Higher Order Derivatives
Implicit Differentiation
Local Linear Approximation and Differentials
Mathematics 53
Institute of Mathematics (UP Diliman)
For today
1
Higher Order Derivatives
2
Implicit Differentiation
3
Local Linear Approximation and Differentials
For today
1
Higher Order Derivatives
2
Implicit Differentiation
3
Local Linear Approximation and Differentials
Higher Order Derivatives
If
f
0
is differentiable, then the derivative of
f
0
is called the
second derivative
of
f
and is denoted
f
00
.
We can continue to obtain the
third derivative
,
f
000
, the
fourth derivative
,
f
(4)
,
and other
higher derivatives
.
Definition
The
n
-th derivative of the function
f
, denoted
f
(
n
)
, is the derivative of the
(
n
−
1)-th derivative of
f
, that is,
f
(
n
)
(
x
) =
lim
∆
x
→
0
f
(
n
−
1)
(
x
+
∆
x
)
−
f
(
n
−
1)
(
x
)
∆
x
Higher Order Derivatives
If
f
0
is differentiable,
then the derivative of
f
0
is called the
second derivative
of
f
and is denoted
f
00
.
We can continue to obtain the
third derivative
,
f
000
, the
fourth derivative
,
f
(4)
,
and other
higher derivatives
.
Definition
The
n
-th derivative of the function
f
, denoted
f
(
n
)
, is the derivative of the
(
n
−
1)-th derivative of
f
, that is,
f
(
n
)
(
x
) =
lim
∆
x
→
0
f
(
n
−
1)
(
x
+
∆
x
)
−
f
(
n
−
1)
(
x
)
∆
x
Higher Order Derivatives
If
f
0
is differentiable, then the derivative of
f
0
is called the
second derivative
of
f
and is denoted
f
00
.
We can continue to obtain the
third derivative
,
f
000
, the
fourth derivative
,
f
(4)
,
and other
higher derivatives
.
Definition
The
n
-th derivative of the function
f
, denoted
f
(
n
)
, is the derivative of the
(
n
−
1)-th derivative of
f
, that is,
f
(
n
)
(
x
) =
lim
∆
x
→
0
f
(
n
−
1)
(
x
+
∆
x
)
−
f
(
n
−
1)
(
x
)
∆
x
Higher Order Derivatives
If
f
0
is differentiable, then the derivative of
f
0
is called the
second derivative
of
f
and is denoted
f
00
.
We can continue to obtain the
third derivative
,
f
000
,
the
fourth derivative
,
f
(4)
,
and other
higher derivatives
.
Definition
The
n
-th derivative of the function
f
, denoted
f
(
n
)
, is the derivative of the
(
n
−
1)-th derivative of
f
, that is,
f
(
n
)
(
x
) =
lim
∆
x
→
0
f
(
n
−
1)
(
x
+
∆
x
)
−
f
(
n
−
1)
(
x
)
∆
x
Higher Order Derivatives
If
f
0
is differentiable, then the derivative of
f
0
is called the
second derivative
of
f
and is denoted
f
00
.
We can continue to obtain the
third derivative
,
f
000
, the
fourth derivative
,
f
(4)
,
and other
higher derivatives
.
Definition
The
n
-th derivative of the function
f
, denoted
f
(
n
)
, is the derivative of the
(
n
−
1)-th derivative of
f
, that is,
f
(
n
)
(
x
) =
lim
∆
x
→
0
f
(
n
−
1)
(
x
+
∆
x
)
−
f
(
n
−
1)
(
x
)
∆
x
Higher Order Derivatives
If
f
0
is differentiable, then the derivative of
f
0
is called the
second derivative
of
f
and is denoted
f
00
.
We can continue to obtain the
third derivative
,
f
000
, the
fourth derivative
,
f
(4)
,
and other
higher derivatives
.
Definition
The
n
-th derivative of the function
f
, denoted
f
(
n
)
, is the derivative of the
(
n
−
1)-th derivative of
f
, that is,
f
(
n
)
(
x
) =
lim
∆
x
→
0
f
(
n
−
1)
(
x
+
∆
x
)
−
f
(
n
−
1)
(
x
)
∆
x
Higher Order Derivatives
If
f
0
is differentiable, then the derivative of
f
0
is called the
second derivative
of
f
and is denoted
f
00
.
We can continue to obtain the
third derivative
,
f
000
, the
fourth derivative
,
f
(4)
,
and other
higher derivatives
.
Definition
The
n
-th derivative of the function
f
, denoted
f
(
n
)
, is the derivative of the
(
n
−
1)-th derivative of
f
,
that is,
f
(
n
)
(
x
) =
lim
∆
x
→
0
f
(
n
−
1)
(
x
+
∆
x
)
−
f
(
n
−
1)
(
x
)
∆
x
Higher Order Derivatives
If
f
0
is differentiable, then the derivative of
f
0
is called the
second derivative
of
f
and is denoted
f
00
.
We can continue to obtain the
third derivative
,
f
000
, the
fourth derivative
,
f
(4)
,
and other
higher derivatives
.
