`f(x)=5xe^(-2x)`, find f'(0.35) with h=0.25 using Richardson extrapolation method

Solution:
Richardson Extrapolation method
Equation is `f(x) = 5xexp(-2x)`.

`h=0.25`

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

`f^'(0.35)=(f(0.35+0.25)-f(0.35-0.25))/(2*0.25)`

`f^'(0.35)=(f(0.6)-f(0.1))/0.5`

`f^'(0.35)=(0.9036-0.4094)/0.5`

`f^'(0.35)=0.9884`

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`h=0.125`

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

`f^'(0.35)=(f(0.35+0.125)-f(0.35-0.125))/(2*0.125)`

`f^'(0.35)=(f(0.475)-f(0.225))/0.25`

`f^'(0.35)=(0.9185-0.7173)/0.25`

`f^'(0.35)=0.8047`

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

`=0.8047 + (0.8047 - 0.9884) / 3`

`=0.7435`

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