Using Two 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^'(1.15) and f^('')(1.15)`
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 |
|---|
Two-point FDF (Forward difference formula)
`f^'(x)=(f(x+h)-f(x))/h`
`f^'(1.15)=(f(1.15+0.05)-f(1.15))/0.05`
`f^'(1.15)=(f(1.2)-f(1.15))/0.05`
`f^'(1.15)=(1.0954-1.0724)/0.05`
`f^'(1.15)=0.4614`
Two-point BDF (Backward difference formula)
`f^'(x)=(f(x)-f(x-h))/h`
`f^'(1.15)=(f(1.15)-f(1.15-0.05))/0.05`
`f^'(1.15)=(f(1.15)-f(1.1))/0.05`
`f^'(1.15)=(1.0724-1.0488)/0.05`
`f^'(1.15)=0.4714`
Two-point CDF (Central difference formula)
`f^'(x)=(f(x+h)-f(x-h))/(2h)`
`f^'(1.15)=(f(1.15+0.05)-f(1.15-0.05))/(2*0.05)`
`f^'(1.15)=(f(1.2)-f(1.1))/0.1`
`f^'(1.15)=(1.0954-1.0488)/0.1`
`f^'(1.15)=0.4664`
This material is intended as a summary. Use your textbook for detail explanation.
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