Definition
The
n
-th derivative of the function
f
, denoted
f
(
n
)
, is the derivative of the
(
n
−
1)-th derivative of
f
, that is,
f
(
n
)
(
x
) =
lim
∆
x
→
0
f
(
n
−
1)
(
x
+
∆
x
)
−
f
(
n
−
1)
(
x
)
∆
x
Higher Order Derivatives
Remarks
1
The
n
in
f
(
n
)
is called the order of the derivative.
2
The derivative of a function
f
is also called the first derivative of
f
.
3The function
f
is sometimes written as
f
(0)
(
x
).
4
Other notations:
D
x
n
[
f
(
x
)]
,
d
n
y
dx
n
,
d
n
dx
n
[
f
(
x
)]
,
y
(
n
)
Higher Order Derivatives
Remarks
1
The
n
in
f
(
n
)
is called the order of the derivative.
2
The derivative of a function
f
is also called the first derivative of
f
.
3
The function
f
is sometimes written as
f
(0)
(
x
).
4Other notations:
D
x
n
[
f
(
x
)]
,
d
n
y
dx
n
,
d
n
dx
n
[
f
(
x
)]
,
y
(
n
)
Higher Order Derivatives
Remarks
1
The
n
in
f
(
n
)
is called the order of the derivative.
2
The derivative of a function
f
is also called the first derivative of
f
.
3The function
f
is sometimes written as
f
(0)
(
x
).
4
Other notations:
D
x
n
[
f
(
x
)]
,
d
n
y
dx
n
,
d
n
dx
n
[
f
(
x
)]
,
y
(
n
)
Higher Order Derivatives
Remarks
1
The
n
in
f
(
n
)
is called the order of the derivative.
2
The derivative of a function
f
is also called the first derivative of
f
.
3The function
f
is sometimes written as
f
(0)
(
x
).
4
Other notations:
D
x
n
[
f
(
x
)]
,
d
n
y
dx
n
,
d
n
dx
n
[
f
(
x
)]
,
y
(
n
)
Higher Order Derivatives
Remarks
1
The
n
in
f
(
n
)
is called the order of the derivative.
2
The derivative of a function
f
is also called the first derivative of
f
.
3The function
f
is sometimes written as
f
(0)
(
x
).
4
Other notations:
D
x
n
[
f
(
x
)]
,
d
n
y
dx
n
,
d
n
dx
n
[
f
(
x
)]
,
y
(
n
)
Higher Order Derivatives
Remarks
1
The
n
in
f
(
n
)
is called the order of the derivative.
2
The derivative of a function
f
is also called the first derivative of
f
.
3The function
f
is sometimes written as
f
(0)
(
x
).
4
Other notations:
D
x
n
[
f
(
x
)]
,
d
n
y
dx
n
,
d
n
dx
n
[
f
(
x
)]
,
y
(
n
)
Higher Order Derivatives
Remarks
1
The
n
in
f
(
n
)
is called the order of the derivative.
2
The derivative of a function
f
is also called the first derivative of
f
.
3The function
f
is sometimes written as
f
(0)
(
x
).
4
Other notations:
D
x
n
[
f
(
x
)]
,
d
n
y
dx
n
,
d
n
dx
n
[
f
(
x
)]
,
y
(
n
)
Higher Order Derivatives
Remarks
1
The
n
in
f
(
n
)
is called the order of the derivative.
2
The derivative of a function
f
is also called the first derivative of
f
.
3The function
f
is sometimes written as
f
(0)
(
x
).
4
Other notations:
D
x
n
[
f
(
x
)]
,
d
n
y
dx
n
,
d
n
dx
n
[
f
(
x
)]
,
y
(
n
)
Higher Order Derivatives
Example
Find
f
(
n
)
(
x
)
for all
n
∈
N
where
f
(
x
) =
x
6
−
x
4
−
3
x
3
+
2
x
2
−
4.
Solution.
We differentiate repeatedly and obtain
f
0
(
x
)
=
6
x
5
−
4
x
3
−
9
x
2
+
4
x
,
f
00
(
x
)
=
30
x
4
−
12
x
2
−
18
x
+
4,
f
000
(
x
)
=
120
x
3
−
24
x
−
18,
f
(4)
(
x
)
=
360
x
2
−
24,
f
(5)
(
x
)
=
720
x
,
f
(6)
(
x
)
=
720,
f
(
n
)
(
x
)
=
0
for all
n
≥
7.
Higher Order Derivatives
Example
Find
f
(
n
)
(
x
)
for all
n
∈
N
where
f
(
x
) =
x
6
−
x
4
−
3
x
3
+
2
x
2
−
4.
Solution.
We differentiate repeatedly and obtain
f
0
(
x
)
=
6
x
5
−
4
x
3
−
9
x
2
+
4
x
,
f
00
(
x
)
=
30
x
4
−
12
x
2
−
18
x
+
4,
f
000
(
x
)
=
120
x
3
−
24
x
−
18,
f
(4)
(
x
)
=
360
x
2
−
24,
f
(5)
(
x
)
=
720
x
,
f
(6)
(
x
)
=
720,
f
(
n
)
(
x
)
=
0
for all
n
≥
7.
Higher Order Derivatives
Example
Find
f
(
n
)
(
x
)
for all
n
∈
N
where
f
(
x
) =
x
6
−
x
4
−
3
x
3
+
2
x
2
−
4.
