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3.2 implicit equations and implicit differentiation

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3.2 implicit equations and implicit differentiation

  1. 1. Implicit Equations and Implicit Differentiation
  2. 2. Implicit Equations and Implicit Differentiation Implicit Equations
  3. 3. Implicit Equations and Implicit Differentiation Implicit EquationsThe function y = 1/x may be written as xy = 1.
  4. 4. Implicit Equations and Implicit Differentiation Implicit EquationsThe function y = 1/x may be written as xy = 1.The y = 1/x is the explicit form because the output y isgiven explicitly as a function of x. The equationxy = 1 is said to be in the implicit form because therelation between x and y are not given as a function.
  5. 5. Implicit Equations and Implicit Differentiation Implicit EquationsThe function y = 1/x may be written as xy = 1.The y = 1/x is the explicit form because the output y isgiven explicitly as a function of x. The equationxy = 1 is said to be in the implicit form because therelation between x and y are not given as a function.The implicit equation x2 + y2 = 1 has the unit circle asits graph.
  6. 6. Implicit Equations and Implicit Differentiation Implicit EquationsThe function y = 1/x may be written as xy = 1.The y = 1/x is the explicit form because the output y isgiven explicitly as a function of x. The equationxy = 1 is said to be in the implicit form because therelation between x and y are not given as a function.The implicit equation x2 + y2 = 1 has the unit circle asits graph. It’s not one function. y = √1 – x2However x2 + y2 = 1may be represented bytwo different functions.y = √1 – x2 or y = –√1 – x2 y = – √1 – x2
  7. 7. Implicit Equations and Implicit DifferentiationThe joint variables x and y in an implicit equation maybe inseparable by algebra.
  8. 8. Implicit Equations and Implicit DifferentiationThe joint variables x and y in an implicit equation maybe inseparable by algebra. sin(xy) = 0,
  9. 9. Implicit Equations and Implicit DifferentiationThe joint variables x and y in an implicit equation maybe inseparable by algebra. In fact for sin(xy) = 0,there are infinitely many solution-branches xy = nπfor n = 0, 1, 2,..
  10. 10. Implicit Equations and Implicit DifferentiationThe joint variables x and y in an implicit equation maybe inseparable by algebra. In fact for sin(xy) = 0,there are infinitely many solution-branches xy = nπfor n = 0, 1, 2,.. Plot xy = 0, xy = π, xy = – π, ..
  11. 11. Implicit Equations and Implicit DifferentiationThe joint variables x and y in an implicit equation maybe inseparable by algebra. In fact for sin(xy) = 0,there are infinitely many solution-branches xy = nπfor n = 0, 1, 2,.. Plot xy = 0, xy = π, xy = – π, .. y x sin(xy) = 0
  12. 12. Implicit Equations and Implicit DifferentiationThe joint variables x and y in an implicit equation maybe inseparable by algebra. In fact for sin(xy) = 0,there are infinitely many solution-branches xy = nπfor n = 0, 1, 2,.. Plot xy = 0, xy = π, xy = – π, ..Any point (x, y) on any of these branch is a solutionfor sin(xy) = 0. y xy = 0 x sin(xy) = 0
  13. 13. Implicit Equations and Implicit DifferentiationThe joint variables x and y in an implicit equation maybe inseparable by algebra. In fact for sin(xy) = 0,there are infinitely many solution-branches xy = nπfor n = 0, 1, 2,.. Plot xy = 0, xy = π, xy = – π, ..Any point (x, y) on any of these branch is a solutionfor sin(xy) = 0. y xy = 0 x xy = –π sin(xy) = 0
