Differential calculus

Differential calculus is concerned with the rate at which things change. For example, the speed of a car is the rate at which the distance it travels changes with time. First we shall review the gradient of a straight line graph, which represents a rate of change.

Gradient of a straight line graph The gradient of the line between two points (x1, y1) and (x2, y2) is where m is a fixed number called a constant. A gradient can be thought of as the rate of change of y with respect to x.

Gradient of a curve A curve does not have a constant gradient. Its direction is continuously changing, so its gradient will continuously change too. y = f(x) The gradient of a curve at any point on the curve is defined as being the gradient of the tangent to the curve at this point.

We cannot calculate the gradient of a tangent directly A tangent is a straight line, which touches, but does not cut, the curve. x y O A Tangent to the curve at A. , as we know only one point on the tangent and we require two points to calculate the gradient of a line. We cannot calculate the gradient of a tangent directly

Using geometry to approximate to a gradient Look at this curve. B1 B2 Tangent to the curve at A. B3 x O Look at the chords AB1, AB2, AB3, . . . For points B1, B2, B3, . . . that are closer and closer to A the sequence of chords AB1, AB2, AB3, . . . move closer to becoming the tangent at A. The gradients of the chords AB1, AB2, AB3, . . . move closer to becoming the gradient of the tangent at A.

A numerical approach to rates of change Here is how the idea can be applied to a real example. Look at the section of the graph of y = x2 for 2 > x > 3. We want to find the gradient of the curve at A(2, 4). Chord x changes from y gradient AB1 AB2 AB3 AB4 AB5 B1 (3, 9) 2 to 3 4 to 9 = 5 B2 (2.5, 6.25) 2 to 2.5 4 to 6.25 = 4.5 B3 (2.1, 4.41) B4 (2.001, 4.004001) 2 to 2.1 4 to 4.41 = 4.1 A (2, 4) Complete the table The gradient of the chord AB1 is 2 to 2.001 4 to 4.004001 = 4.001 2 to 2.00001 4 to 4.0000400001 4.00001

As the points B1, B2, B3, . . . get closer and closer to A the gradient is getting closer to 4. This suggests that the gradient of the curve y = x2 at the point (2, 4) is 4. x y 2 4 y = x2 It looks right to me.

The gradient of the chord is Example (1) Find the gradient of the chord joining the two points with x-coordinates 1 and 1.001 on the graph of y = x2. Make a guess about the gradient of the tangent at the point x = 1. The gradient of the chord is (1.001, ) 1.0012 = 2.001 I’d guess 2. (1, ) 1)

The gradient of the chord is Example (2) Find the gradient of the chord joining the two points with x-coordinates 8 and 8.0001 on the graph of y = x2. Make a guess about the gradient of the tangent at the point x = 8. The gradient of the chord is (8.0001, ) 8.00012 = 16.0001 I’d guess 16. (8, ) 64

Let’s make a table of the results so far: x-coordinate gradient 1 2 4 3 5 6 7 8 16 I think I can sees a pattern but can I prove it?

I will call it ∆x. I need to consider what happens when I increase x by a general increment. I will call it h. (2 + h, (2 + h)2) (2, 4) h

Let y = x2 and let A be the point (2, 4) Let B be the point (2 + h, (2 + h)2) x y A(2, 4) y = x2 O Here we have increased x by a very small amount h. In the early days of calculus h was referred to as an infinitesimal. B(2 + h, (2 + h)2) Draw the chord AB. Gradient of AB If h ≠ 0 we can cancel the h’s. = 4 + h Use a similar method to find the gradient of y = x2 at the points (i) (3, 9) (ii) (4, 16) As h approaches zero, 4 + h approaches 4. So the gradient of the curve at the point (2, 4) is 4.

We can now add to our table: x-coordinate gradient 1 2 4 3 5 6 7 8 16 It looks like the gradient is simply 2x. 6 8

Let’s check this result. y = x2

Let’s check this result. y = x2 Gradient at (3, 9) = 6

Let’s check this result. y = x2 Gradient at (2, 4) = 4

Let’s check this result. y = x2 Gradient at (1, 1) = 2

Let’s check this result. y = x2 Gradient at (0, 0) = 0

Let’s check this result. y = x2 Gradient at (–1, 1) = –2

Let’s check this result. y = x2 Gradient at (–2, 4) = –4

Let’s check this result. y = x2 Gradient at (–3, 9) = –6

Zooming in

Another way of seeing what the gradient is at the point (2, 4) is to plot an accurate graph and ‘zoom in’. y = x2 ZOOM IN