Solution.
We differentiate repeatedly and obtain
f
0
(
x
)
=
6
x
5
−
4
x
3
−
9
x
2
+
4
x
,
f
00
(
x
)
=
30
x
4
−
12
x
2
−
18
x
+
4,
f
000
(
x
)
=
120
x
3
−
24
x
−
18,
f
(4)
(
x
)
=
360
x
2
−
24,
f
(5)
(
x
)
=
720
x
,
f
(6)
(
x
)
=
720,
f
(
n
)
(
x
)
=
0
for all
n
≥
7.
Higher Order Derivatives
Example
Find
f
(
n
)
(
x
)
for all
n
∈
N
where
f
(
x
) =
x
6
−
x
4
−
3
x
3
+
2
x
2
−
4.
Solution.
We differentiate repeatedly and obtain
f
0
(
x
)
=
6
x
5
−
4
x
3
−
9
x
2
+
4
x
,
f
00
(
x
)
=
30
x
4
−
12
x
2
−
18
x
+
4,
f
000
(
x
)
=
120
x
3
−
24
x
−
18,
f
(4)
(
x
)
=
360
x
2
−
24,
f
(5)
(
x
)
=
720
x
,
f
(6)
(
x
)
=
720,
f
(
n
)
(
x
)
=
0
for all
n
≥
7.
Higher Order Derivatives
Example
Find
f
(
n
)
(
x
)
for all
n
∈
N
where
f
(
x
) =
x
6
−
x
4
−
3
x
3
+
2
x
2
−
4.
Solution.
We differentiate repeatedly and obtain
f
0
(
x
)
=
6
x
5
−
4
x
3
−
9
x
2
+
4
x
,
f
00
(
x
)
=
30
x
4
−
12
x
2
−
18
x
+
4,
f
000
(
x
)
=
120
x
3
−
24
x
−
18,
f
(4)
(
x
)
=
360
x
2
−
24,
f
(5)
(
x
)
=
720
x
,
f
(6)
(
x
)
=
720,
f
(
n
)
(
x
)
=
0
for all
n
≥
7.
Higher Order Derivatives
Example
Find
f
(
n
)
(
x
)
for all
n
∈
N
where
f
(
x
) =
x
6
−
x
4
−
3
x
3
+
2
x
2
−
4.
Solution.
We differentiate repeatedly and obtain
f
0
(
x
)
=
6
x
5
−
4
x
3
−
9
x
2
+
4
x
,
f
00
(
x
)
=
30
x
4
−
12
x
2
−
18
x
+
4,
f
000
(
x
)
=
120
x
3
−
24
x
−
18,
f
(4)
(
x
)
=
360
x
2
−
24,
f
(5)
(
x
)
=
720
x
,
f
(6)
(
x
)
=
720,
f
(
n
)
(
x
)
=
0
for all
n
≥
7.
Higher Order Derivatives
Example
Find
f
(
n
)
(
x
)
for all
n
∈
N
where
f
(
x
) =
x
6
−
x
4
−
3
x
3
+
2
x
2
−
4.
Solution.
We differentiate repeatedly and obtain
f
0
(
x
)
=
6
x
5
−
4
x
3
−
9
x
2
+
4
x
,
f
00
(
x
)
=
30
x
4
−
12
x
2
−
18
x
+
4,
f
000
(
x
)
=
120
x
3
−
24
x
−
18,
f
(4)
(
x
)
=
360
x
2
−
24,
f
(5)
(
x
)
=
720
x
,
f
(6)
(
x
)
=
720,
f
(
n
)
(
x
)
=
0
for all
n
≥
7.
Higher Order Derivatives
Example
Find
f
(
n
)
(
x
)
for all
n
∈
N
where
f
(
x
) =
x
6
−
x
4
−
3
x
3
+
2
x
2
−
4.
Solution.
We differentiate repeatedly and obtain
f
0
(
x
)
=
6
x
5
−
4
x
3
−
9
x
2
+
4
x
,
f
00
(
x
)
=
30
x
4
−
12
x
2
−
18
x
+
4,
f
000
(
x
)
=
120
x
3
−
24
x
−
18,
f
(4)
(
x
)
=
360
x
2
−
24,
f
(5)
(
x
)
=
720
x
,
f
(6)
(
x
)
=
720,
f
(
n
)
(
x
)
=
0
for all
n
≥
7.
Higher Order Derivatives
Example
Find
f
(
n
)
(
x
)
for all
n
∈
N
where
f
(
x
) =
x
6
−
x
4
−
3
x
3
+
2
x
2
−
4.
Solution.
We differentiate repeatedly and obtain
f
0
(
x
)
=
6
x
5
−
4
x
3
−
9
x
2
+
4
x
,
f
00
(
x
)
=
30
x
4
−
12
x
2
−
18
x
+
4,
f
000
(
x
)
=
120
x
3
−
24
x
−
18,
f
(4)
(
x
)
=
360
x
2
−
24,
f
(5)
(
x
)
=
720
x
,
f
(6)
(
x
)
=
720,
f
(
n
)
(
x
)
=
0
for all
n
≥
7.
Higher Order Derivatives
Example
Find
f
(4)
(
x
)
if
f
(
x
) =
√
2
x
−
3.
