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Method and examples
Numerical Differentiation using
Newton's Forward, Backward Method
Method

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x =
1.  x 1.4 1.6 1.8 2 2.2 f(x) 4.0552 4.953 6.0496 7.3891 9.025
and x=1.4
2.  x 1.4 1.6 1.8 2 2.2 f(x) 4.0552 4.953 6.0496 7.3891 9.025
and x=2.2
3.  x 0 0.1 0.2 0.3 0.4 f(x) 1 0.9975 0.99 0.9776 0.8604
and x=0.0
4.  x 0 0.1 0.2 0.3 0.4 f(x) 1 0.9975 0.99 0.9776 0.8604
and x=0.1
5.  x 0 0.1 0.2 0.3 0.4 f(x) 1 0.9975 0.99 0.9776 0.8604
and x=0.3
6.  x 0 0.1 0.2 0.3 0.4 f(x) 1 0.9975 0.99 0.9776 0.8604
and x=0.4
 f(x) = x1 = and x2 = x = Step value (h) =  OR  Interval (N) = f(x)=2x^3-4x+1x1 = 2 and x2 = 4x = 3.5Step value (h) = 0.5 or N = 5f(x)=2x^3-4x+1x1 = 2 and x2 = 4x = 2.25Step value (h) = 0.25 or N = 8f(x)=x^3-x+1x1 = 2 and x2 = 4x = 3.5Step value (h) = 0.5 or N = 5f(x)=x^3-x+1x1 = 2 and x2 = 4x = 2.25Step value (h) = 0.25 or N = 8f(x)=x^3+x+2x1 = 2 and x2 = 4x = 3.5Step value (h) = 0.5 or N = 5f(x)=x^3+x+2x1 = 2 and x2 = 4x = 2.25Step value (h) = 0.25 or N = 8
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