Partial Derivatives
Applied Mathematics
MAT538
1
Introduction
In this section, we will learn how to differentiate a function of several independent variables. Let
f
(
x, y
) be the
function of two independent variables, then the
firstorder partial derivatives
or
first partial derivatives
with respect to
x
and
y
at point (
x
0
, y
0
) can be denoted as
∂f
∂x
(
x
0
, y
0
) =
f
x
(
x
0
, y
0
)
,
and
∂f
∂y
(
x
0
, y
0
) =
f
y
(
x
0
, y
0
)
.
In partial derivative, when we differentiate the function with respect to any of independent variable, other
independent variable(s) will be treated as constant(s).
Example 1.
Find the value of
∂f
∂x
and
∂f
∂y
at the point (4
,

5) if
f
(
x, y
) =
x
2
+ 3
xy
+
y

1.
Solution
∂f
∂x
=
∂
∂x
(
x
2
+ 3
xy
+
y

1
)
;
=
2
x
+ 3(1)(
y
) + 0

0;
=
2
x
+ 3
y.
Thus
∂f
∂x
(4
,

5) =
f
x
(4
,

5) = 2(4) + 3(

5) =

7.
∂f
∂y
=
∂
∂y
(
x
2
+ 3
xy
+
y

1
)
;
=
0 + 3(
x
)(1) + 1

0;
=
3
x
+ 1
.
Thus
∂f
∂y
(4
,

5) =
f
y
(4
,

5) = 3(4) + 1 = 13.
Example 2.
Find
f
y
if
f
(
x, y
) =
y
sin(
xy
) + (
x
+
y
)
6
+
y
y
+ 1
.
Solution
f
y
=
∂
∂y
[
y
sin(
xy
)] +
∂
∂y
(
x
+
y
)
6
+
∂
∂y
y
y
+ 1
;
=
y
∂
∂y
[sin(
xy
)] + sin(
xy
)
∂
∂y
[
y
]

{z
}
product rule
+ 6(
x
+
y
)
5
∂
∂y
(
x
+
y
)

{z
}
chain rule
+
(
y
+ 1)
∂
∂y
(
y
)

(
y
)
∂
∂y
(
y
+ 1)
(
y
+ 1)
2

{z
}
quotient rule
;
=
y
cos(
xy
)(
x
) + sin(
xy
)(1) + 6(
x
+
y
)
5
(0 + 1) +
(
y
+ 1)(1)

(
y
)(1 + 0)
(
y
+ 1)
2
;
=
xy
cos(
xy
) + sin(
xy
) + 6(
x
+
y
)
5
+
1
(
y
+ 1)
2
.
Practice
Find all the first partial derivatives for the following functions
1.
f
(
x, y
) = 2
x
2

3
y

4.
ans
f
x
= 4
x, f
y
=

3
2.
f
(
x, y
) =
x/y
.
ans
f
x
=
1
y
, f
y
=

x
y
2
.
3.
f
(
x, y
) =
p
x
2
+
y
2
.
ans
f
x
=
x
p
x
2
+
y
2
, f
y
=
y
p
x
2
+
y
2
.
4.
f
(
x, y
) =
e
xy
ln(
y
).
ans
f
x
=
ye
xy
ln(
y
)
, f
y
=
e
xy
y
+
xe
xy
ln(
y
).
5.
f
(
x, y
) =
2
y

x
x
+
y
+ 1
,
(

1
,
1).
ans
f
x
=

4
, f
y
=

1.
Page 1 of 6
Partial Derivatives
Applied Mathematics
MAT538
1.1
Implicit Differentiation
Example 3.
Let
z
=
f
(
x, y
), find
∂z
∂x
and
∂z
∂y
for the equation
yz
2

ln(
z
) =
x
+
y
.
Solution
∂
∂x
yz
2

∂
∂x
[ln(
z
)]
=
∂
∂x
[
x
] +
∂
∂x
[
y
];
y
(2
z
)
∂z
∂x

1
z
∂z
∂x
=
1 + 0;
2
yz

1
z
∂z
∂x
=
1;
∂z
∂x
=
1
2
yz

1
z
.
∂
∂y
yz
2

∂
∂y
[ln(
z
)]
=
∂
∂y
[
x
] +
∂
∂y
[
y
];
y
(2
z
)
∂z
∂y
+
z
2
(1)

1
z
∂z
∂y
=
0 + 1;
2
yz

1
z
∂z
∂y
=
1

z
2
;
∂z
∂y
=
1

z
2
2
yz

1
z
.
Alternatively, we can use
Implicit Function Theorem
to do the differentiation. The theorem state that if
z
=
f
(
x, y
) is differentiable and
F
(
x, y, z
) = 0 then
z
x
=
∂z
∂x
=

∂F
∂x
/
∂F
∂z
=

F
x
F
z
,
z
y
=
∂z
∂y
=

∂F
∂y
/
∂F
∂z
=

F
y
F
z
.
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 Spring '17