Solution.
f
0
(
x
)
=
1
2
(2
x
−
3)
−
12
·
(2)
=
(2
x
−
3)
−
12f
00
(
x
)
=
−
1
2
(2
x
−
3)
−
32
·
(2)
=
−
(2
x
−
3)
−
32f
000
(
x
)
=
3
2
(2
x
−
3)
−
52
·
(2)
=
3
(2
x
−
3)
−
52f
(4)
(
x
)
=
−
15
2
(2
x
−
3)
−
72
·
(2)
=
−
15
(2
x
−
3)
−
72Higher Order Derivatives
Example
Find
f
(4)
(
x
)
if
f
(
x
) =
√
2
x
−
3.
Solution.
f
0
(
x
)
=
1
2
(2
x
−
3)
−
12
·
(2)
=
(2
x
−
3)
−
12f
00
(
x
)
=
−
1
2
(2
x
−
3)
−
32
·
(2)
=
−
(2
x
−
3)
−
32f
000
(
x
)
=
3
2
(2
x
−
3)
−
52
·
(2)
=
3
(2
x
−
3)
−
52f
(4)
(
x
)
=
−
15
2
(2
x
−
3)
−
72
·
(2)
=
−
15
(2
x
−
3)
−
72Higher Order Derivatives
Example
Find
f
(4)
(
x
)
if
f
(
x
) =
√
2
x
−
3.
Solution.
f
0
(
x
)
=
1
2
(2
x
−
3)
−
1 2·
(2)
=
(2
x
−
3)
−
12f
00
(
x
)
=
−
1
2
(2
x
−
3)
−
32
·
(2)
=
−
(2
x
−
3)
−
32f
000
(
x
)
=
3
2
(2
x
−
3)
−
52
·
(2)
=
3
(2
x
−
3)
−
52f
(4)
(
x
)
=
−
15
2
(2
x
−
3)
−
72
·
(2)
=
−
15
(2
x
−
3)
−
72Higher Order Derivatives
Example
Find
f
(4)
(
x
)
if
f
(
x
) =
√
2
x
−
3.
Solution.
f
0
(
x
)
=
1
2
(2
x
−
3)
−
12
·
(2)
=
(2
x
−
3)
−
12f
00
(
x
)
=
−
1
2
(2
x
−
3)
−
32
·
(2)
=
−
(2
x
−
3)
−
32f
000
(
x
)
=
3
2
(2
x
−
3)
−
52
·
(2)
=
3
(2
x
−
3)
−
52f
(4)
(
x
)
=
−
15
2
(2
x
−
3)
−
72
·
(2)
=
−
15
(2
x
−
3)
−
72Higher Order Derivatives
Example
Find
f
(4)
(
x
)
if
f
(
x
) =
√
2
x
−
3.
Solution.
f
0
(
x
)
=
1
2
(2
x
−
3)
−
12
·
(2)
=
(2
x
−
3)
−
12f
00
(
x
)
=
−
1
2
(2
x
−
3)
−
32
·
(2)
=
−
(2
x
−
3)
−
32f
000
(
x
)
=
3
2
(2
x
−
3)
−
52
·
(2)
=
3
(2
x
−
3)
−
52f
(4)
(
x
)
=
−
15
2
(2
x
−
3)
−
72
·
(2)
=
−
15
(2
x
−
3)
−
72Higher Order Derivatives
Example
Find
f
(4)
(
x
)
if
f
(
x
) =
√
2
x
−
3.
Solution.
f
0
(
x
)
=
1
2
(2
x
−
3)
−
12
·
(2)
=
(2
x
−
3)
−
12f
00
(
x
)
=
−
1
2
(2
x
−
3)
−
3 2·
(2)
=
−
(2
x
−
3)
−
32f
000
(
x
)
=
3
2
(2
x
−
3)
−
52
·
(2)
=
3
(2
x
−
3)
−
52f
(4)
(
x
)
=
−
15
2
(2
x
−
3)
−
72
·
(2)
=
−
15
(2
x
−
3)
−
72Higher Order Derivatives
Example
Find
f
(4)
(
x
)
if
f
(
x
) =
√
2
x
−
3.
Solution.
f
0
(
x
)
=
1
2
(2
x
−
3)
−
12
·
(2)
=
(2
x
−
3)
−
12f
00
(
x
)
=
−
1
2
(2
x
−
3)
−
32
·
(2)
=
−
(2
x
−
3)
−
32f
000
(
x
)
=
3
2
(2
x
−
3)
−
52
·
(2)
=
3
(2
x
−
3)
−
52f
(4)
(
x
)
=
−
15
2
(2
x
−
3)
−
72
·
(2)
=
−
15
(2
x
−
3)
−
72Higher Order Derivatives
Example
Find
f
(4)
(
x
)
if
f
(
x
) =
√
2
x
−
3.
Solution.
f
0
(
x
)
=
1
2
(2
x
−
3)
−
12
·
(2)
=
(2
x
−
3)
−
12f
00
(
x
)
=
−
1
2
(2
x
−
3)
−
32
·
(2)
=
−
(2
x
−
3)
−
32f
000
(
x
)
=
3
2
(2
x
−
3)
−
52
·
(2)
=
3
(2
x
−
3)
−
52f
(4)
(
x
)
=
−
15
2
(2
x
−
3)
−
72
·
(2)
=
−
15
(2
x
−
3)
−
72Higher Order Derivatives
Example
Find
f
(4)
(
x
)
if
f
(
x
) =
√
2
x
−
3.
