From the Taylor series expansion, we can write f(x₁) in terms of f(x₀) and its derivatives as follows:

Now, solve for [ ¶_xf ]₀ :

(x1 - x0)[ ¶xf ]0	=	[ f(x1) - f(x0) ] - { (2!)-1(x1 - x0)2 [ ¶x2 f ]0 + (3!)-1(x1 - x0)3 [ ¶x3 f ]0 + ... }
[ ¶xf ]0	=	[ f(x1) - f(x0) ] (x1 - x0)-1 - { (2!)-1(x1 - x0) [ ¶x2 f ]0 + (3!)-1(x1 - x0)2 [ ¶x3 f ]0 + ... }
[ ¶xf ]0	»	[ f(x1) - f(x0) ] (x1 - x0)-1 + order[ (Dx) ]

With this expression, then, we have a FD expression for [ ¶_xf ]₀ that is said to be accurate to 1^st-order in Dx.