# AS-Level Maths: Core 1 for Edexcel

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AS-Level Maths: Core 1 for Edexcel
C1.7 Differentiation This icon indicates the slide contains activities created in Flash. These activities are not editable. For more detailed instructions, see the Getting Started presentation. 1 of 45 © Boardworks Ltd 2005

Contents Rates of change Rates of change
The gradient of the tangent at a point The gradient of the tangent as a limit Differentiation of polynomials Second order derivatives Tangents and normals Examination-style questions Contents 2 of 45 © Boardworks Ltd 2005

Rates of change This graph shows the distance that a car travels over a period of 5 seconds. The gradient of the graph tells us the rate at which the distance changes with respect to time. time (s) distance (m) 5 40 In other words, the gradient tells us the speed of the car. The car in this example is travelling at a constant speed since the gradient is the same at every point on the graph.

Rates of change In most situations, however, the speed will not be constant and the distance–time graph will be curved. For example, this graph shows the distance–time graph as the car moves off from rest. time (s) distance (m) The speed of the car, and therefore the gradient, changes as you move along the curve. To find the rate of change in speed we need to find the gradient of the curve. The process of finding the rate at which one variable changes with respect to another is called differentiation. In most situations this involves finding the gradient of a curve.

The gradient of the tangent at a point
Rates of change The gradient of the tangent at a point The gradient of the tangent as a limit Differentiation of polynomials Second order derivatives Tangents and normals Examination-style questions Contents 5 of 45 © Boardworks Ltd 2005

The gradient of a curve The gradient of a curve at a point is given by the gradient of the tangent at that point. Look at how the gradient changes as we move along a curve: Explain that the direction of a curve changes as we move along it. Demonstrate this by moving the point along the curve. Observe where the gradient of the tangent at the point is positive, negative and 0.

Finding the gradient of a curve at a point
Suppose we want to find the gradient of the curve y = x2 at the point (3, 9). Use graph paper to carefully plot the curve y = x2 for –5 ≤ x ≤ 5. Use a ruler to draw the tangent at the point (3, 9) and find the gradient of the tangent. Find the gradients of the tangents at four other points and write down what you notice. As you can see, it is quite difficult to do this accurately because we only know a single point that the line passes through. Supply students with graph paper so that they can complete this exercise. Students who have found the gradient sufficiently accurately should notice that the gradient at a given point on the curve y = x2 is double the value of the x-coordinate at that point. To find the gradient of a line we need to know two points on the line (x1, y1) and (x2, y2). We can then find its gradient:

The gradient of the tangent as a limit
Rates of change The gradient of the tangent at a point The gradient of the tangent as a limit Differentiation of polynomials Second order derivatives Tangents and normals Examination-style questions Contents 8 of 45 © Boardworks Ltd 2005

Differentiation from first principles
Suppose we want to find the gradient of a curve at a point A. We can add another point B on the line close to point A. δx represents a small change in x and δy represents a small change in y. As point B moves closer to point A, the gradient of the chord AB gets closer to the gradient of the tangent at A. Explain that the small letter delta, δ, is used to represent a small change in the variable. A capital delta, Δ, can also be used.

Differentiation from first principles
We can write the gradient of the chord AB as: As B gets closer to A, δx gets closer to 0 and gets closer to the value of the gradient of the tangent at A. δx can’t actually be equal to 0 because we would then have division by 0 and the gradient would then be undefined. It is important that students are clear that delta is used to represent a small change in the variable. It is not itself a variable and certainly cannot be cancelled out. You may need to explain the meaning of infinitesimal, perhaps by referring to the infinite. Explain that at the limit, the gradient of the chord will actually be equal to the gradient of the tangent. As the chord approaches the gradient, δx approaches 0, so that we can treat it as zero. The idea of a limit can be a difficult one to grasp. It is, however, central to the understanding of the principles of differential calculus. Differentiation from first principles is not examined in this module and so it is enough for students to understand the basic concept. Instead we must consider the limit as δx tends to 0. This means that δx becomes infinitesimal without actually becoming 0.

