Section 10.3: Partial Derivatives
Definition: Suppose that f (x, y) is a function of two variables. The partial derivatives of
f with respect to x and y are defined by
f (x + h, y) − f (x, y)
,
h→0
h
f (x, y + h) − f (x, y)
fy (x, y) = lim
,
h→0
h
fx (x, y) = lim
provided these limits exist.
Example: Find the partial derivatives of f (x, y) = x2 − 4xy + y 2 using the limit definition.
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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: Find the partial derivatives fx (x, y) and fy (x, y) for each function.
(a) f (x, y) = x3 + x2 y + xy 2 − 3y 3
(b) f (x, y) = ln(3x2 − xy)
(c) f (x, y) = xexy + sin(3x2 + y 3 )
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Example: Find zx and zy if z is defined implicitly by the equation
x2 z + yz 3 = x + y 2 .
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 directions, respectively.
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Note: The definitions of partial derivatives can be extended to functions of three or more
variables.
Example: Find the partial derivatives fx , fy , and fz for each function.
(a) f (x, y, z) = x5 + x4 y 4 z 3 + yz 2
(b) f (x, y, z) = xy tan(x2 + z)
(c) f (x, y, z) = ln(x2 y + y 2 z + 2xz 3 )
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Definition: Suppose that f (x, y) is a function of two variables. The second-order partial
derivatives of f are
∂ ∂f
∂ 2f
fxx = (fx )x =
=
,
∂x ∂x
∂x2
∂ 2f
∂ ∂f
=
fyy = (fy )y =
,
∂y ∂y
∂y 2
∂ ∂f
∂ 2f
fxy = (fx )y =
=
,
∂y ∂x
∂y∂x
∂ ∂f
∂ 2f
fyx = (fy )x =
=
.
∂x ∂y
∂x∂y
The second-order partial derivatives fxy and fyx are called mixed partial derivatives.
Example: Find all second-order partial derivatives of f (x, y) = x2 y + sin(x + e3y ).
Example: Find the mixed partial derivatives of f (x, y) = ln(x2 + 3xy).
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Theorem: (The Mixed-Partials Theorem)
If f (x, y) and its partial derivatives fx , fy , fxy , and fyx are continuous on an open disk
centered at (x0 , y0 ), then
fxy (x0 , y0 ) = fyx (x0 , y0 ).
Example: Suppose that f (x, y) = tan−1 (x2 y). Show that fxy (x, y) = fyx (x, y) for all (x, y)
in the domain of f .
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