16 partial derivatives

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16 partial derivatives

  1. 1. Partial Derivatives
  2. 2. Partial DerivativesLet p = (a, b, c) be a point on the surface z = f(x, y).
  3. 3. Partial DerivativesLet p = (a, b, c) be a point on the surface z = f(x, y).The plane y = b intersects the surface
  4. 4. Partial DerivativesLet p = (a, b, c) be a point on the surface z = f(x, y).The plane y = b intersects the surface z y=b (a, b, c) y x
  5. 5. Partial DerivativesLet p = (a, b, c) be a point on the surface z = f(x, y).The plane y = b intersects the surface at the curvez = f(x, b) y=b z z y=b z = f(x, b) (a, c) (a, b, c) x y x
  6. 6. Partial DerivativesLet p = (a, b, c) be a point on the surface z = f(x, y).The plane y = b intersects the surface at the curvez = f(x, b) which is a y–trace that contains p = (a, b, c). y=b z z y=b z = f(x, b) (a, c) (a, b, c) x y x
  7. 7. Partial DerivativesLet p = (a, b, c) be a point on the surface z = f(x, y).The plane y = b intersects the surface at the curvez = f(x, b) which is a y–trace that contains p = (a, b, c).The slope at p on this trace in the y = b plane, y=b z z y=b z = f(x, b) (a, c) (a, b, c) x y x
  8. 8. Partial DerivativesLet p = (a, b, c) be a point on the surface z = f(x, y).The plane y = b intersects the surface at the curvez = f(x, b) which is a y–trace that contains p = (a, b, c).The slope at p on this trace in the y = b plane, is calledthe partial derivative with respect to x. y=b z z y=b z = f(x, b) (a, c) (a, b, c) x y x
  9. 9. Partial DerivativesLet p = (a, b, c) be a point on the surface z = f(x, y).The plane y = b intersects the surface at the curvez = f(x, b) which is a y–trace that contains p = (a, b, c).The slope at p on this trace in the y = b plane, is calledthe partial derivative with respect to x. Specifically,The partial derivative of z = f(x, y) with respect to x is df dz dx = dx y=b z z y=b z = f(x, b) (a, c) (a, b, c) x y x d z gives the slopes of y traces. dx
  10. 10. Partial DerivativesLet p = (a, b, c) be a point on the surface z = f(x, y).The plane y = b intersects the surface at the curvez = f(x, b) which is a y–trace that contains p = (a, b, c).The slope at p on this trace in the y = b plane, is calledthe partial derivative with respect to x. Specifically,The partial derivative of z = f(x, y) with respect to x is df dz f(x + h, y) – f(x, y) , if it exists. d x = d x = h 0 lim h y=b z z y=b z = f(x, b) (a, c) (a, b, c) x y x d z gives the slopes of y traces. dx
  11. 11. Partial Derivatives dzAlgebraically, to find d x , we treat y as a constantand apply the derivative with respect to x.
  12. 12. Partial Derivatives dzAlgebraically, to find d x , we treat y as a constantand apply the derivative with respect to x.Example A. Find dx dza. z = x + y b. z = xyc. z = x2y d. z = xy2 ye. z = x y f. z = x
  13. 13. Partial Derivatives dzAlgebraically, to find d x , we treat y as a constantand apply the derivative with respect to x.Example A. Find dx dza. z = x + y dz = 1 b. z = xy dx because y = 0.c. z = x2y d. z = xy2 ye. z = x y f. z = x
  14. 14. Partial Derivatives dzAlgebraically, to find d x , we treat y as a constantand apply the derivative with respect to x.Example A. Find dx dza. z = x + y dz = 1 b. z = xy dz = y dx dx because y = 0. because (cx) = c.c. z = x2y d. z = xy2 ye. z = x y f. z = x
  15. 15. Partial Derivatives dzAlgebraically, to find d x , we treat y as a constantand apply the derivative with respect to x.Example A. Find dx dza. z = x + y dz = 1 b. z = xy dz = y dx dx because y = 0. because (cx) = c. dzc. z = x y 2 dx = 2xy d. z = xy2 because (xn) = nxn–1. x ye. z = y f. z = x
  16. 16. Partial Derivatives dzAlgebraically, to find d x , we treat y as a constantand apply the derivative with respect to x.Example A. Find dx dza. z = x + y dz = 1 b. z = xy dz = y dx dx because y = 0. because (cx) = c. dz dzc. z = x y 2 dx = 2xy d. z = xy 2 dx = y2 because (xn) = nxn–1. because (cx) = c. x ye. z = y f. z = x
  17. 17. Partial Derivatives dzAlgebraically, to find d x , we treat y as a constantand apply the derivative with respect to x.Example A. Find dx dza. z = x + y dz = 1 b. z = xy dz = y dx dx because y = 0. because (cx) = c. dz dzc. z = x y 2 dx = 2xy d. z = xy 2 dx = y2 because (xn) = nxn–1. because (cx) = c. x dz 1 ye. z = y dx = y f. z = x because (cx) = c.
