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Higher Mathematics: Unit 1.3 Introduction to Differentiation.

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Presentation on theme: "Higher Mathematics: Unit 1.3 Introduction to Differentiation."— Presentation transcript:

1 Higher Mathematics: Unit 1.3 Introduction to Differentiation

2 Required 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 …..

3 RuleExamples a 0 = = a -m =x -5 = = a m × a n = a m+n 2a 3/2 × 3a 1/2 = = a m  a n = a m-n x 2  x -3 = = (a m ) n = a mn (q 2 ) 3 = = Rules of indices 1

4 What is Differentiation? Differentiation is the process of deriving f ′(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 function the gradient of the tangent to the graph of the function.

5 Tangents to curves 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 1.1 y 2 – y 1 x 2 – x 1 The gradient of AB = m AB A(x 1,y 1 ) B(x 2, y 2 ) C

6 Tangents to curves We 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) = x 2 Click the graphic to investigate this function.

7 Tangents to curves 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 function f(x) = x 2 It appears that the relationship may be f ′(x) = 2x Can we find this algebraically? xf ′ (x)

8 Tangents to curves 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. A is the point (x, f(x)) and B is a point on the function a short distance h from A. This gives B the coordinates (x+h, f(x+h)) 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+h f(x+h) x f(x) x

9 Tangents to curves f(x+h) f(x)x x+h The gradient of the line AB would therefore be

10 Tangents to curves Look 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 A This is the value we were looking for.

11 Putting it together…. Differentiate the functionf(x) = x 2 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 is written as h gets so small its effectively zero.

12 What does the answer mean? For each and every point on the curve f(x)=x 2, the gradient of the tangent to the curve at the point x is given by the formula f ′(x)=2x Value of xGradient of the tangent A table of results might make this clearer Is there an easier way to do this? We can agree the observation graphically and algebraically.

13 f(x)f ′ (x)f(x)f ′ (x)f(x)f ′ (x) x2x2 2x 2 3x 3 x3x3 3x 2 5x 4 x4x4 4x 2 7x 3 x5x5 5x 2 6x 7 xnxn ax 2 ax n Spot the pattern f(x)f ′ (x)f(x)f ′ (x)f(x)f ′ (x) x2x2 2x 2 3x 3 x3x3 3x 2 5x 4 x4x4 4x 2 7x 3 x5x5 5x 2 6x 7 xnxn ax 2 ax n f(x)f ′ (x)f(x)f ′ (x)f(x)f ′ (x) x2x2 2x 2 3x 3 x3x3 3x 2 5x 4 x4x4 4x 2 7x 3 x5x5 5x 2 6x 7 xnxn ax 2 ax n 2x2x 3x23x2 4x34x3 5x45x4 nx n-1 4x4x 6x6x 8x8x 10x 2ax 9x29x2 20x 3 21x 2 42x 6 anx n-1

14 Rules for differentiation There are four rules for differentiating – remember these and you can differentiate anything … RuleExamples f(x) = x n  f ′(x) = nx n-1 f(x) = x 6  f ′ (x) = f(x) = cx n  f ′ (x) = cnx n-1 f(x)= 4x 2  f ′ (x) = f(x) = c  f ′ (x) = 0f(x) = 65  f ′ (x) = f(x) = g(x) + h(x)  f ′ (x) = g′ (x) + h′ (x) f(x)= x 6 + 4x  f ′ (x) = 6 x = 6 x 5 4 x 2 x 2-1 = 8 x 1 or 8 x 0 6 x 5 + 8x

15 Notation There are many different ways of writing f ′ (x): f ′ (x) y′ y ′ (x) The most common of these are: f ′ (x)  functional notation used when function is defined as f(x)  Leibnitz notation used when function is defined as y =

16 The 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 = 3x 3 – 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, ?) )

17 The Gradient of a Tangent Step 1 :Finding the gradient when x = 2 Differentiate the function As the function is given in terms of y we will use the dy / dx notation At the point x = 2, So the gradient of the tangent is 35

18 The Gradient of a Tangent Step 2 :Finding the coordinates at the point At the point x = 2 the function value is given by y = 3 (2) 3 – y = 3(8) + 4 y = 28 The coordinate is therefore the point (2, 28)

19 The Gradient of a Tangent Step 3 :Finding the equation of the tangent The line passing through (2,28) with gradient 35 is : The equation of the tangent at x = 2 is y – 35x + 42 = 0

20 The Gradient of a Tangent You can confirm this by checking on a graphing calculator or by sketch.

21 Sketching 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

22 Sketching 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 1Decide what type of function it is: This looks like a cubic function, i.e.it has an x 3 term in it. When we differentiate the function the x 3 term will become an x 2 term.  f ′ (x) will be a quadratic function and will look like this or

23 Sketching Graphs of Derived Functions Step 2Consider the gradient of the function at key points Tangent is horizontal m = 0 Gradient is positive m > 0 Tangent is horizontal m = 0 Gradient is positive m > 0 Gradient is negative m < 0 This can be summarised on a table

24 Sketching Graphs of Derived Functions x → → 3 → f ′ (x) + ve 0 - ve 0 + ve

25 Sketching Graphs of Derived Functions x→→3→ f ′ (x)+ ve0- ve0+ ve Step 3Sketch this information as a graph f’(x) +ve above the line f’(x) -ve below the line f’(x) = 0 on the line 3


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