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Page 1
?= +
- ?
?
?
+
-+
+
?
?
?
dy
dx h
y
p
y
pp
y
12 1
2
36 2
3
0
2
0
2
3
0
??
?
()
!
()
!
?
?
?
?
?
?
?
dy
dx
at x = x
0
=? -? +? -? +?
?
?
?
?
?
?
11
2
1
3
1
4
1
5
0
2
0
3
0
4
0
5
0
h
yy yy …
(
\ At i = x
0
; p = 0)
And
dy
dx
d
dx
dy
dx
d
dp
dy
dx
dp
dx h
2
2
1
=
?
?
?
?
?
?
=
?
?
?
?
?
?
=
?? ?
2
0
3
0
2
4
0
66
3
12 36 22
4
1
y
p
y
pp
y
h
+
-
+
-+ ?
?
?
?
?
?
!!
dy
dx h
y
p
y
pp
y
2
2
2
0
3
0
2
4
0
16 6
3
12 36 22
1
2
?
?
?
?
?
?
=? +
-
?+
-+ ?
?
?
?
?
?
?
!!
dy
dx
xx
2
2
0
?
?
?
?
?
?
= at
=? -? +? -? +
?
?
?
?
?
?
111
12
5
6
2
2
0
3
0
4
0
5
0
h
yy yy
\ By using Newton’ s forward interpolation formula,
the first and second derivatives of y = f (x) at x = x
0
are
given by
dy
dx
xx
?
?
?
?
?
?
=
0
=? -? +? -? +?
?
?
?
?
?
?
11
2
1
3
1
4
1
5
0
2
0
3
0
4
0
5
0
h
yy yy y…
and
dy
dx h
yy yy
xx
2
22
2
0
3
0
4
0
5
0
0
111
12
5
6
?
?
?
?
?
?
=? -? +? -? +
?
?
?
?
?
?
=
…
2. Derivatives using newton’s backward difference
interpolation formula: We know that the Newton’s
backward difference interpolation formula is y = y
n
+
p? y
n
+
pp ()
!
+1
2
?
2
y
n
+
pp p ()()
!
++ 12
3
+ ?
3
y
n
…
Differentiating on both sides wrt p,
dy
dx
y
p
y
pp
y
nn n
=? +
+
?+
++
?+
21
2
36 2
3
2
2
3
!
As p
xx
h
dp
dx h
n
=
-
=
1
Chapter 06.indd 131 5/31/2017 10:59:42 AM
Page 2
?= +
- ?
?
?
+
-+
+
?
?
?
dy
dx h
y
p
y
pp
y
12 1
2
36 2
3
0
2
0
2
3
0
??
?
()
!
()
!
?
?
?
?
?
?
?
dy
dx
at x = x
0
=? -? +? -? +?
?
?
?
?
?
?
11
2
1
3
1
4
1
5
0
2
0
3
0
4
0
5
0
h
yy yy …
(
\ At i = x
0
; p = 0)
And
dy
dx
d
dx
dy
dx
d
dp
dy
dx
dp
dx h
2
2
1
=
?
?
?
?
?
?
=
?
?
?
?
?
?
=
?? ?
2
0
3
0
2
4
0
66
3
12 36 22
4
1
y
p
y
pp
y
h
+
-
+
-+ ?
?
?
?
?
?
!!
dy
dx h
y
p
y
pp
y
2
2
2
0
3
0
2
4
0
16 6
3
12 36 22
1
2
?
?
?
?
?
?
=? +
-
?+
-+ ?
?
?
?
?
?
?
!!
dy
dx
xx
2
2
0
?
?
?
?
?
?
= at
=? -? +? -? +
?
?
?
?
?
?
111
12
5
6
2
2
0
3
0
4
0
5
0
h
yy yy
\ By using Newton’ s forward interpolation formula,
the first and second derivatives of y = f (x) at x = x
0
are
given by
dy
dx
xx
?
?
?
?
?
?
=
0
=? -? +? -? +?
?
?
?
?
?
?
11
2
1
3
1
4
1
5
0
2
0
3
0
4
0
5
0
h
yy yy y…
and
dy
dx h
yy yy
xx
2
22
2
0
3
0
4
0
5
0
0
111
12
5
6
?
?
?
?
?
?
=? -? +? -? +
?
?
?
?
?
?
=
…
2. Derivatives using newton’s backward difference
interpolation formula: We know that the Newton’s
backward difference interpolation formula is y = y
n
+
p? y
n
+
pp ()
!
+1
2
?
2
y
n
+
pp p ()()
!