Differentiation from first principles
Let’s see how this works for the gradient of the tangent to the curve y = x2 at the point A(3, 9). Move point B slowly towards point A and observe how the gradient of the cord AB gets closer to the gradient of the tangent at A. (When A and B coincide, the gradient δy/δx is shown as undefined.) Students should notice from this activity that the gradient of AB must be closer and closer to the gradient of the tangent at A the closer B is brought to A. But frustratingly, if B is in exactly the same place as A, the gradient is undefined.

Differentiation from first principles
If A is the point (3, 9) on the curve y = x2 and B is another point close to (3, 9) on the curve, we can write the coordinates of B as (3 + δx, (3 + δx)2). The gradient of chord AB is: B(3 + δx, (3 + δx)2) δy When differentiating from first principles, δx is sometimes replaced by the variable h. dy/dx is then given by the limit as h tends to 0. A(3, 9) δx

Differentiation from first principles
At the limit where δx → 0, 6 + δx → 6. We write this as: 6 So the gradient of the tangent to the curve y = x2 at the point (3, 9) is 6. Let’s apply this method to a general point on the curve y = x2. If we let the x-coordinate of a general point A on the curve y = x2 be x, then the y-coordinate will be x2. So, A is the point (x, x2). If B is another point close to A(x, x2) on the curve, we can write the coordinates of B as (x + δx, (x + δx)2).

Differentiation from first principles
The gradient of chord AB is: B(x + δx, (x + δx)2) δy A(x, x2) δx This result tells us that the gradient of any point on the curve y = x2 is 2 times the x-coordinate.

The gradient of y = x2 By moving the point B as close as possible to the point A for a variety of points on the curve, confirm that the gradient of the curve y = x2 is equal to 2x at any given point on the curve.

The gradient function So the gradient of the tangent to the curve y = x2 at the general point (x, y) is 2x. 2x is often called the gradient function or the derived function of y = x2. If the curve is written using function notation as y = f(x), then the derived function can be written as f ′(x). So, if: f(x) = x2 f ′(x) = 2x Then: This notation is useful if we want to find the gradient of f(x) at a particular point. For example, the gradient of f(x) = x2 at the point (5, 25) is: f ′(5) = 2 × 5 = 10

The gradient function of y = x2
Move the point along the curve y = x2 and watch as the corresponding gradient function y = f ’(x) is traced out. In particular draw attention to the fact that when the gradient of y = x2 is negative, the gradient function is below the x-axis; that when the gradient of y = x2 is zero the gradient function crosses the x-axis; and that when the gradient of y = x2 is positive the gradient function is above the x-axis. The gradient of y = x2 changes at a constant rate and so the gradient function of y = x2 is linear.

Differentiating y = x3 from first principles
For the curve y = x3, we can consider the general point A (x, x3) and the point B (x + δx, (x + δx)3) a small distance away from A. The gradient of chord AB is then: The steps of the expansion of (x + δx)3 are not shown here. Students could expand this as a separate exercise if required.

The gradient of y = x3 By moving the point B as close as possible to the point A for a variety of points on the curve confirm that the gradient of the curve y = x3 is equal to 3x2 at any given point on the curve.

Explain why the gradient of f(x) = x3 will always be positive.
The gradient of y = x3 So, the gradient of the tangent to the curve y = x3 at the general point (x, y) is 3x2. So if, f(x) = x3 f ′(x) = 3x2 Then: What is the gradient of the tangent to the curve f(x) = x3 at the point (–2, –8)? f ′(x) = 3x2 Establish that because the gradient function of f(x) = x3 is of the form f ′(x) = 3x2 the gradient will always be positive. (Whatever value x takes 3x2 will always be positive). f(x) = x3 is an example of an increasing function. f ′(–2) = 3(–2)2 = 12 Explain why the gradient of f(x) = x3 will always be positive.