Solution.
f
0
(
x
)
=
1
2
(2
x
−
3)
−
12
·
(2)
=
(2
x
−
3)
−
12f
00
(
x
)
=
−
1
2
(2
x
−
3)
−
32
·
(2)
=
−
(2
x
−
3)
−
32f
000
(
x
)
=
3
2
(2
x
−
3)
−
5 2·
(2)
=
3
(2
x
−
3)
−
52f
(4)
(
x
)
=
−
15
2
(2
x
−
3)
−
72
·
(2)
=
−
15
(2
x
−
3)
−
72Higher Order Derivatives
Example
Find
f
(4)
(
x
)
if
f
(
x
) =
√
2
x
−
3.
Solution.
f
0
(
x
)
=
1
2
(2
x
−
3)
−
12
·
(2)
=
(2
x
−
3)
−
12f
00
(
x
)
=
−
1
2
(2
x
−
3)
−
32
·
(2)
=
−
(2
x
−
3)
−
32f
000
(
x
)
=
3
2
(2
x
−
3)
−
52
·
(2)
=
3
(2
x
−
3)
−
52f
(4)
(
x
)
=
−
15
2
(2
x
−
3)
−
72
·
(2)
=
−
15
(2
x
−
3)
−
72Higher Order Derivatives
Example
Find
f
(4)
(
x
)
if
f
(
x
) =
√
2
x
−
3.
Solution.
f
0
(
x
)
=
1
2
(2
x
−
3)
−
12
·
(2)
=
(2
x
−
3)
−
12f
00
(
x
)
=
−
1
2
(2
x
−
3)
−
32
·
(2)
=
−
(2
x
−
3)
−
32f
000
(
x
)
=
3
2
(2
x
−
3)
−
52
·
(2)
=
3
(2
x
−
3)
−
52f
(4)
(
x
)
=
−
15
2
(2
x
−
3)
−
72
·
(2)
=
−
15
(2
x
−
3)
−
72Higher Order Derivatives
Example
Find
f
(4)
(
x
)
if
f
(
x
) =
√
2
x
−
3.
Solution.
f
0
(
x
)
=
1
2
(2
x
−
3)
−
12
·
(2)
=
(2
x
−
3)
−
12f
00
(
x
)
=
−
1
2
(2
x
−
3)
−
32
·
(2)
=
−
(2
x
−
3)
−
32f
000
(
x
)
=
3
2
(2
x
−
3)
−
52
·
(2)
=
3
(2
x
−
3)
−
52f
(4)
(
x
)
=
−
15
2
(2
x
−
3)
−
7 2·
(2)
=
−
15
(2
x
−
3)
−
72Higher Order Derivatives
Example
Find
f
(4)
(
x
)
if
f
(
x
) =
√
2
x
−
3.
Solution.
f
0
(
x
)
=
1
2
(2
x
−
3)
−
12
·
(2)
=
(2
x
−
3)
−
12f
00
(
x
)
=
−
1
2
(2
x
−
3)
−
32
·
(2)
=
−
(2
x
−
3)
−
32f
000
(
x
)
=
3
2
(2
x
−
3)
−
52
·
(2)
=
3
(2
x
−
3)
−
52f
(4)
(
x
)
=
−
15
2
(2
x
−
3)
−
72
·
(2)
=
−
15
(2
x
−
3)
−
72For today
1
Higher Order Derivatives
2
Implicit Differentiation
3
Local Linear Approximation and Differentials
Equations in two variables are used to define a function explicitly or implicitly.
•
y
=
x
2
−
1
defines
f
(
x
) =
x
2
−
1
(explicitly).
•
y
2
=
x
+
1
defines two functions of
x
: (implicitly)
f
1
(
x
) =
√
x
+
1
f
2
(
x
) =
−
√
x
+
1.
Equations in two variables are used to define a function explicitly or implicitly.
•
y
=
x
2
−
1
defines
f
(
x
) =
x
2
−
1
(explicitly).
•
y
2
=
x
+
1
defines two functions of
x
: (implicitly)
f
1
(
x
) =
√
x
+
1
f
2
(
x
) =
−
√
x
+
1.
Equations in two variables are used to define a function explicitly or implicitly.
•
y
=
x
2
−
1
defines
f
(
x
) =
x
2
−
1
(explicitly).
•
y
2
=
x
+
1
defines two functions of
x
: (implicitly)
f
1
(
x
) =
√
x
+
1
f
2
(
x
) =
−
√
x
+
1.
Equations in two variables are used to define a function explicitly or implicitly.
•
y
=
x
2
−
1
defines
f
(
x
) =
x
2
−
1
(explicitly).
•
y
2
=
x
+
1
defines two functions of
x
: (implicitly)
f
1
(
x
) =
√
x
+
1
f
2
(
x
) =
−
√
x
+
1.
Equations in two variables are used to define a function explicitly or implicitly.
