The document discusses partial derivatives. It defines a partial derivative as the slope of a curve intersecting a surface at a point, where the curve is obtained by fixing one of the variables in the surface equation. The partial derivative with respect to x is the slope of the curve intersecting when y is fixed, and vice versa for the partial derivative with respect to y. Examples are provided to demonstrate calculating partial derivatives algebraically and finding equations of tangent lines using partial derivatives.

1. Partial Derivatives

2. Partial Derivatives
Let p = (a, b, c) be a point on the surface z = f(x, y).

3. Partial Derivatives
Let p = (a, b, c) be a point on the surface z = f(x, y).
The plane y = b intersects the surface

4. Partial Derivatives
Let 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. Partial Derivatives
Let p = (a, b, c) be a point on the surface z = f(x, y).
The plane y = b intersects the surface at the curve
z = f(x, b)
y=b
z
z
y=b
z = f(x, b) (a, c)
(a, b, c)
x
y
x

6. Partial Derivatives
Let p = (a, b, c) be a point on the surface z = f(x, y).
The plane y = b intersects the surface at the curve
z = 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. Partial Derivatives
Let p = (a, b, c) be a point on the surface z = f(x, y).
The plane y = b intersects the surface at the curve
z = 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. Partial Derivatives
Let p = (a, b, c) be a point on the surface z = f(x, y).
The plane y = b intersects the surface at the curve
z = 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 called
the 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. Partial Derivatives
Let p = (a, b, c) be a point on the surface z = f(x, y).
The plane y = b intersects the surface at the curve
z = 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 called
the 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. Partial Derivatives
Let p = (a, b, c) be a point on the surface z = f(x, y).
The plane y = b intersects the surface at the curve
z = 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 called
the 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. Partial Derivatives
dz
Algebraically, to find d x , we treat y as a constant
and apply the derivative with respect to x.

12. Partial Derivatives
dz
Algebraically, to find d x , we treat y as a constant
and apply the derivative with respect to x.
Example A. Find dx dz
a. z = x + y b. z = xy
c. z = x2y d. z = xy2
y
e. z = x
y
f. z =
x

13. Partial Derivatives
dz
Algebraically, to find d x , we treat y as a constant
and apply the derivative with respect to x.
Example A. Find dx dz
a. z = x + y dz = 1 b. z = xy
dx
because y' = 0.
c. z = x2y d. z = xy2
y
e. z = x
y
f. z =
x

14. Partial Derivatives
dz
Algebraically, to find d x , we treat y as a constant
and apply the derivative with respect to x.
Example A. Find dx dz
a. z = x + y dz = 1 b. z = xy dz = y
dx dx
because y' = 0. because (cx)' = c.
c. z = x2y d. z = xy2
y
e. z = x
y
f. z =
x

15. Partial Derivatives
dz
Algebraically, to find d x , we treat y as a constant
and apply the derivative with respect to x.
Example A. Find dx dz
a. z = x + y dz = 1 b. z = xy dz = y
dx dx
because y' = 0. because (cx)' = c.
dz
c. z = x y
2
dx = 2xy d. z = xy2
because (xn)' = nxn–1.
x y
e. z = y f. z =
x

16. Partial Derivatives
dz
Algebraically, to find d x , we treat y as a constant
and apply the derivative with respect to x.
Example A. Find dx dz
a. z = x + y dz = 1 b. z = xy dz = y
dx dx
because y' = 0. because (cx)' = c.
dz dz
c. z = x y
2
dx = 2xy d. z = xy 2
dx = y2
because (xn)' = nxn–1. because (cx)' = c.
x y
e. z = y f. z =
x

17. Partial Derivatives
dz
Algebraically, to find d x , we treat y as a constant
and apply the derivative with respect to x.
Example A. Find dx dz
a. z = x + y dz = 1 b. z = xy dz = y
dx dx
because y' = 0. because (cx)' = c.
dz dz
c. z = x y
2
dx = 2xy d. z = xy 2
dx = y2
because (xn)' = nxn–1. because (cx)' = c.
x dz 1 y
e. z = y dx = y f. z =
x
because (cx)' = c.

18. Partial Derivatives
dz
Algebraically, to find d x , we treat y as a constant
and apply the derivative with respect to x.
Example A. Find dx dz
a. z = x + y dz = 1 b. z = xy dz = y
dx dx
because y' = 0. because (cx)' = c.
dz dz
c. z = x y
2
dx = 2xy d. z = xy 2
dx = y2
because (xn)' = nxn–1. because (cx)' = c.
x dz 1 y dz
e. z = y dx = y f. z = dx = –yx
–2
x
because (cx)' = c. The Quotient Rule

19. Partial Derivatives
dz
Algebraically, to find d x , we treat y as a constant
and apply the derivative with respect to x.
Example A. Find dx dz
a. z = x + y dz = 1 b. z = xy dz = y
dx dx
because y' = 0. because (cx)' = c.
dz dz
c. z = x y
2
dx = 2xy d. z = xy 2
dx = y2
because (xn)' = nxn–1. because (cx)' = c.
x dz 1 y dz
e. z = y dx = y f. z = dx = –yx
–2
x
because (cx)' = c. The Quotient Rule
The plane y = b is parallel to the x–axis, so the partial
dz/dx is also referred to as the partial derivative in the
x–direction (as in the compass direction).

