The finite difference expressions for the first, second and higher derivatives in the first, second or higher order of accuracy can be easily derived from Taylor's expansions. But, numerically, the successive application of the first derivative, in general, is not same as application of the second derivative.

First, a case where it works. Let's say that we want to compute second derivative of function $f$ given on 3-points stencil $(i-1, i, i+1)$. The finite difference formula is: $$\left(\frac{\partial^2 f}{\partial x^2}\right)_i = \frac{1}{h^2}(f_{i-1} - 2f_i + f_{i+1})$$

This result is derived from Taylor's expansions, but it can also be interpreted in the following way. The first derivatives of the first order accuracy at the intervals $(i-1, i)$ and $(i, i+1)$ are: $$\left(\frac{\partial f}{\partial x}\right)_{i-1/2} = \frac{1}{h}(f_i - f_{i-1})$$ and $$\left(\frac{\partial f}{\partial x}\right)_{i+1/2} = \frac{1}{h}(f_{i+1} - f_{i})$$ where I use $i-1/2$ and $i+1/2$ because these derivatives are representative for the cell faces (In the first order I have actually approximated my function as piece-wise linear between the grid points $x_i$. Therefore, in every grid point the slope on the left and the right hand side of it is not the same.) The second derivative in point $i$ is now: $$\left(\frac{\partial^2 f}{\partial x^2}\right)_{i} = \frac{1}{h}(f'_{i+1/2} - f'_{i-1/2}) = \frac{1}{h^2}(f_{i+1} - f_{i} - (f_i - f_{i-1})) $$ And this is identical to the finite difference expression for the second derivative in the second order of accuracy.

I wonder if there is a similar procedure to represent the second derivative in the 4th order accuracy (on 5-points stencil) as successive application of two first order derivative of the lower accuracy (on shorter stencils)?

A naive approach would be to apply first derivatives of the second order accuracy to the stencils $(i-2, i-i, i)$ and $(i, i+1, i+2)$: $$\left(\frac{\partial u}{\partial x}\right)_{i-1} = \frac{1}{2h}(u_i - u_{i-2})$$ and $$\left(\frac{\partial u}{\partial x}\right)_{i+1} = \frac{1}{2h}(u_{i+2} - u_{i})$$ and then to find the second derivative as the first derivative of the previous two: $$\left(\frac{\partial^2 u}{\partial x^2}\right)_{i} = \frac{1}{4h^2}(u_{i+2} - 2u_{i} - u_{i-2})$$ This is obviously not correct or, at least, not the same as application of the second derivative of the 4th order straight away: $$\left(\frac{\partial^2 u}{\partial x^2}\right)_{i} = \frac{1}{12h^2}(-u_{i-2} + 16u_{i-1} + 30 u_i + 16 u_{i+1} - u_{i+2})$$

**So, is there a way to reproduce the last equation as a successive combination of first derivatives of the lower accuracy order? If not, why not?**

Many thanks for help! This is driving me crazy!