•
y
=
x
2
−
1
defines
f
(
x
) =
x
2
−
1
(explicitly).
•
y
2
=
x
+
1
defines two functions of
x
: (implicitly)
f
1
(
x
) =
√
x
+
1
f
2
(
x
) =
−
√
x
+
1.
Equations in two variables are used to define a function explicitly or implicitly.
•
y
=
x
2
−
1
defines
f
(
x
) =
x
2
−
1
(explicitly).
•
y
2
=
x
+
1
defines two functions of
x
: (implicitly)
f
1
(
x
) =
√
x
+
1
f
2
(
x
) =
−
√
x
+
1.
Equations in two variables are used to define a function explicitly or implicitly.
•
y
=
x
2
−
1
defines
f
(
x
) =
x
2
−
1
(explicitly).
•
y
2
=
x
+
1
defines two functions of
x
: (implicitly)
f
1
(
x
) =
√
x
+
1
f
2
(
x
) =
−
√
x
+
1
.
Consider: If
x
3
−
2
xy
−
3
y
−
6
=
0,
how do we find
dy
dx
?
Solution
From
x
3
−
2
xy
−
3
y
−
6
=
0
, we can define
y
explicitly in terms of
x
as
y
=
x
3
−
6
2
x
+
3
.
Thus,
dy
dx
=
(2
x
+
3)(3
x
2
)
−
(
x
3
−
6)(2)
(2
x
+
3)
2
=
4
x
3
+
9
x
2
+
12
(2
x
+
3)
2
.
Consider: If
x
3
−
2
xy
−
3
y
−
6
=
0,
how do we find
dy
dx
?
Solution
From
x
3
−
2
xy
−
3
y
−
6
=
0
, we can define
y
explicitly in terms of
x
as
y
=
x
3
−
6
2
x
+
3
.
Thus,
dy
dx
=
(2
x
+
3)(3
x
2
)
−
(
x
3
−
6)(2)
(2
x
+
3)
2
=
4
x
3
+
9
x
2
+
12
(2
x
+
3)
2
.
Consider: If
x
3
−
2
xy
−
3
y
−
6
=
0,
how do we find
dy
dx
?
Solution
From
x
3
−
2
xy
−
3
y
−
6
=
0
, we can define
y
explicitly in terms of
x
as
y
=
x
3
−
6
2
x
+
3
.
Thus,
dy
dx
=
(2
x
+
3)(3
x
2
)
−
(
x
3
−
6)(2)
(2
x
+
3)
2
=
4
x
3
+
9
x
2
+
12
(2
x
+
3)
2
.
Consider: If
x
3
−
2
xy
−
3
y
−
6
=
0,
how do we find
dy
dx
?
Solution
From
x
3
−
2
xy
−
3
y
−
6
=
0
, we can define
y
explicitly in terms of
x
as
y
=
x
3
−
6
2
x
+
3
.
Thus,
dy
dx
=
(2
x
+
3)(3
x
2
)
−
(
x
3
−
6)(2)
(2
x
+
3)
2
=
4
x
3
+
9
x
2
+
12
(2
x
+
3)
2
.
Consider: If
x
3
−
2
xy
−
3
y
−
6
=
0,
how do we find
dy
dx
?
Solution
From
x
3
−
2
xy
−
3
y
−
6
=
0
, we can define
y
explicitly in terms of
x
as
y
=
x
3
−
6
2
x
+
3
.
Thus,
dy
dx
=
(2
x
+
3)(3
x
2
)
−
(
x
3
−
6)(2)
(2
x
+
3)
2
=
4
x
3
+
9
x
2
+
12
(2
x
+
3)
2
.
Consider: If
x
3
−
2
xy
−
3
y
−
6
=
0,
how do we find
dy
dx
?
Solution
From
x
3
−
2
xy
−
3
y
−
6
=
0
, we can define
y
explicitly in terms of
x
as
y
=
x
3
−
6
2
x
+
3
.
Thus,
dy
dx
=
(2
x
+
3)(3
x
2
)
−
(
x
3
−
6)(2)
(2
x
+
3)
2
=
4
x
3
+
9
x
2
+
12
(2
x
+
3)
2
.
Implicit Differentiation
Question:
Suppose we have
x
4
y
3
−
7
xy
=
7
or
tan(
x
2
−
2
xy
) =
y
.
How do we find
dy
dx
?
Implicit Differentiation
Question:
Suppose we have
x
4
y
3
−
7
xy
=
7
or
tan(
x
2
−
2
xy
) =
y
.
How do we find
dy
dx
?
Implicit Differentiation
To find
dy
dx
using implicit differentiation, we
1
think of the variable
y
as a differentiable function of the variable
x
,
2
differentiate both sides of the equation, using the chain rule where necessary,
and
3
solve for
dy
dx
.
Implicit Differentiation
To find
dy
dx
using implicit differentiation, we
1
think of the variable
y
as a differentiable function of the variable
x
,
2
differentiate both sides of the equation, using the chain rule where necessary,
and
3
solve for
dy
dx
.
Implicit Differentiation
To find
dy
dx
using implicit differentiation, we
1
think of the variable
y
as a differentiable function of the variable
x
,
2
differentiate both sides of the equation, using the chain rule where necessary,
and
3
solve for
dy
dx
.