When we zoom in the curve starts to look like a straight line which makes it easy to estimate the gradient. 0.8 0.2

∆x is the same as h used on the previous slide show. Using a similar approach to that in the previous slide show it is possible to find the gradient at any point (x, y) on the curve y = f(x). Gottfried Leibniz 1646 - 1716 At this point it is useful to introduce some “new” notation due to Leibniz. The Greek letter ∆(delta) is used as an abbreviation for “the increase in”. Thus the “increase in x” is written as ∆x, and the “increase in y” is written as ∆y. So when considering the gradient of a straight line, ∆x is the same as x2 – x1 and ∆y is the same as y2 – y1. P (x1, y1) Q (x2, y2) x y O Note ∆x is the same as h used on the previous slide show. gradient ∆y ∆x

y = x2 Gradient of the curve y = x2 at the point P(x, y) Suppose that the point Q(x + ∆x, y + ∆y) is very close to the point P on the curve. x O y P(x, y) Q(x + ∆x, y + ∆y) y = x2 The small change from P in the value of x is ∆x and the corresponding small change in the value of y is ∆y. x y ∆x ∆y It is important to understand that ∆x is read as “delta x” and is a single symbol. The gradient of the chord PQ is: The coordinates of P can also be written as (x, x2) and the coordinates of Q as [(x + ∆x), (x + ∆x)2]. So the gradient of the chord PQ can be written as: = 2x + ∆x

So = 2x + ∆x As ∆x gets smaller approaches a limit and we start to refer to it in theoretical terms. This limit is the gradient of the tangent at P which is the gradient of the curve at P. It is called the rate of change of y with respect to x at the point P. This is denoted by or . For the curve y = x2, = 2x This is the result we obtained previously.

Gradient of the curve y = f(x) at the point P(x, y) For any function y = f(x) the gradient of the chord PQ is: x O y P(x, y) Q(x + δx, y + δy) y = f(x) δx δy The coordinates of P can also be written as (x, f(x)) and the coordinates of Q as [(x + δx), f(x + δx)]. So the gradient of the chord PQ can be written as:

DEFINITION OF THE DERIVATIVE OF A FUNCTION The symbol is called the derivative or the differential coefficient of y with respect to x. It is defined by In words we say: “dee y by dee x is the limit of as δx tends to zero” “tends to” is another way of saying “approaches”

DEFINITION OF THE DERIVATIVE OF A FUNCTION Sometimes we write h instead of ∆x and so the derivative of f(x) can be written as

If y = f(x) we can also use the notation: In this case f’ is often called the derived function of f. This is also called f-prime. The procedure used to find from y is called differentiating y with respect to x.

Results so far f(x) f ‘ (x) x2 2x x3 3x2 Example (1) Find for the function y = x3. Results so far f(x) f ‘ (x) x2 2x x3 3x2 In this case, f(x) = x3. = 3x2

y = x3

y = x3 Gradient at (2, 8) = 12

Results so far f(x) f ‘(x) x2 2x x3 3x2 x4 4x3 Example (2) Find for the function y = x4. Results so far f(x) f ‘(x) x2 2x x3 3x2 x4 4x3 In this case, f(x) = x4. = 4x3

Find for the function y = . Example (3) In this case, f(x) = .

So we now have the following results: xn –1 2 x2 2x 3 x3 3x2 4 x4 4x3 We also know that if y = x = x1 then and if y = 1 = x0 then These results suggest that if y = xn then It can be proven that this statement is true for all values of n.

Example (1) But of course we knew this already. y = 1 1 y x O Find for the function y = 1. y = 1 can be thought of as y = x0. Using = 0

Example (2) But of course we knew this already. y = x y x O Find for the function y = x. y = x can be thought of as y = x1. Using = x0 = 1

Example (3) Find for the function y = x2. Using = 2x

Example (4) Find for the function y = x3. Using = 3x2

Example (5) Find for the function y = x4. Using = 4x3

Find for the function y = . Example (6) y = can be written as y = x–1. Using = –x–2

Example (7) Find for the function y = . y = can be written as y = x–2. Using = –2x–3

Example (8) Find for the function y = . y = can be written as . Using

Example (9) Find for the function y = . y = can be written as . Using

Most of the functions we meet are not powers of the single variable x but consist of a number of terms such as y = 2x2 + 3x + 5. The following rules can be proven: If you multiply a function by a constant, you multiply its derivative by the same constant. If f(x) = ag(x), then f’ (x) = ag’ (x). This is because multiplying a function by a constant a has the effect of stretching the function in the y direction by a scale factor a which increases the gradient by the factor a. If you add two functions, then the derivative of the sum is the sum of the derivatives. If f(x) = g(x) + h(x), then f’ (x) = g’ (x) + h’ (x). However if you multiply two functions, then the derivative of the product is NOT the product of the derivatives.

