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Chapter 4

Differentiation. Chapter 4. The Derivative and the Tangent Line Problem. Section 4.1. The Tangent Line Problem.

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Chapter 4

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  1. Differentiation Chapter 4

  2. The Derivative and the Tangent Line Problem Section 4.1

  3. The Tangent Line Problem To find the tangent line of a curve at a point P, you need to find the slope of the tangent line at point P. You can approximate this slope using the secant line through the point of tangency and a second point on the curve where (c, f(c)) is the point of tangency and (c + ∆x, f (c + ∆x)) is a second point on the graph of f. Slope formula: m = (y2 –y1)/(x2-x1) = ∆y/ ∆x Slope formula: msec=( f (c + ∆x) – f(c))/(c + ∆x – c) msec = f (c + ∆x) – f(c))/ ∆x

  4. Definition of Tangent Line with Slope m If f is defined on an open interval containing c, and if the limit lim ∆y/ ∆x = lim(f(c + ∆x) – f(c))/ ∆x = m ∆x0 ∆x0 Exists, then the line passing through (c,f(c)) with slope m is the tangent line to the graph of f at the point (c,f(c)) The slope of the tangent line is also called the slope of the graph of f at x =c

  5. Example Find the slope of the graph of f(x) = 2x -3 at the point (2,1) Use the definition of Tangent Line with Slope m Then c = 2 lim ∆y/ ∆x = lim(f(c + ∆x) – f(c))/ ∆x = m ∆x0 ∆x0 lim(f(2 + ∆x) – f(2))/ ∆x = {[2(2 + ∆x) -3]-[2(2) -3]}/ ∆x ∆x0 = lim(4 + 2 ∆x - 3 – 4 + 3)/ ∆x = 2 ∆x / ∆x = 2 ∆x0 m = 2

  6. Example Find the slope of the tangent line to the graph of • g(x) = 3x – 6 at the point (2,0) • g(x) = -4x + 1 at the point (-1,5) • f(x) = x2 + 1 at the point (-1,2) • f(x) = x2 + 1 at the point (0,1)

  7. Definition of the Derivative of a Function The derivative of f at x is given by f′(x)= lim(f(x + ∆x) – f(x))/ ∆x ∆x0 Provided the limit exists. For all x for which this limit exists, f′ is a function of x. f′ is a new function of x which gives the slope of the tangent line to the graph of f at the point (x,f(x))

  8. Differentiation Differentiation the process of finding the derivative of a function. A function is differentiable at x if its derivative exists at x and is differentiable on an open interval (a,b) if it is differentiable at every point in the interval

  9. Notations for Derivatives f′(x) dy/dx y′ d/dx[f(x)] Dx[y]

  10. Derivatives dy/dx = lim ∆y/ ∆x ∆x 0 = lim(f(x + ∆x) – f(x))/ ∆x ∆x0 = f′(x)

  11. Example Find the derivative by the limit process • g(x) = 2x2 - 9 • f(x) = x3 + 2x • f(x) = √x • y = 2/t

  12. Differentiability and Continuity The following alternative limit form of the derivative is useful in investigating the relationship between differentiability and continuity. The derivative of f at c is f′(c) = lim(f(x) –f(c))/(x – c) x c provided the limit exists. The existence of the limit in this alternative form requires that the one-sided limits exist and are equal

  13. Differentiability and Continuity – con’d lim(f(x) –f(c))/(x – c) x c- and lim(f(x) –f(c))/(x – c) x c+ These one-sided limits are called the derivatives from the left and from the right. It follows that f is differentiable on the closed interval [a,b] if it is differentiable on (a,b) and if the derivative from the right at a and the derivative from the left at b both exist.

  14. Differentiability and Continuity – con’d Differentiability implies continuity. • If a function is differentiable at a point, then the function is continuous at that point Continuity does not imply differentiability. • A function may be continuous at a point, but not differentiable at that point • A function that is not continuous at a point is not differentiable at that point

  15. Example Determine differentiability and/or continuity • g(x) = |x – 2| at x = 2 • f(x) = x1/3 at x = 0 • f(x) = [| x |] at x = 0 the greatest integer

  16. Basic Differentiation Rules and Rates of Change Section 4.2

  17. The Constant Rule The derivative of a constant function is 0. That is, if c is a real number, then d/dx[c] = 0

  18. Example Find the derivative of these functions • y = 7 • f(x) = 0 • s(t) = -3 • y = kπ2, where k is constant

  19. The Power Rule If n is a rational number, then the function given by f(x) = xnis differentiable and d/dx[xn] = nxn-1 For f to be differentiable at x=0, n must be a number such that xn-1 is defined on an interval containing 0.

  20. The Power Rule when n = 1 If n =1, then the function given by f(x) = x is differentiable and d/dx[x] = 1

  21. Example Find the derivatives of these functions • f(x) = x3 • f(x) =3√x • s(t) = 1/x2 • y = x4

  22. Finding the Equation of a Tangent Line • First, evaluate the function at x • Second, find the slope using the derivative and evaluate at x • Third, use point slope form to find the tangent line

  23. Example Find the equation of a tangent line for each of these functions • f(x) = x2 when x = -2 • f(x) =3√x when x = 8 • s(t) = 1/x2 when x = -3 • y = x4 when x = ½

  24. The Constant Multiple Rule If f is a differentiable function and c is a real number, then cf is also differentiable and d/dx[cf(x)] = cf′(x)

  25. Example Find the derivatives of these functions • f(x) =2/x • f(x) =4t2/5 • s(t) =2√x • y = -3x/2 • 1/2( 3√x2 ) • y = 7/(3x)-2

