Finite differences of function composition I'm trying to express the following in finite differences:
$$\frac{d}{dx}\left[ A(x)\frac{d\, u(x)}{dx} \right].$$ 
Let $h$ be the step size and $x_{i-1} = x_i - h$ and $x_{i+ 1} = x_i + h$ 
If I take centered differences evaluated in $x_i$, I get:
$\begin{align*}\left\{\frac{d}{dx}\left[ A(x)\frac{d\, u(x)}{dx}\right]\right\}_i 

&= \frac{\left[A(x)\frac{d\, u(x)}{dx}\right]_{i+1/2} - \left[A(x)\frac{d\, u(x)}{dx}\right]_{i-1/2}}{h} \\
&= \frac{A_{i+1/2}\left[\frac{u_{i+1}-u_{i}}{h}\right] - A_{i-1/2}\left[\frac{u_{i}-u_{i-1}}{h}\right]}{h}
\end{align*}$
So, if I use centered differences I would have to have values for $A$ at $i + \frac 12$ and $A$ at $i - \frac 12$; however those nodes don't exist (in my stencil I only have $i \pm$ integer nodes); is that correct? If I use forward or backward differences I need A values at $i$, $i + 1$, $i + 2$ and at $i$, $i -1$, $i -2$ respectively.
Am I on the correct path?
I would really appreciate any hint.
Thanks in advance, 
Federico

$$
\frac{\mathrm{d}}{\mathrm{d}x}\left[A\frac{\mathrm{d}u}{\mathrm{d}x}\right]=\frac{\mathrm{d}A}{\mathrm{d}x}\frac{\mathrm{d}u}{\mathrm{d}x}+A\frac{\mathrm{d}^2u}{\mathrm{d}x^2}
$$
So you might try using $\frac{1}{4}(A_{i+1}-A_{i-1})(u_{i+1}-u_{i-1}) + A_i(u_{i+1}-2u_i+u_{i-1})$