++ 12
3
+ ?
3
y
n
…
Differentiating on both sides wrt p,
dy
dx
y
p
y
pp
y
nn n
=? +
+
?+
++
?+
21
2
36 2
3
2
2
3
!
As p
xx
h
dp
dx h
n
=
-
=
1
Chapter 06.indd 131 5/31/2017 10:59:42 AM
?=
dy
dx
dy
dp
dp
dx
=? +
+
?+
++
?+
?
?
?
?
?
?
y
p
y
pp
y
h
nn n
21
2
36 2
3
1
2
2
3
!!
?= ?+
+
?+
++ ?
?
?
?+
?
?
?
dy
dx h
y
p
y
pp
y
nn
n
12 1
2
36 2
3
2
2
3
!!
dy
dx
xx
n
?
?
?
?
?
?
=
=? +? +? +? +? +
?
?
?
?
?
?
11
2
1
3
1
4
1
5
23 45
h
yy yy y
nn nn n
...
(
\ At x = x
n
; p = 0)
And
dy
dx
d
dx
dy
dx
d
dp
dy
dx
dp
dx
2
2
=
?
?
?
?
?
?
=
?
?
?
?
?
?
=? +
+
?+
++ ?
?
?
?+
?
?
?
16 6
3
12 36 22
4
1
23
2
4
h
y
p
y
pp
y
h
nn
n
!!
?=
dy
dx h
2
22
1
?+
+
?+
++
?+
?
?
?
?
?
?
23
2
4
66
3
12 36 22
4
y
p
y
pp
y
nn n
!!
?
?
?
?
?
?
?
=? +? +? +?
?
?
?
?
?
?
=
dy
dx h
yy yy
xx
nn nn
n
2
22
23 45
111
12
5
6
\ By using Newton’s backward difference
interpolation formula, the first and second derivatives
of y = f (x) at x = x
n
are given by
dy
dx
xx
n
?
?
?
?
?
?
=
=? +? +? +? +? +
?
?
?
?
?
?
11
2
1
3
1
4
1
5
23 45
h
yy yy y
nn nn n
and
dy
dx
xx
n
2
2
?
?
?
?
?
?
=
=? +? +? +? +
?
?
?
?
?
?
111
12
5
6
2
23 45
h
yy yy
nn nn
Example
Using the values of a function y = f (x) given in the following
table, find the first two derivatives of f (x) at x = 7.
x 2 3 4 5 6 7
y = f(x) 4 8 15 27 36 42
Solution
As we have to find the first two derivatives of y = f (x) at x
= 7, (end point of the given data), we use the derivatives’
formulae obtained from Newton’s backward interpolation
formula.
The backward difference table for the given data is
X y = f(x) Dy D 2
y D 3
y D 4
y D 5
y
2
3
4
5
6
7
4
8
15
27
36
42
4
7
12
9
6
3
5
-3
-3
-2
-8
0
6
8
2
By Newton’s backward interpolation formula, we have
dy
dx
xx
n
?
?
?
?
?
?
==7
=? +? +? +? +?
?
?
?
?
?
?
11
2
1
3
1
5
23 45
h
yy yy y
nn nn n
=+ ×- +× +× +×
?
?
?
?
?
?
==
1
1
6
1
2
3
1
3
0
1
4
8
1
5
2
69
10
69 () .
And
dy
dx
xx
n
2
2
7
?
?
?
?
?
?
==
=? +? ++?+ ?
?
?
?
?
?
?
111
12
5
6
2
23 45
h
yy yy
nn nn
=- ++ ×+ ×
?
?
?
?
?
?
=
1
1
30
11
12
8
5
6
26
2
.
Example
Find the first derivative at x = 6 for a function y = f (x) with
the following data.
x 5 7 10 11 13
y = f(x) 100 294 900 1210 2028
Solution
As the given values of x are not equally spaced, to find the
first derivative of f(x), we will make use of the Newton’s
divided difference interpolation formula, which is given by
y = f(x) = f(x
0
) + (x - x
0
) [x
0
, x
1
] + (x - x
0
)(x - x
1
) [x
0
, x
1
, x
2
] +
(x - x
0
)(x - x
1
)(x - x
2
) [x
0
, x
1
, x
2
x
3
] + (x - x
0
)(x - x
1
)(x - x
2
)
(x - x
3
)][x
0
, x
1
, x
2
, x
3
, x
4
] +
…
(1)
The divided differences of various orders for the given data
can be represented as shown below.
Chapter 06.indd 132 5/31/2017 10:59:45 AM
Page 3
?= +
- ?
