= 4x2 Þ f ′ (x) =f(x) = c Þ f ′ (x) = 0f(x) = 65 Þ f ′ (x) =f(x) = g(x) + h(x)Þ f ′ (x) = g′ (x) + h′ (x)f(x)= x6 + 4x Þ f ′ (x) =6 x 6 - 1= 6 x 54 x 2 x 2-1= 8 x 1 or 8 x6 x 5 + 8x
15Notation There are many different ways of writing f ′ (x): f ′ (x) y′ y ′ (x)The most common of these are:f ′ (x) Þ functional notation used when function is defined as f(x)Þ Leibnitz notation used when function is defined as y =
16The Gradient of a Tangent Remember:Differentiation is used to find the gradient of a tangent to a graph.Find the equation of the tangent to the curve y = 3x3 – x + 6 at x = 2.A tangent is a straight line, so we need to use:y – b = m(x – a)To use this we need to know:the gradient of the tangent at the point on the curve(in this case when x = 2)the coordinates of a point on the line (in this case (2, ?) )
17The Gradient of a Tangent Step 1 : Finding the gradient when x = 2Differentiate the functionAs the function is given in terms of y we will use the dy/dx notationAt the point x = 2,So the gradient of the tangent is 35
18The Gradient of a Tangent Step 2 : Finding the coordinates at the pointAt the point x = 2 the function value is given byy = 3 (2)3 – 2 + 6y = 3(8) + 4y = 28The coordinate is therefore the point (2, 28)
19The Gradient of a Tangent Step 3 : Finding the equation of the tangentThe line passing through (2,28) with gradient 35 is :The equation of the tangent at x = 2 is y – 35x + 42 = 0
20The Gradient of a Tangent You can confirm this by checking on a graphing calculator or by sketch.
21Sketching Graphs of Derived Functions We can investigate graphs of derived functions by looking at a dynamic setting of a function, along with the resulting derived function.There are 3 examples shown on separate pages. Look at each one and build up the derived graph as you are prompted. Try to predict the derivative of the third graph before it is drawn.Click the graphic to start your investigation
22Sketching Graphs of Derived Functions We can sketch the graph of the derived function, f ′(x), by considering the graph of f(x).The diagram shows the graph of y = f(x)Sketch the graph of y = f ′ (x)Step 1 Decide what type of function it is:This looks like a cubic function,i.e. it has an x3 term in it.When we differentiate the functionthe x3 term will become an x2 term. f ′ (x) will be a quadratic function and will look like this or
23Sketching Graphs of Derived Functions Step 2 Consider the gradient of the function at key pointsGradient is negativem < 0Tangent is horizontalm = 0Gradient is positivem > 0Gradient is positivem > 0Tangent is horizontalm = 0This can be summarised on a table
24Sketching Graphs of Derived Functions x→-13f ′ (x)+ ve- ve+ ve
25Sketching Graphs of Derived Functions Step 3 Sketch this information as a graphx→-13f ′ (x)+ ve- ve-13f’(x) +ve above the linef’(x) = 0 on the linef’(x) -ve below the line