  18. 18. Partial Derivatives dzAlgebraically, to find d x , we treat y as a constantand apply the derivative with respect to x.Example A. Find dx dza. z = x + y dz = 1 b. z = xy dz = y dx dx because y = 0. because (cx) = c. dz dzc. z = x y 2 dx = 2xy d. z = xy 2 dx = y2 because (xn) = nxn–1. because (cx) = c. x dz 1 y dze. z = y dx = y f. z = dx = –yx –2 x because (cx) = c. The Quotient Rule
  19. 19. Partial Derivatives dzAlgebraically, to find d x , we treat y as a constantand apply the derivative with respect to x.Example A. Find dx dza. z = x + y dz = 1 b. z = xy dz = y dx dx because y = 0. because (cx) = c. dz dzc. z = x y 2 dx = 2xy d. z = xy 2 dx = y2 because (xn) = nxn–1. because (cx) = c. x dz 1 y dze. z = y dx = y f. z = dx = –yx –2 x because (cx) = c. The Quotient RuleThe plane y = b is parallel to the x–axis, so the partialdz/dx is also referred to as the partial derivative in thex–direction (as in the compass direction).
  20. 20. Partial DerivativesExample B. Find an equation for the tangent line forat (2, 3, 6) on the surface of z = f(x, y) = √49 – x2 – y2in the direction of the x-axis.
  21. 21. Partial DerivativesExample B. Find an equation for the tangent line forat (2, 3, 6) on the surface of z = f(x, y) = √49 – x2 – y2in the direction of the x-axis. Reminder: the graph of z = √49 – x2 – y2 is the top half of the hemisphere with radius 7.
  22. 22. Partial DerivativesExample B. Find an equation for the tangent line forat (2, 3, 6) on the surface of z = f(x, y) = √49 – x2 – y2in the direction of the x-axis. z y=3 (2,3,6) y x
  23. 23. Partial DerivativesExample B. Find an equation for the tangent line forat (2, 3, 6) on the surface of z = f(x, y) = √49 – x2 – y2in the direction of the x-axis. z y=3 (2,3,6) y x
  24. 24. Partial DerivativesExample B. Find an equation for the tangent line forat (2, 3, 6) on the surface of z = f(x, y) = √49 – x2 – y2in the direction of the x-axis.We need df/ dx|(2,3). z y=3 (2,3,6) y x
  25. 25. Partial DerivativesExample B. Find an equation for the tangent line forat (2, 3, 6) on the surface of z = f(x, y) = √49 – x2 – y2in the direction of the x-axis.We need df/ dx|(2,3). (i.e. the slope this tangent) z y=3 (2,3,6) y x
  26. 26. Partial DerivativesExample B. Find an equation for the tangent line forat (2, 3, 6) on the surface of z = f(x, y) = √49 – x2 – y2in the direction of the x-axis.We need df/ dx|(2,3). df/dx = –x/√49 – x2 – y2 , z y=3 (2,3,6) y x
  27. 27. Partial DerivativesExample B. Find an equation for the tangent line forat (2, 3, 6) on the surface of z = f(x, y) = √49 – x2 – y2in the direction of the x-axis.We need df/ dx|(2,3). df/dx = –x/√49 – x2 – y2 , at (2, 3),we get df / dx|(2,3) = –2/6 = –1/3 z y=3 (2,3,6) y x
  28. 28. Partial DerivativesExample B. Find an equation for the tangent line forat (2, 3, 6) on the surface of z = f(x, y) = √49 – x2 – y2in the direction of the x-axis.We need df/ dx|(2,3). df/dx = –x/√49 – x2 – y2 , at (2, 3),we get df / dx|(2,3) = –2/6 = –1/3 = slope as shown. z y=3 z y=3 1 (2,3,6) –1/3 x y x
