1 / 93

DIFFERENTIATION

Outcome 3. Using differentiation (Application). DIFFERENTIATION. Finding the gradient for a polynomial. Increasing / Decreasing functions. Max / Min and inflexion Points. Differentiating Brackets ( Type 1 ). Curve Sketching. Differentiating Harder Terms (Type 2).

scassandra
Download Presentation

DIFFERENTIATION

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Outcome 3 Using differentiation (Application) DIFFERENTIATION Finding the gradient for a polynomial Increasing / Decreasing functions Max / Min and inflexion Points Differentiating Brackets ( Type 1 ) Curve Sketching Differentiating Harder Terms (Type 2) Max & Min Values on closed Intervals Differentiating with Leibniz Notation Optimization Equation of a Tangent Line ( Type 3 ) Mind Map of Chapter

  2. Gradients & Curves Outcome 3 Higher On a straight line the gradient remains constant, however with curves the gradient changes continually, and the gradient at any point is in fact the same as the gradient of the tangent at that point. The sides of the half-pipe are very steep(S) but it is not very steep near the base(B). S Demo B

  3. Gradients & Curves Outcome 3 Higher Gradient of tangent = gradient of curve at A A B Gradient of tangent = gradient of curve at B

  4. To find the gradient at any point on a curve we need to modify the gradient formula Gradients & Curves Outcome 3 Higher For the function y = f(x) we do this by taking the point (x, f(x)) and another “very close point” ((x+h), f(x+h)). Then we find the gradient between the two. ((x+h), f(x+h)) Approx gradient (x, f(x)) True gradient

  5. Gradients & Curves Outcome 3 Higher The gradient is not exactly the same but is quite close to the actual value We can improve the approximation by making the value of h smaller This means the two points are closer together. ((x+h), f(x+h)) Approx gradient (x, f(x)) True gradient

  6. Gradients & Curves Outcome 3 Higher We can improve upon this approximation by making the value of h even smaller. So the points are even closer together. ((x+h), f(x+h)) Approx gradient True gradient (x, f(x))

  7. Outcome 3 Higher We have seen that on curves the gradient changes continually and is dependant on the position on the curve. ie the x-value of the given point. Derivative Finding the GRADIENT Differentiating The process of finding the gradient is called Finding the rate of change DIFFERENTIATING or FINDING THE DERIVATIVE (Gradient)

  8. Derivative Outcome 3 Higher If the formula/equation of the curve is given by f(x) Then the derivative is called f '(x) - “f dash x” There is a simple way of finding f '(x) from f(x). f(x) f '(x) 2x2 4x 4x2 8x Have guessed the rule yet ! 5x10 50x9 6x7 42x6 x3 3x2 x5 5x4 x99 99x98

  9. Derivative Rule for Differentiating Outcome 3 Higher It can be given by this simple flow diagram ... multiply by the power reduce the power by 1 If f(x) = axn n n -1 ax then f '(x) = NB: the following terms & expressions mean the same GRADIENT, DERIVATIVE, RATE OF CHANGE, f '(x)

  10. Derivative Rule for Differentiating Outcome 3 Higher To be able to differentiate it is VERY IMPORTANT that you are comfortable using indices rules

  11. Outcome 3 Higher (I) f(x) = ax (Straight line function) Special Points Index Laws x0 = 1 If f(x) = ax = ax1 then f '(x) = 1 X ax0 = a X 1 = a So if g(x) = 12x then g '(x) = 12 Also using y = mx + c The line y = 12x has gradient 12, and derivative = gradient !!

  12. Outcome 3 Higher Special Points (II) f(x) = a, (Horizontal Line) Index Laws x0 = 1 If f(x) = a = a X 1 = ax0 then f '(x) = 0 X ax-1 = 0 So if g(x) = -2 then g '(x) = 0 Also using formula y = c , (see outcome 1 !) The line y = -2 is horizontal so has gradient 0 !