  14. 14. Implicit Equations and Implicit DifferentiationThe joint variables x and y in an implicit equation maybe inseparable by algebra. In fact for sin(xy) = 0,there are infinitely many solution-branches xy = nπfor n = 0, 1, 2,.. Plot xy = 0, xy = π, xy = – π, ..Any point (x, y) on any of these branch is a solutionfor sin(xy) = 0. y xy = 0 x xy = –π xy = –π sin(xy) = 0
  15. 15. Implicit Equations and Implicit DifferentiationThe joint variables x and y in an implicit equation maybe inseparable by algebra. In fact for sin(xy) = 0,there are infinitely many solution-branches xy = nπfor n = 0, 1, 2,.. Plot xy = 0, xy = π, xy = – π, ..Any point (x, y) on any of these branch is a solutionfor sin(xy) = 0.Given a point P(a, b) that satisfies ya given implicit equation, P mustbe on one (or more ) of thebranch which is well defined bysome function y = y(x). xy = 0 x xy = –π xy = –π sin(xy) = 0
  16. 16. Implicit Equations and Implicit DifferentiationThe joint variables x and y in an implicit equation maybe inseparable by algebra. In fact for sin(xy) = 0,there are infinitely many solution-branches xy = nπfor n = 0, 1, 2,.. Plot xy = 0, xy = π, xy = – π, ..Any point (x, y) on any of these branch is a solutionfor sin(xy) = 0.Given a point P(a, b) that satisfies ya given implicit equation, P mustbe on one (or more ) of thebranch which is well defined by y=π/x P(π, 1)some function y = y(x). xFor example, the point (π, 1)satisfies sin(xy) = 0 and it’s on thebranch xy = π or y = y(x) = π/x. sin(xy) = 0
  17. 17. Implicit Equations and Implicit DifferentiationWe assume that the branch yfunction y = y(x) that contains Pis differentiable at the point P as P(π, 1)the case in the example here y=π/xthat y = π/x is differentiable at xP(π, 1). sin(xy) = 0
  18. 18. Implicit Equations and Implicit DifferentiationWe assume that the branch yfunction y = y(x) that contains Pis differentiable at the point P as P(π, 1)the case in the example here y=π/xthat y = π/x is differentiable at xP(π, 1). The slope at P may befound by taking the derivative ofthe branch function y = π/x sin(xy) = 0which may be difficult to identify.
  19. 19. Implicit Equations and Implicit DifferentiationWe assume that the branch yfunction y = y(x) that contains Pis differentiable at the point P as P(π, 1)the case in the example here y=π/xthat y = π/x is differentiable at xP(π, 1). The slope at P may befound by taking the derivative ofthe branch function y = π/x sin(xy) = 0which may be difficult to identify.Instead we may compute a different form of thederivative directly from the implicit equation sin(xy) = 0and use it to compute the slope at the point P(π, 1).
  20. 20. Implicit Equations and Implicit DifferentiationWe assume that the branch yfunction y = y(x) that contains Pis differentiable at the point P as P(π, 1)the case in the example here y=π/xthat y = π/x is differentiable at xP(π, 1). The slope at P may befound by taking the derivative ofthe branch function y = π/x sin(xy) = 0which may be difficult to identify.Instead we may compute a different form of thederivative directly from the implicit equation sin(xy) = 0and use it to compute the slope at the point P(π, 1).This procedure is called implicit differentiation and theform of the derivative is called the implicit derivative.
  21. 21. Implicit Equations and Implicit DifferentiationTo differentiate implicitly, we assume that y is y(x)a function of x, and that we are to solve for y(x).
  22. 22. Implicit Equations and Implicit DifferentiationTo differentiate implicitly, we assume that y is y(x)a function of x, and that we are to solve for y(x).Because y is y(x), all the the y’s in an implicitequation are subjected to the Chain Rule.