?
?
+
-+
+
?
?
?
dy
dx h
y
p
y
pp
y
12 1
2
36 2
3
0
2
0
2
3
0
??
?
()
!
()
!
?
?
?
?
?
?
?
dy
dx
at x = x
0
=? -? +? -? +?
?
?
?
?
?
?
11
2
1
3
1
4
1
5
0
2
0
3
0
4
0
5
0
h
yy yy …
(
\ At i = x
0
; p = 0)
And
dy
dx
d
dx
dy
dx
d
dp
dy
dx
dp
dx h
2
2
1
=
?
?
?
?
?
?
=
?
?
?
?
?
?
=
?? ?
2
0
3
0
2
4
0
66
3
12 36 22
4
1
y
p
y
pp
y
h
+
-
+
-+ ?
?
?
?
?
?
!!
dy
dx h
y
p
y
pp
y
2
2
2
0
3
0
2
4
0
16 6
3
12 36 22
1
2
?
?
?
?
?
?
=? +
-
?+
-+ ?
?
?
?
?
?
?
!!
dy
dx
xx
2
2
0
?
?
?
?
?
?
= at
=? -? +? -? +
?
?
?
?
?
?
111
12
5
6
2
2
0
3
0
4
0
5
0
h
yy yy
\ By using Newton’ s forward interpolation formula,
the first and second derivatives of y = f (x) at x = x
0
are
given by
dy
dx
xx
?
?
?
?
?
?
=
0
=? -? +? -? +?
?
?
?
?
?
?
11
2
1
3
1
4
1
5
0
2
0
3
0
4
0
5
0
h
yy yy y…
and
dy
dx h
yy yy
xx
2
22
2
0
3
0
4
0
5
0
0
111
12
5
6
?
?
?
?
?
?
=? -? +? -? +
?
?
?
?
?
?
=
…
2. Derivatives using newton’s backward difference
interpolation formula: We know that the Newton’s
backward difference interpolation formula is y = y
n
+
p? y
n
+
pp ()
!
+1
2
?
2
y
n
+
pp p ()()
!
++ 12
3
+ ?
3
y
n
…
Differentiating on both sides wrt p,
dy
dx
y
p
y
pp
y
nn n
=? +
+
?+
++
?+
21
2
36 2
3
2
2
3
!
As p
xx
h
dp
dx h
n
=
-
=
1
Chapter 06.indd 131 5/31/2017 10:59:42 AM
?=
dy
dx
dy
dp
dp
dx
=? +
+
?+
++
?+
?
?
?
?
?
?
y
p
y
pp
y
h
nn n
21
2
36 2
3
1
2
2
3
!!
?= ?+
+
?+
++ ?
?
?
?+
?
?
?
dy
dx h
y
p
y
pp
y
nn
n
12 1
2
36 2
3
2
2
3
!!
dy
dx
xx
n
?
?
?
?
?
?
=
=? +? +? +? +? +
?
?
?
?
?
?
11
2
1
3
1
4
1
5
23 45
h
yy yy y
nn nn n
...
(
\ At x = x
n
; p = 0)
And
dy
dx
d
dx
dy
dx
d
dp
dy
dx
dp
dx
2
2
=
?
?
?
?
?
?
=
?
?
?
?
?
?
=? +
+
?+
++ ?
?
?
?+
?
?
?
16 6
3
12 36 22
4
1
23
2
4
h
y
p
y
pp
y
h
nn
n
!!
?=
dy
dx h
2
22
1
?+
+
?+
++
?+
?
?
?
?
?
?
23
2
4
66
3
12 36 22
4
y
p
y
pp
y
nn n
!!
?
?
?
?
?
?
?
=? +? +? +?
?
?
?
?
?
?
=
dy
dx h
yy yy
xx
nn nn
n
2
22
23 45
111
12
5
6
\ By using Newton’s backward difference
interpolation formula, the first and second derivatives
of y = f (x) at x = x
n
are given by
dy
dx
xx
n
?
?
?
?
?
?
=
=? +? +? +? +? +
?
?
?
?
?
?
11
2
1
3
1
4
1
5
23 45
h
yy yy y
nn nn n
and
dy
dx
xx
n
2
2
?
?
?
?
?
?
=
=? +? +? +? +
?
?
?
?
?
?
