Partial Differentiation
1- Functions of two variables
Let D be a set of points (x,y) in the xy-plane
y
D
.(x,y)
x
Definition: A real-valued function of the variables x and y is a rule or correspondence
that assigns a unique real number, denoted by z = f(x,y), to each point (x,y) in D.
The set D is called the domain of the function.
The number z = f(x,y) is called the value of the function at (x,y).
The coordinates x and y of the point (x,y) are called the independent variables, while z is
called the dependent variable.
y
R
D
.(x,y)
x
Example: Let f(x, y) =
x
,
x!y
then
1
= 1,
1! 0
2
2 1
f(2, !2) =
= =
2 ! (!2) 4 2
1
f(1,1) =
!is!undefined
1!1
f(1, 0) =
> - z = f(x,y)
This function is undefined when x – y = 0 or x = y, this means that the domain of f is
! 2 \ {(x, y) : x = y} .
The graph of a function z = f(x,y) is a surface in the space R3
2. Partial Derivatives
Let z = f(x,y) a function of the variables x and y. If we hold the variable y fixed, say
y = y0 and view x as the only variable then we have f(x,yo) is a function of one variable.
If f(x,y0) is a differentiable function at x = x0, then the value of the derivative is denoted
by:
!f
fx (x 0 , y 0 ) =
(x 0 , y 0 )
!x
it is called the partial derivative of f with respect to x at (x0,y0).
Similarly, if we hold x fixed, say x = x0, then f(x0,y) is a function of y alone.
If we differentiate f(x0,y) at y = y0, then the value of this derivative is denoted by:
!f
fy (x 0 , y 0 ) = (x 0 , y 0 )
!y
it is the partial derivative of f with respect to y at (x0,y0).
They are also called first partial derivatives
Remark: To obtain fx(x,y), we differentiate f(x,y) with respect to x, treating y as a
constant.
Similarly, To obtain fy(x,y), we differentiate f(x,y) with respect to y, treating x as a
constant.
Examples: Find the partial derivatives of f(x,y) = 2x3y2 + 2y + 4x
To compute the partial derivative with respect to x, we treat y as a constant and we
consider f as a function of the variable x alone.
!f
fx (x, y) =
(x, y) = 6x 2 y 2 + 4
!x
To compute the partial derivative with respect to y, we treat x as a constant and we
consider f as a function of the variable y alone.
!f
(x, y) = 4x 3y + 2
!y
To evaluate the partial derivatives at the point (1,2), substitute x = 1 and y = 2
fx (1, 2) = 6(1)2 (2)2 + 4 = 28
fy (x, y) =
fy (1, 2) = 4(1)3 (2) + 2 = 10
Notation: If z = f(x,y), then
!z !f
!z !f
=
= fx !!and!! =
= fy
!x !x
!y !y
Example: z = f(x,y) = x4sin(xy3)
!z
= 4x 3 sin(xy 3 ) + x 4 cos(xy 3 )y 3
!x
!z
= x 4 cos(xy 3 )3xy 2
!y
Second–order partial derivative
If we take the partial derivatives of the first partial derivatives we obtain the second
partial derivatives.
! " !f % !2 f
= fxx second partial derivative of f with respect to x twice
$ '=
!x # !x & !x 2
! " !f % !2 f
=
= fyy second partial derivative of f with respect to y twice
!y $# !y '& !y 2
! " !f %
!2 f
=
= fyx
!x $# !y '& !x!y
! " !f %
!2 f
=
= fxy
$ '
!y # !x & !y!x
mixed second partial derivatives
Exercise: Compute the partial derivatives
1- f(x,y) = 3x3y4
2- f(x,y) = e2x sin(y)
3- f(x,y)= ln(x2 + y2)
4- z =(x + y) -1
5- f(x,y) = xe-y + 8x
6- f(x,y) = 3 xy 4 + arctan(xy)