  23. 23. Implicit Equations and Implicit DifferentiationTo differentiate implicitly, we assume that y is y(x)a function of x, and that we are to solve for y(x).Because y is y(x), all the the y’s in an implicitequation are subjected to the Chain Rule.Example A. Assume that y is y(x), expand thefollowing derivative in expressions of x, y and y.a. (x3 y)b. (x y3)
  24. 24. Implicit Equations and Implicit DifferentiationTo differentiate implicitly, we assume that y is y(x)a function of x, and that we are to solve for y(x).Because y is y(x), all the the y’s in an implicitequation are subjected to the Chain Rule.Example A. Assume that y is y(x), expand thefollowing derivative in expressions of x, y and y.a. (x3 y) By the Product Rule, (x3 y) = (x3) y + x3 (y)b. (x y3)
  25. 25. Implicit Equations and Implicit DifferentiationTo differentiate implicitly, we assume that y is y(x)a function of x, and that we are to solve for y(x).Because y is y(x), all the the y’s in an implicitequation are subjected to the Chain Rule.Example A. Assume that y is y(x), expand thefollowing derivative in expressions of x, y and y.a. (x3 y) By the Product Rule, (x3 y) = (x3) y + x3 (y) = 3x2y + x3(y)b. (x y3)
  26. 26. Implicit Equations and Implicit DifferentiationTo differentiate implicitly, we assume that y is y(x)a function of x, and that we are to solve for y(x).Because y is y(x), all the the y’s in an implicitequation are subjected to the Chain Rule.Example A. Assume that y is y(x), expand thefollowing derivative in expressions of x, y and y.a. (x3 y) By the Product Rule, (x3 y) = (x3) y + x3 (y) = 3x2y + x3(y)b. (x y3) By the Product Rule, (x y3) = (x) y3 + x3 (y3)
  27. 27. Implicit Equations and Implicit DifferentiationTo differentiate implicitly, we assume that y is y(x)a function of x, and that we are to solve for y(x).Because y is y(x), all the the y’s in an implicitequation are subjected to the Chain Rule.Example A. Assume that y is y(x), expand thefollowing derivative in expressions of x, y and y.a. (x3 y) By the Product Rule, (x3 y) = (x3) y + x3 (y) = 3x2y + x3(y)b. (x y3) (3y2y) By the Product Rule, (x y3) = (x) y3 + x3 (y3)
  28. 28. Implicit Equations and Implicit DifferentiationTo differentiate implicitly, we assume that y is y(x)a function of x, and that we are to solve for y(x).Because y is y(x), all the the y’s in an implicitequation are subjected to the Chain Rule.Example A. Assume that y is y(x), expand thefollowing derivative in expressions of x, y and y.a. (x3 y) By the Product Rule, (x3 y) = (x3) y + x3 (y) = 3x2y + x3(y)b. (x y3) (3y2y) By the Product Rule, (x y3) = (x) y3 + x3 (y3) = y3 + 3x3y2y
  29. 29. Implicit Equations and Implicit Differentiationd. Expand (x3 y – x y3) and collect the y – term
  30. 30. Implicit Equations and Implicit Differentiationd. Expand (x3 y – x y3) and collect the y – term (x3 y – x y3) = [3x2y + x3 y] – [y3 + 3x3y2 y]
  31. 31. Implicit Equations and Implicit Differentiationd. Expand (x3 y – x y3) and collect the y – term (x3 y – x y3) = [3x2y + x3 y] – [y3 + 3x3y2 y] = 3x2y + x3 y – y3 – 3x3y2 y
  32. 32. Implicit Equations and Implicit Differentiationd. Expand (x3 y – x y3) and collect the y – term (x3 y – x y3) = [3x2y + x3 y] – [y3 + 3x3y2 y] = 3x2y + x3 y – y3 – 3x3y2 y = 3x2y – y3 + (x3 – 3x3y2) y
  33. 33. Implicit Equations and Implicit Differentiationd. Expand (x3 y – x y3) and collect the y – term (x3 y – x y3) = [3x2y + x3 y] – [y3 + 3x3y2 y] = 3x2y + x3 y – y3 – 3x3y2 y = 3x2y – y3 + (x3 – 3x3y2) y
  34. 34. Implicit Equations and Implicit Differentiationd. Expand (x3 y – x y3) and collect the y – term (x3 y – x y3) = [3x2y + x3 y] – [y3 + 3x3y2 y] = 3x2y + x3 y – y3 – 3x3y2 y = 3x2y – y3 + (x3 – 3x3y2) y yLet us go back to theequation sin(xy) = 0. x sin(xy) = 0
  35. 35. Implicit Equations and Implicit Differentiationd. Expand (x3 y – x y3) and collect the y – term (x3 y – x y3) = [3x2y + x3 y] – [y3 + 3x3y2 y] = 3x2y + x3 y – y3 – 3x3y2 y = 3x2y – y3 + (x3 – 3x3y2) y yLet us go back to theequation sin(xy) = 0. P(π, 1)Example B. Use implicit y=π/xdifferentiation to find y xgiven that sin(xy) = 0 and findthe slope and the tangent lineat the point (π, 1) on the graph sin(xy) = 0of sin(xy) = 0.