Implicit Differentiation
To find
dy
dx
using implicit differentiation, we
1
think of the variable
y
as a differentiable function of the variable
x
,
2
differentiate both sides of the equation, using the chain rule where necessary,
and
3
solve for
dy
dx
.
Implicit Differentiation
Example
Find
dy
dx
if
x
3
−
2
xy
−
3
y
−
6
=
0
.
Solution.
D
x
[
x
3
−
2
xy
−
3
y
−
6]
=
D
x
[0]
D
x
[
x
3
]
−
D
x
[2
xy
]
−
D
x
[3
y
]
−
D
x
[6]
=
D
x
[0]
3
x
2
−
2
1
·
y
+
x
·
dy
dx
−
3
·
dy
dx
−
0
=
0
3
x
2
−
2
y
−
2
x
·
dy
dx
−
3
·
dy
dx
=
0
2
x
·
dy
dx
+
3
·
dy
dx
=
3
x
2
−
2
y
dy
dx
(2
x
+
3)
=
3
x
2
−
2
y
Implicit Differentiation
Example
Find
dy
dx
if
x
3
−
2
xy
−
3
y
−
6
=
0
.
Solution.
D
x
[
x
3
−
2
xy
−
3
y
−
6]
=
D
x
[0]
D
x
[
x
3
]
−
D
x
[2
xy
]
−
D
x
[3
y
]
−
D
x
[6]
=
D
x
[0]
3
x
2
−
2
1
·
y
+
x
·
dy
dx
−
3
·
dy
dx
−
0
=
0
3
x
2
−
2
y
−
2
x
·
dy
dx
−
3
·
dy
dx
=
0
2
x
·
dy
dx
+
3
·
dy
dx
=
3
x
2
−
2
y
dy
dx
(2
x
+
3)
=
3
x
2
−
2
y
Implicit Differentiation
Example
Find
dy
dx
if
x
3
−
2
xy
−
3
y
−
6
=
0
.
Solution.
D
x
[
x
3
−
2
xy
−
3
y
−
6]
=
D
x
[0]
D
x
[
x
3
]
−
D
x
[2
xy
]
−
D
x
[3
y
]
−
D
x
[6]
=
D
x
[0]
3
x
2
−
2
1
·
y
+
x
·
dy
dx
−
3
·
dy
dx
−
0
=
0
3
x
2
−
2
y
−
2
x
·
dy
dx
−
3
·
dy
dx
=
0
2
x
·
dy
dx
+
3
·
dy
dx
=
3
x
2
−
2
y
dy
dx
(2
x
+
3)
=
3
x
2
−
2
y
Implicit Differentiation
Example
Find
dy
dx
if
x
3
−
2
xy
−
3
y
−
6
=
0
.
Solution.
D
x
[
x
3
−
2
xy
−
3
y
−
6]
=
D
x
[0]
D
x
[
x
3
]
−
D
x
[2
xy
]
−
D
x
[3
y
]
−
D
x
[6]
=
D
x
[0]
3
x
2
−
2
1
·
y
+
x
·
dy
dx
−
3
·
dy
dx
−
0
=
0
3
x
2
−
2
y
−
2
x
·
dy
dx
−
3
·
dy
dx
=
0
2
x
·
dy
dx
+
3
·
dy
dx
=
3
x
2
−
2
y
dy
dx
(2
x
+
3)
=
3
x
2
−
2
y
Implicit Differentiation
Example
Find
dy
dx
if
x
3
−
2
xy
−
3
y
−
6
=
0
.
Solution.
D
x
[
x
3
−
2
xy
−
3
y
−
6]
=
D
x
[0]
D
x
[
x
3
]
−
D
x
[2
xy
]
−
D
x
[3
y
]
−
D
x
[6]
=
D
x
[0]
3
x
2
−
2
1
·
y
+
x
·
dy
dx
−
3
·
dy
dx
−
0
=
0
3
x
2
−
2
y
−
2
x
·
dy
dx
−
3
·
dy
dx
=
0
2
x
·
dy
dx
+
3
·
dy
dx
=
3
x
2
−
2
y
dy
dx
(2
x
+
3)
=
3
x
2
−
2
y
Implicit Differentiation
Example
Find
dy
dx
if
x
3
−
2
xy
−
3
y
−
6
=
0
.
Solution.
D
x
[
x
3
−
2
xy
−
3
y
−
6]
=
D
x
[0]
D
x
[
x
3
]
−
D
x
[2
xy
]
−
D
x
[3
y
]
−
D
x
[6]
=
D
x
[0]
3
x
2
−
2
1
·
y
+
x
·
dy
dx
−
3
·
dy
dx
−
0
=
0
3
x
2
−
2
y
−
2
x
·
dy
dx
−
3
·
dy
dx
=
0
2
x
·
dy
dx
+
3
·
dy
dx
=
3
x
2
−
2
y
dy
dx
(2
x
+
3)
=
3
x
2
−
2
y
Implicit Differentiation
Example
Find
dy
dx
if
x
3
−
2
xy
−
3
y
−
6
=
0
.
Solution.