The gradient function of y = x3
Move the point along the curve y = x3 and watch as the corresponding gradient function y = f ’(x) is traced out. Notice in particular that the gradient of y = x3 is always positive (or, at one point, 0) and so the gradient function of y = x3 is always above the x-axis. y = x3 is an example of an increasing function.

Using the notation We have shown that for y = x3
dy dx Using the notation We have shown that for y = x3 is usually written as then: So if y = x3 represents the derivative of y with respect to x. You may wish to point out that many other notations are employed to represent derivatives. However, the notations f ′(x) and dy/dy are the only ones used at this level. Stress that dy/dy is the rate of change in y with respect to x. of the tangent. Remember, is the gradient of a chord, while is the gradient

dy dx Using the notation This notation can be adapted for other variables so, for example: represents the derivative of s with respect to t. If s is distance and t is time then we can interpret this as the rate of change in distance with respect to time. In other words, the speed. Also, if we want to differentiate 2x4 with respect to x, for example, we can write: We could work this out by differentiating from first principles, but in practice this is unusual.

Differentiation of polynomials
Rates of change The gradient of the tangent at a point The gradient of the tangent as a limit Differentiation of polynomials Second order derivatives Tangents and normals Examination-style questions Contents 24 of 45 © Boardworks Ltd 2005

Differentiation If we continued the process of differentiating from first principles we would obtain the following results: y x x2 x3 x4 x5 x6 1 2x 3x2 4x3 5x4 6x5 What pattern do you notice? In general: Establish that to differentiate a power of x we multiply by the power and subtract 1 from the original power. These results are expected to be used without proof at this stage. and when xn is preceded by a constant multiplier a we have:

Differentiation For example, if y = 2x4
Suppose we have a function of the form y = c, where c is a constant number. This corresponds to a horizontal line through the point (0, c). y x The gradient of this line is always equal to 0. We could also differentiate y = c by thinking of it as y = cx0. Multiplying by the power of x in this case gives us a derivative that is equal to 0. It is particularly helpful for students to be aware that the gradient of a horizontal line is 0 when considering the gradient of a curve at a turning point. c  If y = c

Find the gradient of the curve y = 3x4 at the point (–2, 48).
Differentiation Find the gradient of the curve y = 3x4 at the point (–2, 48). Differentiating: At the point (–2, 48) x = –2 so: The gradient of the curve y = 3x4 at the point (–2, 48) is –96.

Differentiating polynomials
Suppose we want to differentiate a polynomial function. For example: Differentiate y = x4 + 3x2 – 5x + 2 with respect to x. We can differentiate each term in the function to give So: 4x3 + 6x – 5 In general, if y is made up of the sum or difference of any given number of functions, its derivative will be made up of the derivative of each of the functions. Note that the constant term has disappeared because the derivative of a constant is always equal to 0.

Differentiating polynomials
Find the point on y = 4x2 – x – 5 where the gradient is 15. when 8x – 1 = 15 8x = 16 x = 2 We now substitute this value into the equation y = 4x2 – x – 5 to find the value of y when x = 2 y = 4(2)2 – 2 – 5 = 9 The gradient of y = 4x2 – x – 5 is 15 at the point (2, 9).

Differentiating polynomials
In some cases, a function will have to be written as separate terms containing powers of x before differentiating. For example: Given that f(x) = (2x – 3)(x2 – 5) find f ′(x). Expanding: f(x) = 2x3 – 3x2 – 10x + 15 f ′(x) = 6x2 – 6x –10 Find the gradient of f(x) at the point (–3, –36). When x = –3 we have f ′(–3) = 6(–3)2 – 6(–3) –10 = – 10 = 62

Differentiating polynomials
We can now differentiate: = 6x3 + 2 = 2(3x3 + 1)

Differentiating axn for all rational n
This is true for all negative or fractional values of n. For example: Start by writing this as y = 2x–1 So:

Differentiating axn for all rational n
Start by writing this as So: Remember, in some cases, a function will have to be written as separate terms containing powers of x before differentiating.