  36. 36. Implicit Equations and Implicit DifferentiationDifferentiate both sides of sin(xy) = 0 with respect to x,[sin(xy) ] = (0) y P(π, 1) y=π/x x sin(xy) = 0
  37. 37. Implicit Equations and Implicit DifferentiationDifferentiate both sides of sin(xy) = 0 with respect to x,[sin(xy) ] = (0) by the Sine Chain Rule[cos(xy)] (xy) = 0 y P(π, 1) y=π/x x sin(xy) = 0
  38. 38. Implicit Equations and Implicit DifferentiationDifferentiate both sides of sin(xy) = 0 with respect to x,[sin(xy) ] = (0) by the Sine Chain Rule[cos(xy)] (xy) = 0 by the Product Rulecos(xy) [(x)y + x(y)] = 0 y P(π, 1) y=π/x x sin(xy) = 0
  39. 39. Implicit Equations and Implicit DifferentiationDifferentiate both sides of sin(xy) = 0 with respect to x,[sin(xy) ] = (0) by the Sine Chain Rule[cos(xy)] (xy) = 0 by the Product Rulecos(xy) [(x)y + x(y)] = 0 ycos(xy) [1*y + x(y)] = 0 P(π, 1) y=π/x x sin(xy) = 0
  40. 40. Implicit Equations and Implicit DifferentiationDifferentiate both sides of sin(xy) = 0 with respect to x,[sin(xy) ] = (0) by the Sine Chain Rule[cos(xy)] (xy) = 0 by the Product Rulecos(xy) [(x)y + x(y)] = 0 ycos(xy) [1*y + x(y)] = 0 P(π, 1)Solve for (y), set y + x(y) = 0 y=π/x x sin(xy) = 0
  41. 41. Implicit Equations and Implicit DifferentiationDifferentiate both sides of sin(xy) = 0 with respect to x,[sin(xy) ] = (0) by the Sine Chain Rule[cos(xy)] (xy) = 0 by the Product Rulecos(xy) [(x)y + x(y)] = 0 ycos(xy) [1*y + x(y)] = 0 P(π, 1)Solve for (y), set y + x(y) = 0 y=π/x –y xx(y) = –y or that y = x sin(xy) = 0
  42. 42. Implicit Equations and Implicit DifferentiationDifferentiate both sides of sin(xy) = 0 with respect to x,[sin(xy) ] = (0) by the Sine Chain Rule[cos(xy)] (xy) = 0 by the Product Rulecos(xy) [(x)y + x(y)] = 0 ycos(xy) [1*y + x(y)] = 0 P(π, 1)Solve for (y), set y + x(y) = 0 y=π/x –y xx(y) = –y or that y = xAt P(π, 1), x = π, y = 1y = –1 = the slope at P. π sin(xy) = 0
  43. 43. Implicit Equations and Implicit DifferentiationDifferentiate both sides of sin(xy) = 0 with respect to x,[sin(xy) ] = (0) by the Sine Chain Rule[cos(xy)] (xy) = 0 by the Product Rulecos(xy) [(x)y + x(y)] = 0 ycos(xy) [1*y + x(y)] = 0 P(π, 1)Solve for (y), set y + x(y) = 0 y=π/x –y xx(y) = –y or that y = xAt P(π, 1), x = π, y = 1y = –1 = the slope at P. π sin(xy) = 0Using the the Point–Slope Formula at P (π, 1), theequation of the tangent line is y = –x + 2. π
  44. 44. Implicit Equations and Implicit DifferentiationYour turn: Expand [cos3 (x/y)]
  45. 45. Implicit Equations and Implicit DifferentiationYour turn: Expand [cos3 (x/y)] By the Power Chain Rule,[cos3 (x/y)] = 3cos2(x/y) [cos (x/y)] = 3cos2(x/y) [– sin(x/y)] (x/y) = 3cos2(x/y) [– sin(x/y)] ((y – xy)/y2)