111
12
5
6
2
23 45
h
yy yy
nn nn
Example
Using the values of a function y = f (x) given in the following
table, find the first two derivatives of f (x) at x = 7.
x 2 3 4 5 6 7
y = f(x) 4 8 15 27 36 42
Solution
As we have to find the first two derivatives of y = f (x) at x
= 7, (end point of the given data), we use the derivatives’
formulae obtained from Newton’s backward interpolation
formula.
The backward difference table for the given data is
X y = f(x) Dy D 2
y D 3
y D 4
y D 5
y
2
3
4
5
6
7
4
8
15
27
36
42
4
7
12
9
6
3
5
-3
-3
-2
-8
0
6
8
2
By Newton’s backward interpolation formula, we have
dy
dx
xx
n
?
?
?
?
?
?
==7
=? +? +? +? +?
?
?
?
?
?
?
11
2
1
3
1
5
23 45
h
yy yy y
nn nn n
=+ ×- +× +× +×
?
?
?
?
?
?
==
1
1
6
1
2
3
1
3
0
1
4
8
1
5
2
69
10
69 () .
And
dy
dx
xx
n
2
2
7
?
?
?
?
?
?
==
=? +? ++?+ ?
?
?
?
?
?
?
111
12
5
6
2
23 45
h
yy yy
nn nn
=- ++ ×+ ×
?
?
?
?
?
?
=
1
1
30
11
12
8
5
6
26
2
.
Example
Find the first derivative at x = 6 for a function y = f (x) with
the following data.
x 5 7 10 11 13
y = f(x) 100 294 900 1210 2028
Solution
As the given values of x are not equally spaced, to find the
first derivative of f(x), we will make use of the Newton’s
divided difference interpolation formula, which is given by
y = f(x) = f(x
0
) + (x - x
0
) [x
0
, x
1
] + (x - x
0
)(x - x
1
) [x
0
, x
1
, x
2
] +
(x - x
0
)(x - x
1
)(x - x
2
) [x
0
, x
1
, x
2
x
3
] + (x - x
0
)(x - x
1
)(x - x
2
)
(x - x
3
)][x
0
, x
1
, x
2
, x
3
, x
4
] +
…
(1)
The divided differences of various orders for the given data
can be represented as shown below.
Chapter 06.indd 132 5/31/2017 10:59:45 AM
x y = f(x) First divided differences Second divided differences Third divided differences Fourth divided differences
5
7
10
11
13
100
294
900
1,210
2,028
294 100
75
97
-
-
=
900 294
10 7
202
-
-
=
1210 900
11 10
310
-
-
=
2028 1210
13 11
409
-
-
=
202 97
10 5
21
-
-
=
310 202
11 7
27
-
-
=
409 310
13 10
33
-
-
=
27 21
11 5
1
-
-
=
33 27
13 7
1
-
-
=
11
13 5
0
-
-
=
Substituting these in Eq. (1), we get
y = f(x) = 100 + (x - 5) × 97 + (x - 5)(x - 7) 21 + (x - 5) (x
- 7)(x - 10) × 1 + (x - 5)(x - 7)(x - 10)(x - 11) × 0
\ f (x) = 100 + 97(x - 5) + 21(x
2
- 12x + 35) + (x
3
- 22x
2
+ 155x - 350)
?=
dy
dx
97 + 21(2x - 12) + (3x
2
- 44x + 155)
?
?
?
?
?
?
?
=
dy
dx
x 6
= 97 + 21(2 × 6 - 12) + (3 × 6
2
- 44 × 6 +155)
= 97 + 0 + (- 1) = 96.
0
0
+
?
?
?
y
n
y
nn
0 0
23
+ +
- ?
?
(
Chapter 06.indd 133 5/31/2017 10:59:48 AM
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Document Description: Numerical Differentiation for Civil Engineering (CE) 2024 is part of Engineering Mathematics preparation.
The notes and questions for Numerical Differentiation have been prepared according to the Civil Engineering (CE) exam syllabus. Information about Numerical Differentiation covers topics
like and Numerical Differentiation Example, for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Numerical Differentiation.
Introduction of Numerical Differentiation in English is available as part of our Engineering Mathematics
for Civil Engineering (CE) & Numerical Differentiation in Hindi for Engineering Mathematics course.
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Full syllabus notes, lecture & questions for Numerical Differentiation | Engineering Mathematics - Civil Engineering (CE) - Civil Engineering (CE) | Plus excerises question with solution to help you revise complete syllabus for Engineering Mathematics | Best notes, free PDF download
Information about Numerical Differentiation
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Additional Information about Numerical Differentiation for Civil Engineering (CE) Preparation
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The content of Numerical Differentiation is prepared as per the latest Civil Engineering (CE) syllabus.
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