1. `f(x)=xexp(x)`, find f'(2) with h=0.2 using Richardson extrapolation method

Solution:
Richardson Extrapolation method
Equation is `f(x) = xexp(x)`.

`h=0.2`

`f^'(x)=(f(x+h)-f(x-h))/(2h)` (CDD)

`f^'(2)=(f(2+0.2)-f(2-0.2))/(2*0.2)`

`f^'(2)=(f(2.2)-f(1.8))/0.4`

`f^'(2)=(19.855-10.8894)/0.4`

`f^'(2)=22.4142`

---

`h=0.1`

`f^'(x)=(f(x+h)-f(x-h))/(2h)`

`f^'(2)=(f(2+0.1)-f(2-0.1))/(2*0.1)`

`f^'(2)=(f(2.1)-f(1.9))/0.2`

`f^'(2)=(17.149-12.7032)/0.2`

`f^'(2)=22.2288`

---

`TV=(AV)_(h/2)+((AV)_(h/2)-(AV)_(h))/3`

`=22.2288 + (22.2288 - 22.4142) / 3`

`=22.167`