The second-order, centered-approximation, finite difference of the derivative of $f(x)$ at $x$ is: $$ f'(x)=\frac{-f(x-\Delta x)+f(x+\Delta x)}{2\Delta x} $$ Question: Suppose we have the value of $f'(x)$ throughout our finite-difference grid with grid spacing $\Delta x$. Let $f'(x_0)$ be such a value at $x_0$. Is there a way to find $f'(x)$ at an arbitray location between $(x_0-\Delta x,x_0+\Delta x)$ based on the value of $f'(x_0)$?
Attempt: I tried linearly interpolating the value of $f'(x_0)$ between $(x_0-\Delta x,x_0+\Delta x)$. However, the result was incorrect.
You asked for an approximation to $\ f'(x). \ $ Using simple algebra the answer is $$ f'(x) \approx \frac{ f(x_0+\Delta x) - f(x_0-\Delta x) }{2 \Delta x} + 2(x - x_0)\frac{ f(x_0+\Delta x) - f(x_0) + f(x_0-\Delta x) }{ \Delta x^2} $$ where the two fractions are approximations to the first and second derivatives of $\ f(x) \ $ at $\ x = x_0. \ $ You can check that when $\ f(x) \ $ is a quadratic polynomial, then the approximation is exact.