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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 change in y y2 y1 m change in x x2 x1 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. A tangent is a straight line, which touches, but does not cut, the curve. y A O Tangent to the curve at A. x We cannot calculate the gradient of a tangent directly, as we know only one point on the tangent and we require two points to calculate the gradient of a line. Using geometry to approximate to a gradient Look at this curve. y B1 B2 B3 Tangent to the curve at A. A O x 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). x changes from y changes from gradient AB1 2 to 3 4 to 9 94 =5 3 2 AB2 2 to 2.5 4 to 6.25 6.25 4 = 4.5 2.5 2 AB3 2 to 2.1 4 to 4.41 4.41 4 = 4.1 2.1 2 AB4 2 to 2.001 4 to 4.004001 AB5 2 to 2.00001 4 to 4.0000400001 B1 (3, 9) Chord B2 (2.5, 6.25) B3 (2.1, 4.41) B4 (2.001, 4.004001) Complete the table A (2, 4) The gradient of the chord AB1 is 94 5 3 2 4.004001 4 = 4.001 2.001 2 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. y = x2 y It looks right to me. 4 2 x 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 1.0012 1 1.0020011 1.0011 1.0011 0.002001 0.001 (1.001, 1.0012) = 2.001 I’d guess 2. (1, 1)) 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 8.00012 64 64.00160001 64 8.0001 8 8.00018 0.00160001 0.0001 (8.0001, 8.00012 ) = 16.0001 I’d guess 16. (8, 64) Let’s make a table of the results so far: I think I can sees a pattern but can I prove it? x-coordinate gradient 1 2 2 4 3 4 5 6 7 8 16 I need to consider what happens when I increase x by a general increment. I will call it h. I will call it ∆x. (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) y Here we have increased x by a very small amount h. In the early days of calculus h was referred to as an infinitesimal. y = x2 B(2 + h, (2 + h)2) Draw the chord AB. A(2, 4) 2 2 ( 2 h ) 4 4 4 h h 4 Gradient of AB 2 h2 2 h2 4h h h h( 4 h) h =4+h 2 O x If h ≠ 0 we can cancel the h’s. As h approaches zero, 4 + h approaches 4. So the gradient of the curve at the point (2, 4) is 4. Use a similar method to find the gradient of y = x2 at the points (i) (3, 9) (ii) (4, 16) We can now add to our table: It looks like the gradient is simply 2x. x-coordinate gradient 1 2 2 4 3 6 4 8 5 6 7 8 16 Let’s check this result. y y = x2 15 10 5 x −4 −3 −2 −1 1 2 3 4 Let’s check this result. y y = x2 15 10 Gradient at (3, 9) = 6 5 x −4 −3 −2 −1 1 2 3 4 Let’s check this result. y y = x2 15 10 Gradient at (2, 4) = 4 5 x −4 −3 −2 −1 1 2 3 4 Let’s check this result. y y = x2 15 10 Gradient at (1, 1) = 2 5 x −4 −3 −2 −1 1 2 3 4 Let’s check this result. y y = x2 15 10 Gradient at (0, 0) = 0 5 x −4 −3 −2 −1 1 2 3 4 Let’s check this result. y y = x2 15 10 Gradient at (–1, 1) = –2 5 x −4 −3 −2 −1 1 2 3 4 Let’s check this result. y y = x2 15 10 Gradient at (–2, 4) = –4 5 x −4 −3 −2 −1 1 2 3 4 Let’s check this result. y y = x2 15 10 Gradient at (–3, 9) = –6 5 x −4 −3 −2 −1 1 2 3 4 Another way of seeing what the gradient is at the point (2, 4) is to plot an accurate graph and ‘zoom in’. 9 y y = x2 8 7 6 ZOOM IN 5 4 3 2 1 x −4 −2 2 4 5 y 4.5 When we zoom in the curve starts to look like a straight line which makes it easy to estimate the gradient. 0.8 4 Gradient 0.8 4 0.2 0.2 3.5 x 3 1 1.5 2 2.5 3 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). 