`f(x)=2x^3+x^2-4` and `h = 0.5`, estimate `f^'(2.5) and f^('')(2.5)`
using Five point Forward difference, Backward difference, Central difference formula numerical differentiation
Also find exact value of f', f'' and error for each estimation

Solution:
Equation is `f(x) = 2x^3+x^2-4`.

`:. f^'(x) = 6x^2+2x`

`:. f^('')(x) = 12x+2`

The value of table for `x` and `y`

x	0.5	1	1.5	2	2.5	3	3.5	4	4.5
y	-3.5	-1	5	16	33.5	59	94	140	198.5

Five-point FDF (Forward difference formula)
`f^'(x)=1/(12h)[-25f(x)+48f(x+h)-36f(x+2h)+16f(x+3h)-3f(x+4h)]`

`f^'(2.5)=1/(12*0.5)[-25f(2.5)+48f(2.5+0.5)-36f(2.5+2*0.5)+16f(2.5+3*0.5)-3f(2.5+4*0.5)]`

`f^'(2.5)=1/(6)[-25f(2.5)+48f(3)-36f(3.5)+16f(4)-3f(4.5)]`

`f^'(2.5)=1/(6)[-25(33.5)+48(59)-36(94)+16(140)-3(198.5)]`

`f^'(2.5)=42.5`

Absolute Error:`|"exact value of " f^'(2.5)-(42.5)|=|42.5 -42.5|=0`

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Five-point CDF (Central difference formula)
`f^'(x)=1/(12h)[f(x-2h)-8f(x-h)+8f(x+h)-f(x+2h)]`

`f^'(2.5)=1/(12*0.5)[f(2.5-2*0.5)-8f(2.5-0.5)+8f(2.5+0.5)-f(2.5+2*0.5)]`

`f^'(2.5)=1/6[f(1.5)-8f(2)+8f(3)-f(3.5)]`

`f^'(2.5)=1/6[5-8(16)+8(59)-94]`

`f^'(2.5)=42.5`

Absolute Error:`|"exact value of " f^'(2.5)-(42.5)|=|42.5 -42.5|=0`

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Five-point CDF (Central difference formula) for second derivatives
`f^('')(x)=1/(12h^2)[-f(x-2h)+16f(x-h)-30f(x)+16f(x+h)-f(x+2h)]`

`f^('')(2.5)=1/(12*(0.5)^2)[-f(2.5-2*0.5)+16f(2.5-0.5)-30f(2.5)+16f(2.5+0.5)-f(2.5+2*0.5)]`

`f^('')(2.5)=1/3[-f(1.5)+16f(2)-30f(2.5)+16f(3)-f(3.5)]`

`f^('')(2.5)=1/3[-5+16(16)-30(33.5)+16(59)-94]`

`f^('')(2.5)=32`

Absolute Error:`|"exact value of " f^('')(2.5)-(32)|=|32 -32|=0`

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