  13. Differentiation techniques Differentiation = Gradient Differentiation = Rate of change Name :

  14. Calculus Revision Differentiate

  15. Calculus Revision Differentiate

  16. Calculus Revision Differentiate

  17. Derivative Outcome 3 Higher Example 1 A curve has equation f(x) = 3x4 Find the formula for its gradient and find the gradient when x = 2 Its gradient is f '(x) = 12x3 f '(2) = 12 X 23 = 12 X 8 = 96 Example 2 A curve has equation f(x) = 3x2 Find the formula for its gradient and find the gradient when x = -4 Its gradient is f '(x) = 6x At the point where x = -4 the gradient is f '(-4) = 6 X -4 = -24

  18. Derivative Outcome 3 Higher Example 3 If g(x) = 5x4 - 4x5 then find g '(2) . g '(x) = 20x3 - 20x4 g '(2) = 20 X 23 - 20 X 24 = 160 - 320 = -160

  19. Derivative Outcome 3 Higher Example 4 h(x) = 5x2 - 3x + 19 so h '(x) = 10x - 3 and h '(-4) = 10 X (-4) - 3 = -40 - 3 = -43 Example 5 k(x) = 5x4 - 2x3 + 19x - 8, find k '(10) . k '(x) = 20x3 - 6x2 + 19 So k '(10) = 20 X 1000 - 6 X 100 + 19 = 19419

  20. Derivative Outcome 3 Higher Example 6 : Find the points on the curve f(x) = x3 - 3x2 + 2x + 7 where the gradient is 2. NB: gradient = derivative = f '(x) Now using original formula We need f '(x) = 2 ie 3x2 - 6x + 2 = 2 f(0) = 7 or 3x2 - 6x = 0 ie 3x(x - 2) = 0 f(2) = 8 -12 + 4 + 7 ie 3x = 0 or x - 2 = 0 = 7 Points are (0,7) & (2,7) so x = 0 or x = 2

  21. Calculus Revision Differentiate

  22. Calculus Revision Differentiate Straight line form Differentiate

  23. Calculus Revision Differentiate Straight line form Differentiate

  24. Calculus Revision Differentiate Straight line form Chain Rule Simplify

  25. Calculus Revision Differentiate Straight line form Differentiate

  26. Calculus Revision Differentiate Straight line form Differentiate

  27. Calculus Revision Differentiate Straight line form Differentiate

  28. Outcome 3 Higher Brackets Basic Rule: Break brackets before you differentiate ! Example h(x) = 2x(x + 3)(x -3) = 2x(x2 - 9) = 2x3 - 18x So h'(x) = 6x2 -18

  29. Calculus Revision Differentiate Multiply out Differentiate

  30. Calculus Revision Differentiate multiply out differentiate

  31. Calculus Revision Differentiate Straight line form multiply out Differentiate

  32. Calculus Revision Differentiate multiply out Differentiate

  33. Calculus Revision Differentiate multiply out Simplify Straight line form Differentiate

  34. Calculus Revision Differentiate Multiply out Straight line form Differentiate

  35. Outcome 3 Higher Fractions Reversing the above we get the following “rule” ! This can be used as follows …..

  36. Fractions Outcome 3 Higher Example f(x) = 3x3 - x + 2 x2 = 3x3 - x + 2 x2 x2x2 = 3x - x-1 + 2x-2 f '(x) = 3 + x-2 - 4x-3 = 3 + 1 - 4 x2 x3

  37. Calculus Revision Differentiate Split up Straight line form Differentiate

  38. Outcome 3 Higher Leibniz Notation is an alternative way of expressing derivatives to f'(x) , g'(x) , etc. Leibniz Notation If y is expressed in terms of x then the derivative is written as dy/dx . eg y = 3x2 - 7x so dy/dx = 6x - 7 . Example 19 Q = 9R2 - 15 R3 Find dQ/dR = 18R + 45 R4 NB: Q = 9R2 - 15R-3 So dQ/dR = 18R + 45R-4

  39. Leibniz Notation Outcome 3 Higher Example 20 A curve has equation y = 5x3 - 4x2 + 7 . Find the gradient where x = -2 ( differentiate ! ) gradient = dy/dx = 15x2 - 8x if x = -2 then gradient = 15 X (-2)2 - 8 X (-2) = 60 - (-16) = 76