  29. 29. Partial DerivativesExample B. Find an equation for the tangent line forat (2, 3, 6) on the surface of z = f(x, y) = √49 – x2 – y2in the direction of the x-axis.We need df/ dx|(2,3). df/dx = –x/√49 – x2 – y2 , at (2, 3),we get df / dx|(2,3) = –2/6 = –1/3 = slope as shown. So adirections vector for the tangent is v = <1, 0, –1/3> z y=3 z y=3 1 (2,3,6) –1/3 x y x
  30. 30. Partial DerivativesExample B. Find an equation for the tangent line forat (2, 3, 6) on the surface of z = f(x, y) = √49 – x2 – y2in the direction of the x-axis.We need df/ dx|(2,3). df/dx = –x/√49 – x2 – y2 , at (2, 3),we get df / dx|(2,3) = –2/6 = –1/3 = slope as shown. So adirections vector for the tangent is v = <1, 0, –1/3> andan equation for the tangent is L = <t + 2, 3, –t/3 + 6>. z y=3 z y=3 1 (2,3,6) –1/3 x y x
  31. 31. Partial DerivativesThe geometry of the partial derivative with respect to yis similar to partial derivative with respect to x.
  32. 32. Partial DerivativesThe geometry of the partial derivative with respect to yis similar to partial derivative with respect to x.Let (a, b, c) be a point on the surface of z = f(x, y),the plane x = a intersects the surface at a curve that isthe graph of the equation z = f(a, y), in the plane x = a. x=a (a,b,c) y x
  33. 33. Partial DerivativesThe geometry of the partial derivative with respect to yis similar to partial derivative with respect to x.Let (a, b, c) be a point on the surface of z = f(x, y),the plane x = a intersects the surface at a curve that isthe graph of the equation z = f(a, y), in the plane x = a. dz dy |P = slope of the tangent of z = f(a, y) in the plane x = a at the point y= b . x=a (a,b,c) y x
  34. 34. Partial DerivativesThe geometry of the partial derivative with respect to yis similar to partial derivative with respect to x.Let (a, b, c) be a point on the surface of z = f(x, y),the plane x = a intersects the surface at a curve that isthe graph of the equation z = f(a, y), in the plane x = a. dz dy |P = slope of the tangent of z = f(a, y) in the plane x = a at the point y= b . x=a x=a z (a,b,c) (b, c) z = f(a,y) y y x
  35. 35. Partial DerivativesThe geometry of the partial derivative with respect to yis similar to partial derivative with respect to x.Let (a, b, c) be a point on the surface of z = f(x, y),the plane x = a intersects the surface at a curve that isthe graph of the equation z = f(a, y), in the plane x = a. dz dy |P = slope of the tangent of z = f(a, y) in the plane x = a at the point y= b . x=a x=a z (a,b,c) (b, c) z = f(a,y) y y x d z gives the slopes of x traces. dy
  36. 36. Partial Derivatives dzTo find d y , treat x as a constant in the formula.
  37. 37. Partial Derivatives dzTo find d y , treat x as a constant in the formula. dz dzExample C. a. Find dx , dy if z = cos(x2y).
  38. 38. Partial Derivatives dzTo find d y , treat x as a constant in the formula. dz dzExample C. a. Find dx , dy if z = cos(x2y).We need to use the chain–rule for both partials.