20. Partial Derivatives
Example B. Find an equation for the tangent line for
at (2, 3, 6) on the surface of z = f(x, y) = √49 – x2 – y2
in the direction of the x-axis.

21. Partial Derivatives
Example B. Find an equation for the tangent line for
at (2, 3, 6) on the surface of z = f(x, y) = √49 – x2 – y2
in 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. Partial Derivatives
Example B. Find an equation for the tangent line for
at (2, 3, 6) on the surface of z = f(x, y) = √49 – x2 – y2
in the direction of the x-axis.
z
y=3
(2,3,6)
y
x

23. Partial Derivatives
Example B. Find an equation for the tangent line for
at (2, 3, 6) on the surface of z = f(x, y) = √49 – x2 – y2
in the direction of the x-axis.
z
y=3
(2,3,6)
y
x

24. Partial Derivatives
Example B. Find an equation for the tangent line for
at (2, 3, 6) on the surface of z = f(x, y) = √49 – x2 – y2
in the direction of the x-axis.
We need df/ dx|(2,3).
z
y=3
(2,3,6)
y
x

25. Partial Derivatives
Example B. Find an equation for the tangent line for
at (2, 3, 6) on the surface of z = f(x, y) = √49 – x2 – y2
in 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. Partial Derivatives
Example B. Find an equation for the tangent line for
at (2, 3, 6) on the surface of z = f(x, y) = √49 – x2 – y2
in 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. Partial Derivatives
Example B. Find an equation for the tangent line for
at (2, 3, 6) on the surface of z = f(x, y) = √49 – x2 – y2
in 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. Partial Derivatives
Example B. Find an equation for the tangent line for
at (2, 3, 6) on the surface of z = f(x, y) = √49 – x2 – y2
in 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. Partial Derivatives
Example B. Find an equation for the tangent line for
at (2, 3, 6) on the surface of z = f(x, y) = √49 – x2 – y2
in 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 a
directions 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. Partial Derivatives
Example B. Find an equation for the tangent line for
at (2, 3, 6) on the surface of z = f(x, y) = √49 – x2 – y2
in 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 a
directions vector for the tangent is v = <1, 0, –1/3> and
an 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. Partial Derivatives
The geometry of the partial derivative with respect to y
is similar to partial derivative with respect to x.

32. Partial Derivatives
The geometry of the partial derivative with respect to y
is 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 is
the graph of the equation z = f(a, y), in the plane x = a.
x=a
(a,b,c)
y
x

33. Partial Derivatives
The geometry of the partial derivative with respect to y
is 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 is
the 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. Partial Derivatives
The geometry of the partial derivative with respect to y
is 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 is
the 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. Partial Derivatives
The geometry of the partial derivative with respect to y
is 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 is
the 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. Partial Derivatives
dz
To find d y , treat x as a constant in the formula.

37. Partial Derivatives
dz
To find d y , treat x as a constant in the formula.
dz dz
Example C. a. Find dx , dy if z = cos(x2y).

38. Partial Derivatives
dz
To find d y , treat x as a constant in the formula.
dz dz
Example C. a. Find dx , dy if z = cos(x2y).
We need to use the chain–rule for both partials.

39. Partial Derivatives
dz
To find d y , treat x as a constant in the formula.
dz dz
Example 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. Partial Derivatives
dz
To find d y , treat x as a constant in the formula.
dz dz
Example 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. Partial Derivatives
dz
To find d y , treat x as a constant in the formula.
dz dz
Example 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. Partial Derivatives
dz
To find d y , treat x as a constant in the formula.
dz dz
Example 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. Partial Derivatives
dz
To find d y , treat x as a constant in the formula.
dz dz
Example 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
b. Find dz , dz at the point P = (1, π/2, 0)
dx dy

44. Partial Derivatives
dz
To find d y , treat x as a constant in the formula.
dz dz
Example 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
b. Find dz , dz at the point P = (1, π/2, 0)
dx dy
dz
dx |(1,π/2, 0)

45. Partial Derivatives
dz
To find d y , treat x as a constant in the formula.
dz dz
Example 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
b. 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. Partial Derivatives
dz
To find d y , treat x as a constant in the formula.
dz dz
Example 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
b. 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. Partial Derivatives
dz
To find d y , treat x as a constant in the formula.
dz dz
Example 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
b. 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. Partial Derivatives
Given a function of one variable, we say the function is
"differentiable" at a point P if the derivative exists at P.

49. Partial Derivatives
Given 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. Partial Derivatives
Given 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. Partial Derivatives
Given 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) doesn't exist

52. Partial Derivatives
Given 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) doesn't exist
However, for a function f(x, y) of two variables
(or more), the existence of the partial derivatives
doesn't mean the surface z = f(x, y) is smooth,
it may have creases (i.e. folds) even

53. Partial Derivatives
Given 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) doesn't exist
However, for a function f(x, y) of two variables
(or more), the existence of the partial derivatives
doesn't mean the surface z = f(x, y) is smooth,
it may have creases (i.e. folds) even
if the partials exist.