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”. Leibniz Thus the “increase in x” is written as ∆x, and the “increase in y” is written as ∆y.Gottfried 1646 - 1716 So when considering the gradient of a straight line, ∆x is the same as x2 – x1 and ∆y is the same as y2 – y1. y Note ∆x is the same as h used on the previous slide show. Q (x2, y2) ∆y y2 y1 P (x1, y1) O gradient = Dy Dx ∆x x2 x1 x Gradient of the curve y = x2 at the point P(x, y) Suppose that the point Q(x + ∆x, y + ∆y) y is very close to the point P on the curve. Q(x + ∆x, y + ∆y) The small change from P in the value of x is ∆x and the corresponding small change in the value of y is ∆y. ∆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: (y + Dy)- y = Dy (x + Dx)- x Dx P(x, y) y y = x2 ∆x y x O 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: (x + Dx)2 - x 2= x 2 + 2xDx +(Dx)2 - x 2= 2xDx +(Dx)2 = 2x + ∆x Dx Dx (x + Dx)- x x So Dy = 2x + ∆x Dx As ∆x gets smaller Dy approaches a limit and we start to refer to it in theoretical Dx 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 dy or d y . dx dx dy lim δy dx δx 0 δx For the curve y = x2, dy lim 2x δx dx δx 0 = 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: y Q(x + δx, y + δy) ( y δy) y δy ( x δx) 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)]. P(x, y) So the gradient of the chord PQ can be written as: f ( x δx) f ( x) ( x δx) x dy lim δy dx δx 0 δx O f ( x δx) f ( x) x 0 ( x δx) x lim lim f ( x δx) f ( x) x 0 y = f(x) x δx y δx y x DEFINITION OF THE DERIVATIVE OF A FUNCTION dy The symbol dx is called the derivative or the differential coefficient of y with respect to x. It is defined by dy lim f ( x δx) f ( x) dx δx 0 δx In words we say: δy “dee y by dee x is the limit of δx 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 dy lim f ( x h) f ( x) dx h 0 h If y = f(x) we can also use the notation: dy = f’ (x) dx In this case f’ is often called the derived function of f. This is also called f-prime. dy The procedure used to find dx from y is called differentiating y with respect to x. Example (1) Find dy for the function y = x3. dx Results so far f(x) f‘(x) x2 2x x3 3x2 In this case, f(x) = x3. dy dx f ( x h) f ( x) h h 0 lim 3 3 ( x h) x h h 0 lim 3 2 2 3 3 x 3x h 3xh h x h h 0 lim 2 2 3 3x h 3xh h h h 0 lim lim 3x 2 3xh h 2 h 0 = 3x2 y = x3 y 15 10 5 x −4 −3 −2 −1 1 2 3 4 y = x3 y 15 10 Gradient at (2, 8) = 12 5 x −4 −3 −2 −1 1 2 3 4 Example (2) Find dy for the function y = x4. dx Results so far f(x) f‘(x) x2 2x x3 3x2 x4 4x3 In this case, f(x) = x4. dy dx f ( x h) f ( x) h h 0 lim 4 4 ( x h) x h h 0 lim 4 3 2 2 3 4 4 x 4 x h 6 x h 4 xh h x h h 0 lim 3 2 2 3 4 4 x h 6 x h 4 xh h h h 0 lim lim 4x3 6x 2h 4xh2 h3 h 0 = 4x3 Example (3) Find dy for the function y = 1x . dx In this case, f(x) = 1x . dy lim f ( x h) f ( x) dx h 0 h h h 0 xh( x h) lim 1 h 0 x( x h) lim 12 x h 0 lim x ( x h) h 1 1 x h x lim x( x h) lim x( x h) h h h h 0 h 0 So we now have the following results: n xn dy dx –1 1 (or x1) x 12 (or x 2 ) x 2 x2 2x 3 x3 3x2 4 x4 4x3 We also know that if y = x = x1 then dy 1 (or 1x0 ) dx dy and if y = 1 = x0 then dx 0 (or 0 x 1) dy These results suggest that if y = xn then dx nx n 1 It can be proven that this statement is true for all values of n. Example (1) Find dy for the function y = 1. dx But of course we knew this already. y y = 1 can be thought of as y = x0. Using dy nx n 1 dx 1 y=1 dy 0 x 0 1 dx =0 O x Example (2) Find dy for the function y = x. dx But of course we knew this already. y y=x y = x can be thought of as y = x1. Using dy nx n 1 dx dy 1x11 dx = x0 =1 O x Example (3) Find dy for the function y = x2. dx Using dy nx n 1 dx dy 2 x 2 1 dx = 2x Example (4) Find dy for the function y = x3. dx Using dy nx n 1 dx dy 3 x 3 1 dx = 3x2 Example (5) Find dy for the function y = x4. dx Using dy nx n 1 dx dy 4 x 4 1 dx = 4x3 Example (6) Find dy for the function y = 1x . dx y = 1x can be written as y = x–1. Using dy nx n 1 dx dy 1x11 dx = –x–2 12 x Example (7) 1 Find dy for the function y = 2 . dx x y= 1 can be written as y = x–2. 