Section 11.3
Partial Derivatives
(1) Tangent Lines on Surfaces,
(2) Partial Derivatives,
(3) Notation and Higher Order Derivatives.
MATH 122 (Section 11.3)
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Derivatives and Rates of Change
Recall that, for functions of one variable, y = f (x), the derivative f 0 (a) at
the point (a, f (a)) is the rate of change of values f (x) at x = a.
|f 0 (a)| is a measure of how fast the y -value changes near x = a.
For functions of two variables z = f (x, y ) and near a point (a, b), there
are infinitely many directions/ways that a point (x, y ) can vary near (a, b).
How do we characterize the “rate” of change of f (x, y ) at a point (a, b)?
MATH 122 (Section 11.3)
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It turns out the rate of change in two directions encodes the information
for the rate of change in any direction.
Let (a, b) be a point in the domain D of f (x, y ). If we fix y = b and vary
x, we get a function g (x) = f (x, b) one variable that lies on the surface of
f (x, y ).
The derivative of g 0 (a) is the slope of the tangent line to the graph of
g (x). The graph of g (x) is parameterized by (x, b, g (x)), so the vector
h1, 0, g 0 (a)i is a tangent vector to the graph.
The derivative g 0 (a) is the partial derivative of f (x, y ) with respect to
the x-variable at (a, b).
MATH 122 (Section 11.3)
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The partial derivative of f (x, y ) with respect to x at (a, b) is denoted
∂f
f (a + h, b) − f (a, b)
(a, b) = fx (a, b) = lim
h→0
∂x
h
The partial derivative of f (x, y ) with respect to y at (a, b) is denoted
∂f
f (a, b + h) − f (a, b)
(a, b) = fy (a, b) = lim
h→0
∂y
h
MATH 122 (Section 11.3)
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To compute the partial derivative with respect to x, fx , treat the
y -variable as a constant and apply the ordinary rules for differentiation.
Partial differentiation is not implicit differentiation.
Compute the partial derivative with respect to y , fy , in an analogous way.
Example: z = xy + ye xy
fx (x, y ) =
fy (x, y ) =
MATH 122 (Section 11.3)
∂z
(x, y ) = y + y 2 e xy
∂x
∂z
(x, y ) = x + (1 + xy )e xy
∂y
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The planes x = a and y = b intersect the surface z = f (x, y ) as curves
z = f (a, y ) and z = f (x, b) (respectively). The partial derivatives are the
slopes of the tangent lines to the two curves.
f (x, b) is the intersection of f and y = b. The tangent line to x = a
has point (a, b, f (a, b)) and directional vector h1, 0, fx (a, b)i.
f (a, y ) is the intersection of f and x = a. The tangent line to y = b
has point (a, b, f (a, b)) and directional vector h0, 1, fy (a, b)i.
Example: f (x, y ) = x(y + 1) − x 2 at (−2, 2).
fx (x, y ) = y + 1 − 2x
fy (x, y ) = x
The tangent line to the intersection The tangent line to the intersection
curve of f (x, y ) and the plane y = 2 curve of f (x, y ) and the plane
x = −2 at (−2, 2, −10) is
at (−2, 2, f (−2, 2)) is
r~x (t) = h−2 + t, 2, −10 + 7ti
MATH 122 (Section 11.3)
r~y (t) = h−2, 2 + t, −10 − 2ti
Partial Derivatives
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Implicit Partial Derivatives
∂z
∂z
and
implicitly from xy + xz + y 2 z − e yz = 1.
∂x
∂y
∂
through the equation:
Solution: Take the partial derivative
∂x
Find
∂
∂
∂ 2
∂ yz
∂
(xy ) +
(xz) +
(y z) −
(e ) =
(1)
∂x
∂x
∂x
∂x
∂x
Note that z = z(x, y ) is a function of (x, y ) implicitly. Thus
∂z
∂z
∂z
− e yz y
=0
y + z +x
+ y2
∂x
∂x
∂x
∂z
−(y + z)
=
∂x
x + y 2 − ye yz
In a similar way,
MATH 122 (Section 11.3)
∂z
ze yz − x − 2yz
=
∂y
x + y 2 − ye yz
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The extension of partial derivatives of more than two variables is
straightforward. In a three variable function t = f (x, y , z), the partial
derivative fz is calculated by differentiating with respect to z, treating x
and y as constants.
∂f ∂f
∂f
Example: Find
,
, and
for f (x, y , z) = z ln(x 2 + y 2 ) + sin(xz).
∂x ∂y
∂z
2xz
∂f
(x, y , z) = fx (x, y , z) = 2
+ z cos(xz)
∂x
x + y2
∂f
2yz
(x, y , z) = fy (x, y , z) = 2
∂y
x + y2
∂f
(x, y , z) = fz (x, y , z) = ln(x 2 + y 2 ) + x cos(xz)
∂z
MATH 122 (Section 11.3)
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Higher Order Partial Derivatives
For z = f (x, y ), the partial derivatives fx (x, y ) and fy (x, y ) are again
functions of (x, y ). We can define partial derivatives partial derivatives of
these functions; second order partial derivatives of f (x, y ).
∂
∂2
∂
f (x, y ) =
f (x, y ) = fxx (x, y )
∂x ∂x
∂x 2
∂
∂
∂2
f (x, y ) =
f (x, y ) = fyx (x, y )
∂x ∂y
∂x ∂y
∂
∂
∂2
f (x, y ) =
f (x, y ) = fxy (x, y )
∂y ∂x
∂y ∂x
∂
∂
∂2
f (x, y ) =
f (x, y ) = fyy (x, y )
∂y ∂y
∂y 2
In general,
∂nf
= fxn ...x2 x1
∂x1 ∂x2 . . . ∂xn
MATH 122 (Section 11.3)
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Compute fxx , fxy , fyx , fyy for
f (x, y ) = xe xy + y 2 − y (x 2 + 1)
Solution:
fx (x, y ) = e xy + xye xy − 2xy
fxx (x, y ) = 2ye xy + xy 2 e xy − 2y
fy (x, y ) = x 2 e xy + 2y − (x 2 + 1)
fxy (x, y ) = 2xe xy + x 2 ye xy − 2x
fxy (x, y ) = 2xe xy + x 2 ye xy − 2x
fyy (x, y ) = x 3 e xy + 2
Note: Here fxy = fyx .
If fxy (x, y ) and fyx (x, y ) are continuous, then fxy = fyx .
MATH 122 (Section 11.3)
Partial Derivatives
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