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Numerical Differentiation using Newton's Forward, Backward method calculator
Numerical Differentiation using
Newton's Forward, Backward Method
Method
1. Find Numerical Differentiation for x & f(x) table data
2. Find Numerical Differentiation for f(x) = x^3+x+2 & step value (h)
Type your data in either horizontal or verical format,
for seperator you can use '-' or ',' or ';' or space or tab
for sample click random button
OR
Rows :
Click On Generate
x =
x
1.4
1.6
1.8
2.0
2.2
f(x)
4.0552
4.9530
6.0496
7.3891
9.0250
and x=1.4
x
1.4
1.6
1.8
2.0
2.2
f(x)
4.0552
4.9530
6.0496
7.3891
9.0250
and x=2.2
x
0.0
0.1
0.2
0.3
0.4
f(x)
1.0000
0.9975
0.9900
0.9776
0.8604
and x=0.0
x
0.0
0.1
0.2
0.3
0.4
f(x)
1.0000
0.9975
0.9900
0.9776
0.8604
and x=0.1
x
0.0
0.1
0.2
0.3
0.4
f(x)
1.0000
0.9975
0.9900
0.9776
0.8604
and x=0.3
x
0.0
0.1
0.2
0.3
0.4
f(x)
1.0000
0.9975
0.9900
0.9776
0.8604
and x=0.4
f(x) =
x
_{1}
=
and x
_{2}
=
x =
Step value (h) =
OR Interval (N) =
`f(x)=2x^3-4x+1`
x1 = 2 and x2 = 4
x = 3.5
Step value (h) = 0.5
or N = 5
`f(x)=2x^3-4x+1`
x1 = 2 and x2 = 4
x = 2.25
Step value (h) = 0.25
or N = 8
`f(x)=x^3-x+1`
x1 = 2 and x2 = 4
x = 3.5
Step value (h) = 0.5
or N = 5
`f(x)=x^3-x+1`
x1 = 2 and x2 = 4
x = 2.25
Step value (h) = 0.25
or N = 8
`f(x)=x^3+x+2`
x1 = 2 and x2 = 4
x = 3.5
Step value (h) = 0.5
or N = 5
`f(x)=x^3+x+2`
x1 = 2 and x2 = 4
x = 2.25
Step value (h) = 0.25
or N = 8
Solution
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