  39. 39. Partial Derivatives dzTo find d y , treat x as a constant in the formula. dz dzExample C. a. Find dx , dy if z = cos(x2y).We need to use the chain–rule for both partials. dz d(x2y) dx = –sin(x2y)* dx
  40. 40. Partial Derivatives dzTo find d y , treat x as a constant in the formula. dz dzExample C. a. Find dx , dy if z = cos(x2y).We need to use the chain–rule for both partials. dz d(x2y) dx = –sin(x2y)* dx = –sin(x2y)*2xy
  41. 41. Partial Derivatives dzTo find d y , treat x as a constant in the formula. dz dzExample C. a. Find dx , dy if z = cos(x2y).We need to use the chain–rule for both partials. dz d(x2y) dz d(x2y) dy = –sin(x y)* dy 2 dx = –sin(x2y)* dx = –sin(x2y)*2xy
  42. 42. Partial Derivatives dzTo find d y , treat x as a constant in the formula. dz dzExample C. a. Find dx , dy if z = cos(x2y).We need to use the chain–rule for both partials. dz d(x2y) dz d(x2y) dy = –sin(x y)* dy 2 dx = –sin(x2y)* dx = –sin(x y)*2xy 2 = –sin(x2y)*x2
  43. 43. Partial Derivatives dzTo find d y , treat x as a constant in the formula. dz dzExample C. a. Find dx , dy if z = cos(x2y).We need to use the chain–rule for both partials. dz d(x2y) dz d(x2y) dy = –sin(x y)* dy 2 dx = –sin(x2y)* dx = –sin(x y)*2xy 2 = –sin(x2y)*x2b. Find dz , dz at the point P = (1, π/2, 0) dx dy
  44. 44. Partial Derivatives dzTo find d y , treat x as a constant in the formula. dz dzExample C. a. Find dx , dy if z = cos(x2y).We need to use the chain–rule for both partials. dz d(x2y) dz d(x2y) dy = –sin(x y)* dy 2 dx = –sin(x2y)* dx = –sin(x y)*2xy 2 = –sin(x2y)*x2b. Find dz , dz at the point P = (1, π/2, 0) dx dy dz dx |(1,π/2, 0)
  45. 45. Partial Derivatives dzTo find d y , treat x as a constant in the formula. dz dzExample C. a. Find dx , dy if z = cos(x2y).We need to use the chain–rule for both partials. dz d(x2y) dz d(x2y) dy = –sin(x y)* dy 2 dx = –sin(x2y)* dx = –sin(x y)*2xy 2 = –sin(x2y)*x2b. Find dz , dz at the point P = (1, π/2, 0) dx dy dz dx |(1,π/2, 0) = –sin(x2y)*2xy|(1,π/2,0) = –π
  46. 46. Partial Derivatives dzTo find d y , treat x as a constant in the formula. dz dzExample C. a. Find dx , dy if z = cos(x2y).We need to use the chain–rule for both partials. dz d(x2y) dz d(x2y) dy = –sin(x y)* dy 2 dx = –sin(x2y)* dx = –sin(x y)*2xy 2 = –sin(x2y)*x2b. Find dz , dz at the point P = (1, π/2, 0) dx dy dz dz dx |(1,π/2, 0) dy |(1,π/2, 0) = –sin(x2y)*2xy|(1,π/2,0) = –π
  47. 47. Partial Derivatives dzTo find d y , treat x as a constant in the formula. dz dzExample C. a. Find dx , dy if z = cos(x2y).We need to use the chain–rule for both partials. dz d(x2y) dz d(x2y) dy = –sin(x y)* dy 2 dx = –sin(x2y)* dx = –sin(x y)*2xy 2 = –sin(x2y)*x2b. Find dz , dz at the point P = (1, π/2, 0) dx dy dz dz dx |(1,π/2, 0) dy |(1,π/2, 0) = –sin(x2y)*2xy|(1,π/2,0) = –sin(x2y)*x2|(1,π/2,0) = –π = –1
  48. 48. Partial DerivativesGiven a function of one variable, we say the function is"differentiable" at a point P if the derivative exists at P.
  49. 49. Partial DerivativesGiven a function of one variable, we say the function is"differentiable" at a point P if the derivative exists at P. f(x) exists means the curve is smooth, not a corner, at P, so its tangent is well defined.