  46. 46. Implicit Equations and Implicit DifferentiationYour turn: Expand [cos3 (x/y)] By the Power Chain Rule,[cos3 (x/y)] = 3cos2(x/y) [cos (x/y)] = 3cos2(x/y) [– sin(x/y)] (x/y) = 3cos2(x/y) [– sin(x/y)] ((y – xy)/y2) exy x2+1 = y2 find y implicitly.Example C. Given
  47. 47. Implicit Equations and Implicit DifferentiationYour turn: Expand [cos3 (x/y)] By the Power Chain Rule,[cos3 (x/y)] = 3cos2(x/y) [cos (x/y)] = 3cos2(x/y) [– sin(x/y)] (x/y) = 3cos2(x/y) [– sin(x/y)] ((y – xy)/y2) exy x2+1 = y2 find y implicitly.Example C. GivenWe are to differentiate with respect to x on both sidesof the equation.
  48. 48. Implicit Equations and Implicit DifferentiationYour turn: Expand [cos3 (x/y)] By the Power Chain Rule,[cos3 (x/y)] = 3cos2(x/y) [cos (x/y)] = 3cos2(x/y) [– sin(x/y)] (x/y) = 3cos2(x/y) [– sin(x/y)] ((y – xy)/y2) exy x2+1 = y2 find y implicitly.Example C. GivenWe are to differentiate with respect to x on both sidesof the equation. However it’s easier if we clear thedenominators first.y2 exy = x2+1
  49. 49. Implicit Equations and Implicit DifferentiationYour turn: Expand [cos3 (x/y)] By the Power Chain Rule,[cos3 (x/y)] = 3cos2(x/y) [cos (x/y)] = 3cos2(x/y) [– sin(x/y)] (x/y) = 3cos2(x/y) [– sin(x/y)] ((y – xy)/y2) exy x2+1 = y2 find y implicitly.Example C. GivenWe are to differentiate with respect to x on both sidesof the equation. However it’s easier if we clear thedenominators first.Hence we differentiate both sides ofy2 exy = x2+1
  50. 50. Implicit Equations and Implicit Differentiation[y2 exy ] = [x2+1]
  51. 51. Implicit Equations and Implicit Differentiation[y2 exy ] = [x2+1] Product Rule(y2) exy + y2(exy) =
  52. 52. Implicit Equations and Implicit Differentiation[y2 exy ] = [x2+1] Product Rule Derivative of x2 with respect to x(y2) exy + y2(exy) = 2x
  53. 53. Implicit Equations and Implicit Differentiation[y2 exy ] = [x2+1] Product Rule Derivative of x2 with respect to x(y2) exy + y2(exy) = 2x Chain Rule2y(y) exy
  54. 54. Implicit Equations and Implicit Differentiation[y2 exy ] = [x2+1] Product Rule Derivative of x2 with respect to x(y2) exy + y2(exy) = 2x Chain Rule2y(y) exy + y2 exy (xy) = 2x
  55. 55. Implicit Equations and Implicit Differentiation[y2 exy ] = [x2+1] Product Rule Derivative of x2 with respect to x(y2) exy + y2(exy) = 2x Chain Rule2y(y) exy + y2 exy (xy) = 2x Product Chain Rule2y(y) exy + y2 exy (y + xy) = 2x
  56. 56. Implicit Equations and Implicit Differentiation[y2 exy ] = [x2+1] Product Rule Derivative of x2 with respect to x(y2) exy + y2(exy) = 2x Chain Rule2y(y) exy + y2 exy (xy) = 2x Product Chain Rule2y(y) exy + y2 exy (y + xy) = 2xWe are done with differentiating.Next is to extract the y.