Example (1) Find for the function y = 3x2. = 6x

Example (2) Find for the function y = 8x3. = 24x2

Example (3) Find for the function y = x4 + x3.

Example (4) Find for the function y = x2 + x + 1 = 2x + 1

Example (5) Find the gradient of the curve y = 2x2 + 3x – 7 at the point (2, 7). At the point (2, 7), So the gradient = 11

Example (6) Find for the function

Example (7) Find the coordinates of the points on the graph of y = 2x3 – 3x2 – 36x + 10 at which the gradient is zero. Let f(x) = 2x3 – 3x2 – 36x + 10 Then f’ (x) = 6x2 – 6x – 36 The gradient is zero when f’ (x) = 0 That is when 6x2 – 6x – 36 = 0 This simplifies to x2 – x – 6 = 0 In factor form this is (x – 3)(x + 2) = 0 So x = –2 or x = 3 Substituting these values into y = 2x3 – 3x2 – 36x + 10 to find the y-coordinates gives y = 54 and y = –71 The coordinates of the required points are therefore (–2, 54) and (3, –71)

The tangent at the point A(a, f(a)) has gradient f’ (a). We can use the formula for the equation of a straight line, y – y1 = m(x – x1) to obtain the equation of the tangent at (a, f(a)). y O A Tangent Normal The equation of the tangent to a curve at a point (a, f(a)) is y – f(a) = f’(a)(x – a) x The normal to the curve at the point A is defined as being the straight line through A which is perpendicular to the tangent at A. The gradient of the normal is as the product of the gradients of perpendicular lines is –1. The equation of the normal to a curve at a point (a, f(a)) is

Example (1) The curve C has equation y = 2x3 + 3x2 + 2. The point A with coordinates (1, 7) lies on C. Find the equation of the tangent to C at A, giving your answer in the form y = mx + c, where m and c are constants. = 6x2 + 6x At x = 1, the gradient of C = 12 Equation of the tangent at A is y – 7 = 12(x – 1) which simplifies to y = 12x – 5

(a) Find the coordinates of A and B. Example (2) The diagram shows the curve C with the equation y = x3 + 3x2 – 4x and the straight line l. The curve C crosses the x-axis at the origin, O, and at the points A and B. (a) Find the coordinates of A and B. A B x y O C l The line l is the tangent to C at O. (b) Find an equation for l. (c) Find the coordinates of the point where l intersects C again. (a) y = x3 + 3x2 – 4x (c) x3 + 3x2 – 4x = –4x = x(x2 + 3x – 4) x3 + 3x2 = 0 = x(x + 4)(x – 1) x2(x + 3) = 0 So A is (–4, 0) and B is (1, 0) So l intersects C again when x = –3 and y = 12 (b) So the coordinates of the point of intersection are (–3, 12) At O, So an equation for l is y = –4x

Example (3) For the curve C with equation y = x4 – 9x2 + 2, (a) find The point A, on the curve C, has x-coordinate 2. (b) Find an equation for the normal to C at A, giving your answer in the form ax + by + c = 0, where a, b and c are integers. (a) y = x4 – 9x2 + 2 = 4x3 – 18x (b) When x = 2, y = –18. So the coordinates of A are (2, –18). When x = 2, gradient of C = 4 • 8 – 18 • 2 = –4 So the gradient of the normal = Equation of the normal to C at A is which simplifies to x – 4y – 74 = 0

Example (4) A curve C has equation y = 2x2 – 6x + 5. The point A, on the curve C, has x-coordinate 1. (a) Find an equation for the normal to C at A, giving your answer in the form ax + by + c = 0, where a, b and c are integers. The normal at A cuts C again at the point B. (b) Find the coordinates of the point B. (a) = 4x – 6 When x = 1, y = 2 – 6 + 5 = 1 Gradient of the tangent to the curve at A = 4 – 6 = –2 Gradient of the normal to the curve at A = Equation of the normal to the curve at A is which simplifies to x – 2y + 1 = 0 (b) 2x2 – 6x + 5 = (x – 1)(4x – 9) = 0 So x-coordinate of B = 4x2 – 12x + 10 = x + 1 y-coordinate of B = 4x2 – 13x + 9 = 0