2 x Using dy nx n 1 dx dy 2x 2 1 dx = –2x–3 23 x Example (8) Find dy for the function y = x . dx y = x can be written as y Using dy nx n 1 dx dy 1 x 12 1 dx 2 1 1x 2 2 1 2 x 1 x2 . Example (9) Find dy for the function y = 1 . x dx 1 1 y = x can be written as y x 2 . Using dy nx n 1 dx dy 1 x121 2 dx 3 1 x 2 2 1 2 x3 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 dy for the function y = 3x2. dx dy 3 2 x dx = 6x Example (2) Find dy for the function y = 8x3. dx dy 8 3 x 2 dx = 24x2 Example (3) Find dy for the function y = x4 + x3. dx dy 4x3 3x2 dx Example (4) Find dy for the function y = x2 + x + 1 dx dy 2 x 1 0 dx = 2x + 1 Example (5) Find the gradient of the curve y = 2x2 + 3x – 7 at the point (2, 7). 15 y dy 4 x 3 dx At the point (2, 7), dy 4 2 3 dx 10 So the gradient = 11 5 x −4 −3 −2 −1 1 −5 2 3 4 Example (6) 2 Find dy for the function y 2x3 6 x 4x 2 4x dx x 2 y 2x3 6 x 4x 2 4x x 2 4 x 2x 6 x 2 4x2 x x 3 2 x3 6 x 4 4 x 2x 3 1 6 x 2 4 4 x1 dy 6 x2 3x12 4 x2 dx 6x2 3 42 x x 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 y 50 – 6x – 36 The gradient is zero when f’ (x) = 0 That is when 6x2 – 6x – 36 = 0 This simplifies to x2 –x–6=0 x −6 −4 −2 2 In factor form this is (x – 3)(x + 2) = 0 So x = –2 or x = 3 −50 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) 4 6 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 Tangent Normal The equation of the tangent to a curve at a point (a, f(a)) is A y – f(a) = f’(a)(x – a) 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. O x 1 The gradient of the normal is as the product of the gradients of perpendicular lines is –1. f (a ) The equation of the normal to a curve at a point (a, f(a)) is y f (a) 1 ( x a) f (a) 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. 15 dy 2 dx = 6x + 6x y At x = 1, the gradient of C = 12 10 Equation of the tangent at A is y – 7 = 12(x – 1) which simplifies to 5 y = 12x – 5 x −3 −2 −1 1 −5 2 3 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. y (a) Find the coordinates of A and B. l C 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 = x(x2 + 3x – 4) = x(x + 4)(x – 1) So A is (–4, 0) and B is (1, 0) (b) dy 3x2 6x 4 dx dy At O, 4 dx So an equation for l is y = –4x A O B x (c) x3 + 3x2 – 4x = –4x x3 + 3x2 = 0 x2(x + 3) = 0 So l intersects C again when x = –3 and y = 12 So the coordinates of the point of intersection are (–3, 12) Example (3) For the curve C with equation y = x4 – 9x2 + 2, dy (a) find dx . 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 dy = 4x3 – 18x dx (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 1 So the gradient of the normal = 4 1 Equation of the normal to C at A is y 18 4 ( x 2) 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) dy = 4x – 6 dx 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 = 12 Equation of the normal to the curve at A is y 1 12 ( x 1) which simplifies to x – 2y + 1 = 0 x 1 2 4x2 – 12x + 10 = x + 1 (b) 2x2 – 6x + 5 = 4x2 – 13x + 9 = 0 (x – 1)(4x – 9) = 0 9 So x-coordinate of B = 4 13 y-coordinate of B = 8