`f(x)=2x^3+x^2-4` and `h = 0.5`, estimate `f^'(2.5) and f^('')(2.5)`
using Three 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	1.5	2	2.5	3	3.5
y	5	16	33.5	59	94

Three-point FDF (Forward difference formula)
`f^'(x)=1/(2h)[-3f(x)+4f(x+h)-f(x+2h)]`

`f^'(2.5)=1/(2*0.5)[-3f(2.5)+4f(2.5+0.5)-f(2.5+2*0.5)]`

`f^'(2.5)=1/1[-3f(2.5)+4f(3)-f(3.5)]`

`f^'(2.5)=1/1[-3(33.5)+4(59)-94]`

`f^'(2.5)=41.5`

Absolute Error:`|"exact value of " f^'(2.5)-(41.5)|=|42.5 -41.5|=1`

---

Three-point BDF (Backward difference formula)
`f^'(x)=1/(2h)[f(x-2h)-4f(x-h)+3f(x)]`

`f^'(2.5)=1/(2*0.5)[f(2.5-2*0.5)-4f(2.5-0.5)+3f(2.5)]`

`f^'(2.5)=1/1[f(1.5)-4f(2)+3f(2.5)]`

`f^'(2.5)=1/1[5-4(16)+3(33.5)]`

`f^'(2.5)=41.5`

Absolute Error:`|"exact value of " f^'(2.5)-(41.5)|=|42.5 -41.5|=1`

---

Three-point CDF (Central difference formula)
`f^'(x)=(f(x+h)-f(x-h))/(2h)`

`f^'(2.5)=(f(2.5+0.5)-f(2.5-0.5))/(2*0.5)`

`f^'(2.5)=(f(3)-f(2))/1`

`f^'(2.5)=(59-16)/1`

`f^'(2.5)=43`

Absolute Error:`|"exact value of " f^'(2.5)-(43)|=|42.5 -43|=0.5`

---

Three-point FDF (Forward difference formula) for second derivatives
`f^('')(x)=(f(x)-2f(x+h)+f(x+2h))/(h^2)`

`f^('')(2.5)=(f(2.5)-2f(2.5+0.5)+f(2.5+2*0.5))/((0.5)^2)`

`f^('')(2.5)=(f(2.5)-2f(3)+f(3.5))/(0.25)`

`f^('')(2.5)=(33.5-2(59)+94)/(0.25)`

`f^('')(2.5)=38`

Absolute Error:`|"exact value of " f^('')(2.5)-(38)|=|32 -38|=6`

---

Three-point BDF (Backward difference formula) for second derivatives
`f^('')(x)=(f(x-2h)-2f(x-h)+f(x))/(h^2)`

`f^('')(2.5)=(f(2.5-2*0.5)-2f(2.5-0.5)+f(2.5))/((0.5)^2)`

`f^('')(2.5)=(f(1.5)-2f(2)+f(2.5))/(0.25)`

`f^('')(2.5)=(5-2(16)+33.5)/(0.25)`

`f^('')(2.5)=26`

Absolute Error:`|"exact value of " f^('')(2.5)-(26)|=|32 -26|=6`

---

Three-point CDF (Central difference formula) for second derivatives
`f^('')(x)=(f(x-h)-2f(x)+f(x+h))/(h^2)`

`f^('')(2.5)=(f(2.5-0.5)-2f(2.5)+f(2.5+0.5))/(0.5)^2`

`f^('')(2.5)=(f(2)-2f(2.5)+f(3))/(0.25)`

`f^('')(2.5)=(16-2(33.5)+59)/(0.25)`

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

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

---

---