1. Using Five 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

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^'(1.10)=1/(12*0.05)[-25f(1.10)+48f(1.10+0.05)-36f(1.10+2*0.05)+16f(1.10+3*0.05)-3f(1.10+4*0.05)]`

`f^'(1.10)=1/(0.6)[-25f(1.10)+48f(1.15)-36f(1.2)+16f(1.25)-3f(1.3)]`

`f^'(1.10)=1/(0.6)[-25(1.0488)+48(1.0724)-36(1.0954)+16(1.118)-3(1.1402)]`

`f^'(1.10)=0.4762`

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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^'(1.10)=1/(12*0.05)[f(1.10-2*0.05)-8f(1.10-0.05)+8f(1.10+0.05)-f(1.10+2*0.05)]`

`f^'(1.10)=1/0.6[f(1)-8f(1.05)+8f(1.15)-f(1.2)]`

`f^'(1.10)=1/0.6[1-8(1.0247)+8(1.0724)-1.0954]`

`f^'(1.10)=0.4767`

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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^('')(1.10)=1/(12*(0.05)^2)[-f(1.10-2*0.05)+16f(1.10-0.05)-30f(1.10)+16f(1.10+0.05)-f(1.10+2*0.05)]`

`f^('')(1.10)=1/0.03[-f(1)+16f(1.05)-30f(1.10)+16f(1.15)-f(1.2)]`

`f^('')(1.10)=1/0.03[-1+16(1.0247)-30(1.0488)+16(1.0724)-1.0954]`

`f^('')(1.10)=-0.2157`