  50. 50. Partial DerivativesGiven a function of one variable, we say the function is"differentiable" at a point P if the derivative exists at P. f(x) exists means the curve is smooth, not a corner, at P, so its tangent is well defined. p f(x) exists
  51. 51. Partial DerivativesGiven a function of one variable, we say the function is"differentiable" at a point P if the derivative exists at P. f(x) exists means the curve is smooth, not a corner, at P, so its tangent is well defined. p p f(x) exists f(x) doesnt exist
  52. 52. Partial DerivativesGiven a function of one variable, we say the function is"differentiable" at a point P if the derivative exists at P. f(x) exists means the curve is smooth, not a corner, at P, so its tangent is well defined. p p f(x) exists f(x) doesnt existHowever, for a function f(x, y) of two variables(or more), the existence of the partial derivativesdoesnt mean the surface z = f(x, y) is smooth,it may have creases (i.e. folds) even
  53. 53. Partial DerivativesGiven a function of one variable, we say the function is"differentiable" at a point P if the derivative exists at P. f(x) exists means the curve is smooth, not a corner, at P, so its tangent is well defined. p p f(x) exists f(x) doesnt existHowever, for a function f(x, y) of two variables(or more), the existence of the partial derivativesdoesnt mean the surface z = f(x, y) is smooth,it may have creases (i.e. folds) evenif the partials exist.
  54. 54. Partial DerivativesGiven a function of one variable, we say the function is"differentiable" at a point P if the derivative exists at P. f(x) exists means the curve is smooth, not a corner, at P, so its tangent is well defined. p p f(x) exists f(x) doesnt existHowever, for a function f(x, y) of two variables(or more), the existence of the partial derivativesdoesnt mean the surface z = f(x, y) is smooth,it may have creases (i.e. folds) even P yif the partials exist. For example,the surface shown here has dx dz |P = dz |P = 0, x dybut the surface is folded, i.e. not smooth at P.
  55. 55. Partial DerivativesIf we impose the conditions that 1. both partials withrespect to x and y exist in a small circle with P as thecenter, z p y x (a, b)
  56. 56. Partial DerivativesIf we impose the conditions that 1. both partials withrespect to x and y exist in a small circle with P as thecenter, and 2. that both partial derivatives arecontinuous in this circle, z p y x (a, b)
  57. 57. Partial DerivativesIf we impose the conditions that 1. both partials withrespect to x and y exist in a small circle with P as thecenter, and 2. that both partial derivatives arecontinuous in this circle, then the surface is smooth(no crease) at P and it has a well defined tangentplane at P as shown. A smooth point P and its tangent plane. z p y x (a, b)
  58. 58. Partial DerivativesIf we impose the conditions that 1. both partials withrespect to x and y exist in a small circle with P as thecenter, and 2. that both partial derivatives arecontinuous in this circle, then the surface is smooth(no crease) at P and it has a well defined tangentplane at P as shown. A smooth point P and its tangent plane.Hence we say z = f(x,y) is z"differentiable" or "smooth" at P p dzif both partials dz , dy exist, dx yand are continuous in a xneighborhood of P. (a, b)
  59. 59. Partial Derivatives If we impose the conditions that 1. both partials with respect to x and y exist in a small circle with P as the center, and 2. that both partial derivatives are continuous in this circle, then the surface is smooth (no crease) at P and it has a well defined tangent plane at P as shown. A smooth point P and its tangent plane. Hence we say z = f(x,y) is z "differentiable" or "smooth" at P p dz if both partials dz , dy exist, dx y and are continuous in a x neighborhood of P. (a, b)For any elementary functions z = f(x, y), if a small circlethat centered at P is in the domain, that all the partialsexist in this circle then the surface is smooth at P.
  60. 60. Partial DerivativesWe also write df as fx and df as dx dy fy .
  61. 61. Partial DerivativesWe also write df as fx and df as dx dyIts easy to talk about highery . f derivatives with thisnotation. So, fxx means to take the partial derivativewith respect to x twice, fxy means to to take thederivative with respect to x first, then with respect to y.