  26. The Sum and Difference Rules The sum (or difference) of two differentiable functions f and g is itself differentiable. Moreover, the derivative of f + g (or f - g) is the sum (or difference) of the derivatives of f and g. d/dx[f(x) + g(x)] = f′ (x)+ g′ (x) Sum Rule d/dx[f(x) - g(x)] = f′ (x) - g′ (x) DifferenceRule

  27. Example Find the derivatives of these functions • f(x) =x3 – 4x + 5 • f(x) =3x2 - 4 • s(t) =-2x3 + 9x2 - 2 • y = -x4/2 + 3x3 – 2x

  28. Rates of Change The derivative can be used to determine the rate of change of one variable with respect to another. A common use of rate of change is to describe the motion of an object moving in a straight line. The function sthat gives the position (relative to the origin) of an object as a function of timetis called a position function. Then rate = distance/time or

  29. Rates of Change – cont’d Average velocity is Change in distance/change in time = ∆s/ ∆t which is the derivative

  30. Example – Finding Average Velocity If a billiard ball is dropped from a height of 100 feet, its height s at time t is given by the position function s = -16t2 +100 Find the average velocity over each time interval • [1,2] • [1,1.5] • [1,1.1] What do the negative answers indicate?

  31. Example – Using Derivative to Find Velocity At time t = 0, a diver jumps from a platform diving board that is 32 feet above the water. The position of the diver is given by s(t) = -16t2 + 16t + 32 where s is measured in feet and t is measured in seconds • When does the diver hit the water? • What is the diver’s velocity at impact?

  32. The Product and Quotient Rules and Higher-Order Derivatives Section 4.3

  33. The Product Rule The product of two differentiable functions f and g is itself differentiable. Moreover, the derivative of fg is the first function times the derivative of the second, plus the second function times the derivative of the first. d/dx[f(x)g(x)] = f(x)g′(x) + g(x)f′(x)

  34. Example Find the derivative of each function • h(x) =(3x-2x2)(5 + 4x) • y = (3x -1)(2x + 5) • g(x) = (x3 +2x2 -5)(x4 – 4x2) • f(x) = x2(x+1)(2x-3) • y= (2x3 + 5x)(7)

  35. The Quotient Rule The product of two differentiable functions f and g is itself differentiable at all values of x for which g(x) ≠ 0. Moreover, the derivative of f/g is given by the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator. d/dx[f(x)/g(x)] = g(x)f′(x) - f(x)g′(x)] ÷ [g(x)]2

  36. Example Find the derivative of each function • h(x) = (5x -2) ÷ (x2 + 1) • y = (x2 + 3x) ÷ 6 • g(x) = (12x3 - 3x2) ÷ (6x2) • f(x) = (x+1) ÷ (x-1) • y = (x-3) ÷ (x2-4)

  37. Example – Find an Equation of the Tangent Line f(x) = (3 – 1/x) ÷ (x+5) at (-1,1) • Find f′(x) • Evaluate f′(x) • Use point-slope form to find equation

  38. Higher-Order Derivatives s(t) Position function v(t) = s′(t) Velocity Function a(t) = v′(t) = s″(t) Acceleration Function

  39. Example – Finding Acceleration A car’s velocity starting from rest is v(t) = 100t ÷ (2t+15) where v is measured in feet per second. Find the acceleration at each of the following times. • 5 seconds • 10 seconds • 20 seconds

  40. The Chain Rule Section 4.4

  41. The Chain Rule The chain rule is a very powerful differentiation rule that deals with composite functions

  42. The Chain Rule Without the Chain Rule With the Chain Rule • f(x) =x2 + 1 • g(x)=3x + 2 • y =x + 2 • y = (x2 +1)½ • f(x) = 3x +2)5 • g(x) = (x + 2)1/3

  43. The Chain Rule If y =f(u) is a differentiable function of u and u =g(x) is a differentiable function of x, then y =f(g(x)) is a differentiable function of x and dy/dx = dy/du · du/dx or d/dx[f(g(x))] = f′(g(x))g′(x)

  44. Example – Applying the Chain Rule • Write the function as a composite of functions • Apply Chain Rule y = f(g(x)) u = g(x) y =f(u) y= 1/(x+1) u = x + 1 1/u y = (3x2 – x +1)½ u = 3x2 – x +1 u½

  45. Example – Applying the Chain Rule • Find dy/dx for y = (x2 +1)3 y = f(g(x)) u = g(x) y =f(u) y = (x2 +1)3 u = x2 +1 u3 dy/dx = dy/du · du/dx 3(x2 + 1)2(2x) = 6x(x2 +1)2 dy/du du/dx

  46. The General Power Rule If y = [u(x)]n where u is a differentiable function of x and n is a rational number, then dy/dx = n[u(x)]n-1du/dxor d/dx[un] = nun-1u′

  47. Example Differentiate the functions • f(x) = (3x -2x2)3 • f(x) = [(x2 -1)2 ]1/3 • g(t) = -7/(2t-3)2 • y = [(3x -1)÷ (x2 +3)]2

  48. Implication Differentiation Section 4.5

  49. Implicit and Explicit Functions Up to this point, most in explicit form. For example in the equation y = 3x2 - 5 the variable y is explicitly written as a function of x Implicit Form Explicit Form Derivative xy = 1 y = 1/x -1/x2 Sometimes, this procedure doesn’t work, so you would use implicit differentiation

  50. Implicit Differentiation To understand how to find dy/dx implicitly, you must realize that the differentiation is taking place with respect to x. This means when you differentiate terms involving x alone, you can differentiate as usual, but when you differentiate terms involving y, you must apply the Chain Rule, because you are assuming that y is defined implicitly as a differentiable function of x.

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