From the Taylor series expansion, we can write f(x_{1}) in terms of f(x_{0}) and its derivatives as follows:
Now, solve for [ ¶_{x}f ]_{0} :
(x_{1} - x_{0})[ ¶_{x}f ]_{0} | = | [ f(x_{1}) - f(x_{0}) ] - { (2!)^{-1}(x_{1} - x_{0})^{2} [ ¶_{x}^{2} f ]_{0} + (3!)^{-1}(x_{1} - x_{0})^{3} [ ¶_{x}^{3} f ]_{0} + ... } |
[ ¶_{x}f ]_{0} | = | [ f(x_{1}) - f(x_{0}) ] (x_{1} - x_{0})^{-1} - { (2!)^{-1}(x_{1} - x_{0}) [ ¶_{x}^{2} f ]_{0} + (3!)^{-1}(x_{1} - x_{0})^{2} [ ¶_{x}^{3} f ]_{0} + ... } |
[ ¶_{x}f ]_{0} | » | [ f(x_{1}) - f(x_{0}) ] (x_{1} - x_{0})^{-1} + order[ (Dx) ] |
With this expression, then, we have a FD expression for [ ¶_{x}f ]_{0} that is said to be accurate to 1^{st}-order in Dx.