D
x
[
x
3
−
2
xy
−
3
y
−
6]
=
D
x
[0]
D
x
[
x
3
]
−
D
x
[2
xy
]
−
D
x
[3
y
]
−
D
x
[6]
=
D
x
[0]
3
x
2
−
2
1
·
y
+
x
·
dy
dx
−
3
·
dy
dx
−
0
=
0
3
x
2
−
2
y
−
2
x
·
dy
dx
−
3
·
dy
dx
=
0
2
x
·
dy
dx
+
3
·
dy
dx
=
3
x
2
−
2
y
dy
dx
(2
x
+
3)
=
3
x
2
−
2
y
Implicit Differentiation
Example
Find
dy
dx
if
x
3
−
2
xy
−
3
y
−
6
=
0
.
Solution.
D
x
[
x
3
−
2
xy
−
3
y
−
6]
=
D
x
[0]
D
x
[
x
3
]
−
D
x
[2
xy
]
−
D
x
[3
y
]
−
D
x
[6]
=
D
x
[0]
3
x
2
−
2
1
·
y
+
x
·
dy
dx
−
3
·
dy
dx
−
0
=
0
3
x
2
−
2
y
−
2
x
·
dy
dx
−
3
·
dy
dx
=
0
2
x
·
dy
dx
+
3
·
dy
dx
=
3
x
2
−
2
y
dy
dx
(2
x
+
3)
=
3
x
2
−
2
y
Implicit Differentiation
Example
Find
dy
dx
if
x
3
−
2
xy
−
3
y
−
6
=
0
.
Solution.
D
x
[
x
3
−
2
xy
−
3
y
−
6]
=
D
x
[0]
D
x
[
x
3
]
−
D
x
[2
xy
]
−
D
x
[3
y
]
−
D
x
[6]
=
D
x
[0]
3
x
2
−
2
1
·
y
+
x
·
dy
dx
−
3
·
dy
dx
−
0
=
0
3
x
2
−
2
y
−
2
x
·
dy
dx
−
3
·
dy
dx
=
0
2
x
·
dy
dx
+
3
·
dy
dx
=
3
x
2
−
2
y
dy
dx
(2
x
+
3)
=
3
x
2
−
2
y
Implicit Differentiation
Example
Find
dy
dx
if
x
3
−
2
xy
−
3
y
−
6
=
0
.
Solution.
D
x
[
x
3
−
2
xy
−
3
y
−
6]
=
D
x
[0]
D
x
[
x
3
]
−
D
x
[2
xy
]
−
D
x
[3
y
]
−
D
x
[6]
=
D
x
[0]
3
x
2
−
2
1
·
y
+
x
·
dy
dx
−
3
·
dy
dx
−
0
=
0
3
x
2
−
2
y
−
2
x
·
dy
dx
−
3
·
dy
dx
=
0
2
x
·
dy
dx
+
3
·
dy
dx
=
3
x
2
−
2
y
dy
dx
(2
x
+
3)
=
3
x
2
−
2
y
Implicit Differentiation
Example
Find
dy
dx
if
x
3
−
2
xy
−
3
y
−
6
=
0
.
Solution.
D
x
[
x
3
−
2
xy
−
3
y
−
6]
=
D
x
[0]
D
x
[
x
3
]
−
D
x
[2
xy
]
−
D
x
[3
y
]
−
D
x
[6]
=
D
x
[0]
3
x
2
−
2
1
·
y
+
x
·
dy
dx
−
3
·
dy
dx
−
0
=
0
3
x
2
−
2
y
−
2
x
·
dy
dx
−
3
·
dy
dx
=
0
2
x
·
dy
dx
+
3
·
dy
dx
=
3
x
2
−
2
y
dy
dx
(2
x
+
3)
=
3
x
2
−
2
y
Implicit Differentiation
Example
Find
dy
dx
if
x
3
−
2
xy
−
3
y
−
6
=
0
.
Solution.
D
x
[
x
3
−
2
xy
−
3
y
−
6]
=
D
x
[0]
D
x
[
x
3
]
−
D
x
[2
xy
]
−
D
x
[3
y
]
−
D
x
[6]
=
D
x
[0]
3
x
2
−
2
1
·
y
+
x
·
dy
dx
−
3
·
dy
dx
−
0
=
0
3
x
2
−
2
y
−
2
x
·
dy
dx
−
3
·
dy
dx
=
0
2
x
·
dy
dx
+
3
·
dy
dx
=
3
x
2
−
2
y
dy
dx
(2
x
+
3)
=
3
x
2
−
2
y
Implicit Differentiation
Example
Find
dy
dx
if
x
3
−
2
xy
−
3
y
−
6
=
0
.
Solution.
D
x
[
x
3
−
2
xy
−
3
y
−
6]
=
D
x
[0]
D
x
[
x
3
]
−
D
x
[2
xy
]
−
D
x
[3
y
]
−
D
x
[6]
=
D
x
[0]
3
x
2
−
2
1
·
y
+
x
·
dy
dx
−
3
·
dy
dx
−
0
=
0
3
x
2
−
2
y
−
2
x
·
dy
dx
−
3
·
dy
dx
=
0
2
x
·
dy
dx
+
3
·
dy
dx
=
3
x
2
−
2
y
dy
dx
(2
x
+
3)
=
3
x
2
−
2
y