  62. 62. Partial DerivativesWe also write df as fx and df as dx dyIts easy to talk about highery . f derivatives with thisnotation. So, fxx means to take the partial derivativewith respect to x twice, fxy means to to take thederivative with respect to x first, then2 with respect to y.In the quotient form fxx is written as d f dxdx
  63. 63. Partial DerivativesWe also write df as fx and df as dx dyIts easy to talk about highery . f derivatives with thisnotation. So, fxx means to take the partial derivativewith respect to x twice, fxy means to to take thederivative with respect to x first, then2 with respect to y.In the quotient form fxx is written as d f dxdxIn the quotient form fxy is written as d2f dydx
  64. 64. Partial DerivativesWe also write df as fx and df as dx dyIts easy to talk about highery . f derivatives with thisnotation. So, fxx means to take the partial derivativewith respect to x twice, fxy means to to take thederivative with respect to x first, then2 with respect to y.In the quotient form fxx is written as d f dxdxIn the quotient form fxy is written as d2f dydx Note the reversed orders in these notations
  65. 65. Partial DerivativesWe also write df as fx and df as dx dyIts easy to talk about highery . f derivatives with thisnotation. So, fxx means to take the partial derivativewith respect to x twice, fxy means to to take thederivative with respect to x first, then2 with respect to y.In the quotient form fxx is written as d f dxdxIn the quotient form fxy is written as d2f dydx Note the reversed orders in these notationsHence fxyy = d f and that f = d f . 3 3 dydydx yxx dxdxdy
  66. 66. Partial DerivativesWe also write df as fx and df as dx dyIts easy to talk about highery . f derivatives with thisnotation. So, fxx means to take the partial derivativewith respect to x twice, fxy means to to take thederivative with respect to x first, then2 with respect to y.In the quotient form fxx is written as d f dxdxIn the quotient form fxy is written as d2f dydx Note the reversed orders in these notationsHence fxyy = d f and that f = d f . 3 3 dydydx yxx dxdxdyIn most cases, the order of the partial is notimportant.
  67. 67. Partial DerivativesWe also write df as fx and df as dx dyIts easy to talk about highery . f derivatives with thisnotation. So, fxx means to take the partial derivativewith respect to x twice, fxy means to to take thederivative with respect to x first, then2 with respect to y.In the quotient form fxx is written as d f dxdxIn the quotient form fxy is written as d2f dydx Note the reversed orders in these notationsHence fxyy = d f and that f = d f . 3 3 dydydx yxx dxdxdyIn most cases, the order of the partial is notimportant.Theorem: If fxy and fyx exist and are continuous in a
  68. 68. Partial DerivativesAgain, for elementary functions, we have f xy = fyxfor all points inside the domain (may not be truefor points on the boundary of the domain.)
  69. 69. Partial DerivativesAgain, for elementary functions, we have f xy = fyxfor all points inside the domain (may not be truefor points on the boundary of the domain.)Example D. Find fxx, fyy, fxy, and fyx if f(x, y) = x2y – 2x3y.
  70. 70. Partial DerivativesAgain, for elementary functions, we have f xy = fyxfor all points inside the domain (may not be truefor points on the boundary of the domain.)Example D. Find fxx, fyy, fxy, and fyx if f(x, y) = x2y – 2x3y.The first order partials are:fx = 2xy – 6x2y,
  71. 71. Partial DerivativesAgain, for elementary functions, we have f xy = fyxfor all points inside the domain (may not be truefor points on the boundary of the domain.)Example D. Find fxx, fyy, fxy, and fyx if f(x, y) = x2y – 2x3y.The first order partials are:fx = 2xy – 6x2y, fy = x2 – 2x3
  72. 72. Partial DerivativesAgain, for elementary functions, we have f xy = fyxfor all points inside the domain (may not be truefor points on the boundary of the domain.)Example D. Find fxx, fyy, fxy, and fyx if f(x, y) = x2y – 2x3y.The first order partials are:fx = 2xy – 6x2y, fy = x2 – 2x3Hence,fxx = 2y – 12xy,
  73. 73. Partial DerivativesAgain, for elementary functions, we have f xy = fyxfor all points inside the domain (may not be truefor points on the boundary of the domain.)Example D. Find fxx, fyy, fxy, and fyx if f(x, y) = x2y – 2x3y.The first order partials are:fx = 2xy – 6x2y, fy = x2 – 2x3Hence,fxx = 2y – 12xy, fyy = 0
  74. 74. Partial DerivativesAgain, for elementary functions, we have f xy = fyxfor all points inside the domain (may not be truefor points on the boundary of the domain.)Example D. Find fxx, fyy, fxy, and fyx if f(x, y) = x2y – 2x3y.The first order partials are:fx = 2xy – 6x2y, fy = x2 – 2x3Hence,fxx = 2y – 12xy, fyy = 0fxy = 2x – 6x2 = fyx

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