  57. 57. Implicit Equations and Implicit Differentiation[y2 exy ] = [x2+1] Product Rule Derivative of x2 with respect to x(y2) exy + y2(exy) = 2x Chain Rule2y(y) exy + y2 exy (xy) = 2x Product Chain Rule2y(y) exy + y2 exy (y + xy) = 2xWe are done with differentiating.Next is to extract the y.Pull out the inherited exy factor on the left first.exy [2y(y) + y2 (y + xy)] = 2x
  58. 58. Implicit Equations and Implicit Differentiationexy [2y(y) + y2 (y + xy)] = 2x
  59. 59. Implicit Equations and Implicit Differentiationexy [2y(y) + y2 (y + xy)] = 2x2y(y) + y2 (y + xy) = 2x exy
  60. 60. Implicit Equations and Implicit Differentiationexy [2y(y) + y2 (y + xy)] = 2x2y(y) + y2 (y + xy) = 2x exy2y y + y3 + xy2 y = 2xexy
  61. 61. Implicit Equations and Implicit Differentiationexy [2y(y) + y2 (y + xy)] = 2x2y(y) + y2 (y + xy) = 2x exy2y y + y3 + xy2 y = 2xexy2yy + xy2 y = 2x – y3 exy
  62. 62. Implicit Equations and Implicit Differentiationexy [2y(y) + y2 (y + xy)] = 2x2y(y) + y2 (y + xy) = 2x exy2y y + y3 + xy2 y = 2xexy2yy + xy2 y = 2x – y3 exyy (2y+ xy2 ) = 2x – y3 exy
  63. 63. Implicit Equations and Implicit Differentiationexy [2y(y) + y2 (y + xy)] = 2x2y(y) + y2 (y + xy) = 2x exy2y y + y3 + xy2 y = 2xexy2yy + xy2 y = 2x – y3 exyy (2y+ xy2 ) = 2x – y3 exy 2x exy – y3y = (2y+ xy2 )
  64. 64. Implicit Equations and Implicit Differentiationexy [2y(y) + y2 (y + xy)] = 2x2y(y) + y2 (y + xy) = 2x exy2y y + y3 + xy2 y = 2xexy2yy + xy2 y = 2x – y3 exyy (2y+ xy2 ) = 2x – y3 exy 2x exy – y3y = multiply exy up and down, we get (2y+ xy2 )y = 2x – y3exy (2y+ xy2 )exy
  65. 65. Log DifferentiationThe technique of log–differentiation takes advantageof the fact that a “messy” function or an equation maybe disassembled into simpler terms after logging.
  66. 66. Log DifferentiationThe technique of log–differentiation takes advantageof the fact that a “messy” function or an equation maybe disassembled into simpler terms after logging.To use log–differentiation to find derivative y,
  67. 67. Log DifferentiationThe technique of log–differentiation takes advantageof the fact that a “messy” function or an equation maybe disassembled into simpler terms after logging.To use log–differentiation to find derivative y,i. takes the log of the function or equation or bothsides and disassemble the formulas as much aspossible in terms of In(#)’s.
  68. 68. Log DifferentiationThe technique of log–differentiation takes advantageof the fact that a “messy” function or an equation maybe disassembled into simpler terms after logging.To use log–differentiation to find derivative y,i. takes the log of the function or equation or bothsides and disassemble the formulas as much aspossible in terms of In(#)’s.ii. takes the derivative of the disassembled functionor equation implicitly.
  69. 69. Log DifferentiationThe technique of log–differentiation takes advantageof the fact that a “messy” function or an equation maybe disassembled into simpler terms after logging.To use log–differentiation to find derivative y,i. takes the log of the function or equation or bothsides and disassemble the formulas as much aspossible in terms of In(#)’s.ii. takes the derivative of the disassembled functionor equation implicitly.iii. then collect and isolate the y.
