Section 12.3: Partial Derivatives
It is often useful to know how a function of several variables changes with respect to one
of its variables.
Definition: If f is a function of two variables, its partial derivatives with respect to x
and y are
f (x + h, y) − f (x, y)
,
h
f (x, y + h) − f (x, y)
,
fy (x, y) = lim
h→0
h
fx (x, y) = lim
h→0
provided these limits exist.
Example: Find the partial derivatives of f (x, y) = x2 − 4xy + y 2 using the limit definition.
Using the definition,
f (x + h, y) − f (x, y)
h→0
h
(x + h)2 − 4(x + h)y + y 2 − (x2 − 4xy + y 2 )
lim
h→0
h
2
2
x + 2xh + h − 4xy − 4hy + y 2 − (x2 − 4xy + y 2 )
lim
h→0
h
2xh + h2 − 4hy
lim
h→0
h
lim (2x + h − 4y)
fx (x, y) = lim
=
=
=
=
h→0
= 2x − 4y.
Similarly,
f (x, y + h) − f (x, y)
h→0
h
2
x − 4x(y + h) + (y + h)2 − (x2 − 4xy + y 2 )
lim
h→0
h
x2 − 4xy − 4xh + y 2 + 2yh + h2 − (x2 − 4xy + y 2 )
lim
h→0
h
2
−4xh + 2yh + h
lim
h→0
h
lim (−4x + 2y + h)
fy (x, y) = lim
=
=
=
=
h→0
= −4x + 2y.
How to find the partial derivatives of f (x, y):
1. To find fx (x, y), treat y as a constant and differentiate f with respect to x.
2. To find fy (x, y), treat x as a constant and differentiate f with respect to y.
Notation: If z = f (x, y), the partial derivatives of f with respect to x and y are denoted by
∂z
∂f
=
= zx = Dx f,
∂x
∂x
∂f
∂z
fy =
=
= zy = Dy f.
∂y
∂y
fx =
Example: If f (x, y) = x3 + x2 y + xy 2 − 3y 3 , find fx (x, y) and fy (x, y).
The partial derivatives are
fx (x, y) = 3x2 + 2xy + y 2 ,
fy (x, y) = x2 + 2xy − 9y 2 .
Example: If z = xexy + sin(3x2 + y 3 ), find zx and zy .
The partial derivatives are
zx = exy + xyexy + 6x cos(3x2 + y 3 ),
zy = x2 exy + 3y 2 cos(3x2 + y 3 ).
Example: Find zx and zy if z is defined implicitly by the equation
x2 z + yz 3 = x + y 2 .
Differentiating with respect to x,
2xz + x2 zx + 3yz 2 zx = 1
zx (x2 + 3yz 2 ) = 1 − 2xz
1 − 2xz
zx = 2
.
x + 3yz 2
Differentiating with respect to y,
x2 zy + z 3 + 3yz 2 zy = 2y
zy (x2 + 3yz 2 ) = 2y − z 3
2y − z 3
zy = 2
.
x + 3yz 2
Note: The concept of partial derivatives can be extended to functions of three or more
variables.
Example: If f (x, y, z) = x5 + x4 y 4 z 3 + yz 2 , find fx , fy , and fz .
The partial derivatives are
fx (x, y, z) = 5x4 + 4x3 y 4 z 3 ,
fy (x, y, z) = 4x4 y 3 z 3 + z 2 ,
fz (x, y, z) = 3x4 y 4 z 2 + 2yz.
Example: If f (x, y, z) = ln(x2 y + y 2 z + 2xz 3 ), find fx , fy , and fz .
The partial derivatives are
2xy + 2z 3
,
x2 y + y 2 z + 2xz 3
x2 + 2yz
fy (x, y, z) = 2
,
x y + y 2 z + 2xz 3
y 2 + 6xz 2
.
fz (x, y, z) = 2
x y + y 2 z + 2xz 3
fx (x, y, z) =
Geometric Interpretation of Partial Derivatives:
The equation z = f (x, y) defines a surface S. If f (x0 , y0 ) = z0 , then the point (x0 , y0 , z0 )
lies on S. The vertical plane x = x0 intersects S in a curve C1 parallel to the yz-plane.
An equation for this curve is z = f (x0 , y). The slope of the tangent line to this curve at
(x0 , y0 , z0 ) is fy (x0 , y0 ). Similarly, the vertical plane y = y0 intersects S in a curve C2 parallel
to the xz-plane. An equation for C2 is z = f (x, y0 ). The slope of the tangent line to C2 at
(x0 , y0 , z0 ) is fx (x0 , y0 ). Thus, partial derivatives can be interpreted as rates of change in the
x and y direction.
Figure 1: Geometric interpretation of partial derivatives.
Definition: If f is a function of two variables, then the second-order partial derivatives
of f are
∂ ∂f
∂ 2f
fxx = (fx )x =
=
,
∂x ∂x
∂x2
∂ ∂f
∂ 2f
fyy = (fy )y =
=
,
∂y ∂y
∂y 2
∂ 2f
∂ ∂f
fxy = (fx )y =
=
,
∂y ∂x
∂y∂x
∂ 2f
∂ ∂f
=
.
fyx = (fy )x =
∂x ∂y
∂x∂y
Example: Find the second-order partial derivatives of f (x, y) = x2 y + sin(x + y).
The partial derivatives are
fx (x, y) = 2xy + cos(x + y)
fxx (x, y) = 2y − sin(x + y)
fxy (x, y) = 2x − sin(x + y)
fy (x, y) = x2 + cos(x + y)
fyy (x, y) = − sin(x + y)
fyx (x, y) = 2x − sin(x + y).
Theorem: (Clairaut’s Theorem)
Suppose f is defined on a disk D containing (a, b). If fxy and fyx are continuous on D, then
fxy (a, b) = fyx (a, b).
Example: Prove that fxy = fyx for f (x, y) = tan−1 (x2 y).
The partial derivatives are
2xy
fx (x, y) =
1 + x4 y 2
2x(1 + x4 y 2 ) − 2xy(2x4 y)
fxy (x, y) =
(1 + x4 y 2 )2
2x(1 + x4 y 2 − 4x5 y 2
fxy (x, y) =
(1 + x4 y 2 )2
x2
fy (x, y) =
1 + x4 y 2
2x(1 + x4 y 2 ) − x2 (4x3 y 2 )
fyx (x, y) =
(1 + x4 y 2 )2
2x(1 + x4 y 2 − 4x5 y 2
fyx (x, y) =
.
(1 + x4 y 2 )2
Thus, fxy = fyx .
Example: If f (x, y, z) = exyz , find fyzy .
The partial derivatives are
fy (x, y, z) = xzexyz
fyz (x, y, z) = xexyz + x2 yzexyz
fyzy (x, y, z) = x2 zexyz + x2 zexyz + x3 yz 2 exyz .
Definition: The equation
fxx + fyy = 0
is called Laplace’s equation. Functions that satisfy this equation are called harmonic.
Example: Show that f (x, y) = ex cos(y) is harmonic.
The partial derivatives are
fx (x, y) = ex cos y
fxx (x, y) = ex cos y
Then fxx + fyy = 0 and f is harmonic.
fy (x, y) = −ex sin y
fyy (x, y) = −ex cos y.