  40. Real Life Example Physics Outcome 3 Higher Newton’s 2ndLaw of Motion s = ut + 1/2at2 where s = distance & t = time. Finding ds/dt means “diff in dist”  “diff in time” ie speed or velocity so ds/dt = u + at but ds/dt = v so we get v = u + at and this is Newton’s 1st Law of Motion

  41. Equation of Tangents y = mx +c Outcome 3 Higher y = f(x) A(a,b) tangent NB: at A(a, b) gradient of line = gradient of curve gradient of line = m (from y = mx + c ) gradient of curve at (a, b) = f (a) it follows that m = f (a)

  42. Straight line so we need a point plus the gradient then we can use the formula y - b = m(x - a) . Equation of Tangents Outcome 3 Higher Example 21 Find the equation of the tangent line to the curve y = x3 - 2x + 1 at the point where x = -1. Point: if x = -1 then y = (-1)3 - (2 X -1) + 1 = -1 - (-2) + 1 = 2 point is (-1,2) Gradient:dy/dx = 3x2 - 2 when x = -1 dy/dx = 3 X (-1)2 - 2 m = 1 = 3 - 2 = 1

  43. Equation of Tangents Outcome 3 Higher Now using y - b = m(x - a) point is (-1,2) m = 1 we get y - 2 = 1( x + 1) or y - 2 = x + 1 or y = x + 3

  44. Equation of Tangents Outcome 3 Higher Example 22 Find the equation of the tangent to the curve y = 4 x2 at the point where x = -2. (x  0) Also find where the tangent cuts the X-axis and Y-axis. Point: when x = -2 then y = 4 (-2)2 = 4/4 = 1 point is (-2, 1) Gradient: y = 4x-2 so dy/dx = -8x-3 = -8 x3 when x = -2 then dy/dx = -8 (-2)3 = -8/-8 = 1 m = 1

  45. Equation of Tangents Outcome 3 Higher Now using y - b = m(x - a) we get y - 1 = 1( x + 2) or y - 1 = x + 2 or y = x + 3 Axes Tangent cuts Y-axis when x = 0 so y = 0 + 3 = 3 at point (0, 3) Tangent cuts X-axis when y = 0 so 0 = x + 3 or x = -3 at point (-3, 0)

  46. Equation of Tangents Outcome 3 Higher Example 23 - (other way round) Find the point on the curve y = x2 - 6x + 5 where the gradient of the tangent is 14. gradient of tangent = gradient of curve dy/dx = 2x - 6 so 2x - 6 = 14 2x = 20 x = 10 Put x = 10 into y = x2 - 6x + 5 Point is (10,45) Giving y = 100 - 60 + 5 = 45

  47. Outcome 3 Higher Increasing & Decreasing Functions and Stationary Points Consider the following graph of y = f(x) ….. y = f(x) + 0 0 + - + + a b c d e f - X + 0

  48. Increasing & Decreasing Functions and Stationary Points Outcome 3 Higher In the graph of y = f(x) The function is increasing if the gradient is positive i.e. f  (x) > 0 when x < b or d < x < f or x > f . The function is decreasing if the gradient is negative and f  (x) < 0 when b < x < d . The function is stationary if the gradient is zero and f  (x) = 0 when x = b or x = d or x = f . These are called STATIONARY POINTS. At x = a, x = c and x = e the curve is simply crossing the X-axis.

  49. Increasing & Decreasing Functions and Stationary Points Outcome 3 Higher Example 24 For the function f(x) = 4x2 - 24x + 19 determine the intervals when the function is decreasing and increasing. f  (x) = 8x - 24 so 8x - 24 < 0 f(x) decreasing when f  (x) < 0 8x < 24 Check: f  (2) = 8 X 2 – 24 = -8 x < 3 f(x) increasing when f  (x) > 0 so 8x - 24 > 0 8x > 24 Check: f  (4) = 8 X 4 – 24 = 8 x > 3

  50. Increasing & Decreasing Functions and Stationary Points Outcome 3 Higher Example 25 For the curve y = 6x – 5/x2 Determine if it is increasing or decreasing when x = 10. y = 6x - 5 x2 = 6x - 5x-2 so dy/dx = 6 + 10x-3 = 6 + 10 x3 when x = 10 dy/dx = 6 + 10/1000 = 6.01 Since dy/dx > 0 then the function is increasing.

More Related