Finite difference applied to a product

Question

I believe that the finite difference

$$\frac{f(x_0 + \frac12 \Delta x) - f(x_0 - \frac12 \Delta x)}{\Delta x}$$

approximates $f'$, and has limit $f'(x_0)$ as $\Delta x \to 0$. Am I correct in reasoning from this that given two functions $\alpha (x)$, $\beta(x)$ differentiable at $x_0$, we have that

$$\lim_{\Delta x \to 0}\frac{\alpha(x_0 - \Delta x)\beta(x_0 - \Delta x) - \alpha(x_0 + \Delta x)\beta(x_0 + \Delta x)}{\Delta x}\\ = -2(\alpha \beta)'(x_0) = -2\alpha'(x_0)\beta(x_0) - 2\alpha(x_0) \beta'(x_0)?$$

Thanks

Answer 1

Note that $\frac{f(x+\frac{1}{2}\delta) - f(x-\frac{1}{2}\delta)}{\delta} = \frac{f(x+\frac{1}{2}\delta) - f(x)+f(x)-f(x-\frac{1}{2}\delta)}{\delta} =\frac{1}{2} \frac{f(x+\frac{1}{2}\delta) - f(x)}{\frac{1}{2}\delta} + \frac{1}{2} \frac{f(x-\frac{1}{2}\delta) -f(x)}{-\frac{1}{2}\delta}$.

It follows that $\lim_{\delta \to 0} \frac{f(x+\frac{1}{2}\delta) - f(x-\frac{1}{2}\delta)}{\delta} = f'(x)$.

The formula for the product $\alpha \beta$ follows from the product rule $(\alpha \beta)' = \alpha' \beta + \alpha \beta'$ (and scaling by $2$ in the formulation above).