Differentiating axn for all rational n

Second order derivatives
Rates of change The gradient of the tangent at a point The gradient of the tangent as a limit Differentiation of polynomials Second order derivatives Tangents and normals Examination-style questions Contents 35 of 45 © Boardworks Ltd 2005

Second order derivatives
Differentiating a function y = f(x) gives us the derivative or f ′(x) Differentiating the function a second time gives us the second order derivative. This can be written as or f ′′(x) The second order derivative gives us the rate of change of the gradient of a function. We can think of it as the gradient of the gradient.

Using second order derivatives
Move the point through the maximum point and explain that moving through a positive gradient to 0 to a negative gradient means that the gradient is decreasing. The second derivative (the gradient of the gradient) will therefore be negative (below the x-axis) for a maximum turning point. Similarly, by moving the point through the minimum point we can see that moving through a negative gradient to 0 to a positive gradient means that the gradient is increasing. The second derivative (the gradient of the gradient) will therefore be positive (above the x-axis) for a minimum turning point.

Contents Tangents and normals Rates of change
The gradient of the tangent at a point The gradient of the tangent as a limit Differentiation of polynomials Second order derivatives Tangents and normals Examination-style questions Contents 38 of 45 © Boardworks Ltd 2005

Tangents and normals Remember, the tangent to a curve at a point is a straight line that just touches the curve at that point. The normal to a curve at a point is a straight line that is perpendicular to the tangent at that point. You may wish to ask students how we can verify that the curve y = x2 – 5x + 8 passes through the point (3, 2). Substituting x = 3 into y = x2 – 5x + 8 gives y = 2 as required. We can use differentiation to find the equation of the tangent or the normal to a curve at a given point. For example: Find the equation of the tangent and the normal to the curve y = x2 – 5x + 8 at the point P(3, 2).

Tangents and normals y = x2 – 5x + 8 At the point P(3, 2) x = 3 so: 1
The gradient of the tangent at P is therefore 1. Using y – y1 = m(x – x1), give the equation of the tangent at the point P(3, 2): y – 2 = x – 3 y = x – 1

Tangents and normals The normal to the curve at the point P(3, 2) is perpendicular to the tangent at that point. The gradient of the tangent at P is 1 and so the gradient of the normal is –1. Using y – y1 = m(x – x1) give the equation of the tangent at the point P(3, 2):

Examination-style questions
Rates of change The gradient of the tangent at a point The gradient of the tangent as a limit Differentiation of polynomials Second order derivatives Tangents and normals Examination-style questions Contents 42 of 45 © Boardworks Ltd 2005

Examination-style question
The curve f(x) = x2 + 2x – 8 cuts the x-axes at the points A(–a, 0) and B(b, 0) where a and b are positive integers. Find the coordinates of points A and B. Find the gradient function of the curve. Find the equations of the normals to the curve at points A and B. Given that these normals intersect at the point C, find the coordinates of C. a) The equation of the curve can be written in factorized form as y = (x + 4)(x – 2) so the coordinates of A are (–4, 0) and the coordinates of B are (2, 0).

Examination-style question
b) f(x) = x2 + 2x – 8  f’(x) = 2x + 2 c) At (–4, 0) the gradient of the curve is –8 + 2 = –6  The gradient of the normal at (–4, 0) is . Using y – y1 = m(x – x1), the equation of the normal at (–4, 0) is: y – 0 = (x + 4) 1 6y = x + 4 At (2, 0) the gradient of the curve is = 6  The gradient of the normal at (2, 0) is – . Using y – y1 = m(x – x1), the equation of the normal at (2, 0) is: y – 0 = – (x – 2) 2 6y = 2 – x

Examination-style question
d) Equating and : 1 2 x + 4 = 2 – x 2x = –2 x = 1 When x = –1, y = . So the coordinates of C are (–1, ).

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