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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 Engineering Mathematics 2026 is part of Engineering Mathematics preparation. The notes and questions for Numerical Differentiation have been prepared according to the Engineering Mathematics exam syllabus. Information about Numerical Differentiation covers topics like and Numerical Differentiation Example, for Engineering Mathematics 2026 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 Engineering Mathematics & Numerical Differentiation in Hindi for Engineering Mathematics course. Download more important topics related with notes, lectures and mock test series for Engineering Mathematics Exam by signing up for free. Engineering Mathematics: Numerical Differentiation - Engineering Mathematics - Engineering Mathematics
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Full syllabus notes, lecture & questions for Numerical Differentiation - Engineering Mathematics - Engineering Mathematics - Engineering Mathematics | Plus exercises question with solution to help you revise complete syllabus for Engineering Mathematics | Best notes, free PDF download
Information about Numerical Differentiation
In this doc you can find the meaning of Numerical Differentiation defined & explained in the simplest way possible. Besides explaining types of Numerical Differentiation theory, EduRev gives you an ample number of questions to practice Numerical Differentiation tests, examples and also practice Engineering Mathematics tests
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