1. Using Three point Forward difference, Backward difference, Central difference formula numerical differentiation to find solution

x	1	1.05	1.10	1.15	1.20	1.25	1.30
f(x)	1	1.02470	1.04881	1.07238	1.09545	1.11803	1.14018

`f^'(1.10) and f^('')(1.10)`

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

x	1	1.05	1.1	1.15	1.2	1.25	1.3
y	1	1.0247	1.0488	1.0724	1.0954	1.118	1.1402

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

`f^'(1.10)=1/(2*0.05)[-3f(1.10)+4f(1.10+0.05)-f(1.10+2*0.05)]`

`f^'(1.10)=1/0.1[-3f(1.10)+4f(1.15)-f(1.2)]`

`f^'(1.10)=1/0.1[-3(1.0488)+4(1.0724)-1.0954]`

`f^'(1.10)=0.4764`

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

`f^'(1.10)=1/(2*0.05)[f(1.10-2*0.05)-4f(1.10-0.05)+3f(1.10)]`

`f^'(1.10)=1/0.1[f(1)-4f(1.05)+3f(1.10)]`

`f^'(1.10)=1/0.1[1-4(1.0247)+3(1.0488)]`

`f^'(1.10)=0.4763`

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

`f^'(1.10)=(f(1.10+0.05)-f(1.10-0.05))/(2*0.05)`

`f^'(1.10)=(f(1.15)-f(1.05))/0.1`

`f^'(1.10)=(1.0724-1.0247)/0.1`

`f^'(1.10)=0.4768`

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

`f^('')(1.10)=(f(1.10)-2f(1.10+0.05)+f(1.10+2*0.05))/((0.05)^2)`

`f^('')(1.10)=(f(1.10)-2f(1.15)+f(1.2))/(0.0025)`

`f^('')(1.10)=(1.0488-2(1.0724)+1.0954)/(0.0025)`

`f^('')(1.10)=-0.2`

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

`f^('')(1.10)=(f(1.10-2*0.05)-2f(1.10-0.05)+f(1.10))/((0.05)^2)`

`f^('')(1.10)=(f(1)-2f(1.05)+f(1.10))/(0.0025)`

`f^('')(1.10)=(1-2(1.0247)+1.0488)/(0.0025)`

`f^('')(1.10)=-0.236`

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

`f^('')(1.10)=(f(1.10-0.05)-2f(1.10)+f(1.10+0.05))/(0.05)^2`

`f^('')(1.10)=(f(1.05)-2f(1.10)+f(1.15))/(0.0025)`

`f^('')(1.10)=(1.0247-2(1.0488)+1.0724)/(0.0025)`

`f^('')(1.10)=-0.216`