54. Partial Derivatives
Given 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) doesn't exist
However, for a function f(x, y) of two variables
(or more), the existence of the partial derivatives
doesn't mean the surface z = f(x, y) is smooth,
it may have creases (i.e. folds) even P
y
if the partials exist. For example,
the surface shown here has dx dz |P = dz |P = 0, x
dy
but the surface is folded, i.e. not smooth at P.

55. 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,
z
p
y
x
(a, b)

56. 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,
z
p
y
x
(a, b)

57. 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.
z
p
y
x
(a, b)

58. 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)

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 circle
that centered at P is in the domain, that all the partials
exist in this circle then the surface is smooth at P.

60. Partial Derivatives
We also write df as fx and df as
dx dy
fy .

61. Partial Derivatives
We also write df as fx and df as
dx dy
Its easy to talk about highery .
f derivatives with this
notation. So, fxx means to take the partial derivative
with respect to x twice, fxy means to to take the
derivative with respect to x first, then with respect to y.

62. Partial Derivatives
We also write df as fx and df as
dx dy
Its easy to talk about highery .
f derivatives with this
notation. So, fxx means to take the partial derivative
with respect to x twice, fxy means to to take the
derivative with respect to x first, then2 with respect to y.
In the quotient form fxx is written as d f
dxdx

63. Partial Derivatives
We also write df as fx and df as
dx dy
Its easy to talk about highery .
f derivatives with this
notation. So, fxx means to take the partial derivative
with respect to x twice, fxy means to to take the
derivative with respect to x first, then2 with respect to y.
In the quotient form fxx is written as d f
dxdx
In the quotient form fxy is written as d2f
dydx

64. Partial Derivatives
We also write df as fx and df as
dx dy
Its easy to talk about highery .
f derivatives with this
notation. So, fxx means to take the partial derivative
with respect to x twice, fxy means to to take the
derivative with respect to x first, then2 with respect to y.
In the quotient form fxx is written as d f
dxdx
In the quotient form fxy is written as d2f
dydx
Note the reversed orders in these notations

65. Partial Derivatives
We also write df as fx and df as
dx dy
Its easy to talk about highery .
f derivatives with this
notation. So, fxx means to take the partial derivative
with respect to x twice, fxy means to to take the
derivative with respect to x first, then2 with respect to y.
In the quotient form fxx is written as d f
dxdx
In the quotient form fxy is written as d2f
dydx
Note the reversed orders in these notations
Hence fxyy = d f and that f = d f .
3 3
dydydx yxx dxdxdy

66. Partial Derivatives
We also write df as fx and df as
dx dy
Its easy to talk about highery .
f derivatives with this
notation. So, fxx means to take the partial derivative
with respect to x twice, fxy means to to take the
derivative with respect to x first, then2 with respect to y.
In the quotient form fxx is written as d f
dxdx
In the quotient form fxy is written as d2f
dydx
Note the reversed orders in these notations
Hence fxyy = d f and that f = d f .
3 3
dydydx yxx dxdxdy
In most cases, the order of the partial is not
important.

67. Partial Derivatives
We also write df as fx and df as
dx dy
Its easy to talk about highery .
f derivatives with this
notation. So, fxx means to take the partial derivative
with respect to x twice, fxy means to to take the
derivative with respect to x first, then2 with respect to y.
In the quotient form fxx is written as d f
dxdx
In the quotient form fxy is written as d2f
dydx
Note the reversed orders in these notations
Hence fxyy = d f and that f = d f .
3 3
dydydx yxx dxdxdy
In most cases, the order of the partial is not
important.
Theorem: If fxy and fyx exist and are continuous in a

68. Partial Derivatives
Again, for elementary functions, we have f xy = fyx
for all points inside the domain (may not be true
for points on the boundary of the domain.)

69. Partial Derivatives
Again, for elementary functions, we have f xy = fyx
for all points inside the domain (may not be true
for points on the boundary of the domain.)
Example D. Find fxx, fyy, fxy, and fyx if f(x, y) = x2y – 2x3y.

70. Partial Derivatives
Again, for elementary functions, we have f xy = fyx
for all points inside the domain (may not be true
for 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. Partial Derivatives
Again, for elementary functions, we have f xy = fyx
for all points inside the domain (may not be true
for 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. Partial Derivatives
Again, for elementary functions, we have f xy = fyx
for all points inside the domain (may not be true
for 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
Hence,
fxx = 2y – 12xy,

73. Partial Derivatives
Again, for elementary functions, we have f xy = fyx
for all points inside the domain (may not be true
for 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
Hence,
fxx = 2y – 12xy, fyy = 0

74. Partial Derivatives
Again, for elementary functions, we have f xy = fyx
for all points inside the domain (may not be true
for 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
Hence,
fxx = 2y – 12xy, fyy = 0
fxy = 2x – 6x2 = fyx