`f(x)=sinx` and `h = 0.1`, estimate `f^'(0.8) and f^('')(0.8)`
using Five 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) = sin(x)`.

`:. f^'(x) = cos(x)`

`:. f^('')(x) = -sin(x)`

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

x	0.4	0.5	0.6	0.7	0.8	0.9	1	1.1	1.2
y	0.3894	0.4794	0.5646	0.6442	0.7174	0.7833	0.8415	0.8912	0.932

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^'(0.8)=1/(12*0.1)[-25f(0.8)+48f(0.8+0.1)-36f(0.8+2*0.1)+16f(0.8+3*0.1)-3f(0.8+4*0.1)]`

`f^'(0.8)=1/(1.2)[-25f(0.8)+48f(0.9)-36f(1)+16f(1.1)-3f(1.2)]`

`f^'(0.8)=1/(1.2)[-25(0.7174)+48(0.7833)-36(0.8415)+16(0.8912)-3(0.932)]`

`f^'(0.8)=0.6967`

Absolute Error:`|"exact value of " f^'(0.8)-(0.6967)|=|0.6967 -0.6967|=0`

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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^'(0.8)=1/(12*0.1)[f(0.8-2*0.1)-8f(0.8-0.1)+8f(0.8+0.1)-f(0.8+2*0.1)]`

`f^'(0.8)=1/1.2[f(0.6)-8f(0.7)+8f(0.9)-f(1)]`

`f^'(0.8)=1/1.2[0.5646-8(0.6442)+8(0.7833)-0.8415]`

`f^'(0.8)=0.6967`

Absolute Error:`|"exact value of " f^'(0.8)-(0.6967)|=|0.6967 -0.6967|=0`

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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^('')(0.8)=1/(12*(0.1)^2)[-f(0.8-2*0.1)+16f(0.8-0.1)-30f(0.8)+16f(0.8+0.1)-f(0.8+2*0.1)]`

`f^('')(0.8)=1/0.12[-f(0.6)+16f(0.7)-30f(0.8)+16f(0.9)-f(1)]`

`f^('')(0.8)=1/0.12[-0.5646+16(0.6442)-30(0.7174)+16(0.7833)-0.8415]`

`f^('')(0.8)=-0.7174`

Absolute Error:`|"exact value of " f^('')(0.8)-(-0.7174)|=|-0.7174 +0.7174|=0`

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