  70. 70. Log DifferentiationThe technique of log–differentiation takes advantageof the fact that a “messy” function or an equation maybe disassembled into simpler terms after logging.To use log–differentiation to find derivative y,i. takes the log of the function or equation or bothsides and disassemble the formulas as much aspossible in terms of In(#)’s.ii. takes the derivative of the disassembled functionor equation implicitly.iii. then collect and isolate the y.Example C.Use log differentiation to find y given that 3 2(x)√x2+1y = sin e2x– 1
  71. 71. Log DifferentiationTakes the natural–log on both sides we have 3 2(x) √x2+1 ln(y) = ln sin e2x– 1
  72. 72. Log DifferentiationTakes the natural–log on both sides we have 3 2(x) √x2+1 ln(y) = ln sin e2x– 1ln(y) = In(sin2(x)) + In( (x2+1)3/2) – In(e2x–1)
  73. 73. Log DifferentiationTakes the natural–log on both sides we have 3 2(x) √x2+1 ln(y) = ln sin e2x– 1ln(y) = In(sin2(x)) + In( (x2+1)3/2) – In(e2x–1)ln(y) = 2In(sin (x)) + 3 In(x2+1) – 2x + 1 2
  74. 74. Log DifferentiationTakes the natural–log on both sides we have 3 2(x) √x2+1 ln(y) = ln sin e2x– 1ln(y) = In(sin2(x)) + In( (x2+1)3/2) – In(e2x–1)ln(y) = 2In(sin (x)) + 3 In(x2+1) – 2x + 1 Differentiate 2 both sides
  75. 75. Log DifferentiationTakes the natural–log on both sides we have 3 2(x) √x2+1 ln(y) = ln sin e2x– 1ln(y) = In(sin2(x)) + In( (x2+1)3/2) – In(e2x–1)ln(y) = 2In(sin (x)) + 3 In(x2+1) – 2x + 1 Differentiate 2 both sides y y =
  76. 76. Log DifferentiationTakes the natural–log on both sides we have 3 2(x) √x2+1 ln(y) = ln sin e2x– 1ln(y) = In(sin2(x)) + In( (x2+1)3/2) – In(e2x–1)ln(y) = 2In(sin (x)) + 3 In(x2+1) – 2x + 1 Differentiate 2 both sides y 2cos(x) y = sin(x)
  77. 77. Log DifferentiationTakes the natural–log on both sides we have 3 2(x) √x2+1 ln(y) = ln sin e2x– 1ln(y) = In(sin2(x)) + In( (x2+1)3/2) – In(e2x–1)ln(y) = 2In(sin (x)) + 3 In(x2+1) – 2x + 1 Differentiate 2 both sides y 2cos(x) 3 2x – 2 y = sin(x) + 2 x2+1
  78. 78. Log DifferentiationTakes the natural–log on both sides we have 3 2(x) √x2+1 ln(y) = ln sin e2x– 1ln(y) = In(sin2(x)) + In( (x2+1)3/2) – In(e2x–1)ln(y) = 2In(sin (x)) + 3 In(x2+1) – 2x + 1 Differentiate 2 both sides y 2cos(x) 3 2x – 2 y = sin(x) + 2 x2+1Hencey = 2cos(x) + 3 2+1 –2 y sin(x) x
  79. 79. Log DifferentiationTakes the natural–log on both sides we have 3 2(x) √x2+1 ln(y) = ln sin e2x– 1 ln(y) = In(sin2(x)) + In( (x2+1)3/2) – In(e2x–1) ln(y) = 2In(sin (x)) + 3 In(x2+1) – 2x + 1 Differentiate 2 both sides y 2cos(x) 3 2x – 2 y = sin(x) + 2 x2+1Hencey = 2cos(x) + 3 2+1 –2 y sin(x) x 3 2cos(x) 2(x)√x2+1y = sin(x) + x2+1 – 2 sin 3 e2x– 1

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