Definition of a derivative

This may seem like a very stupid question but I know the definition of a derivative is $f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$, is an equivalent definition of the derivative: $f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h/2)-f(x-h/2)}{h}$? If I draw a graph, it appears to me that they should be the same, but how can I show it algebraically?

Solution 1:

They're not equivalent in general. But, if the limit $$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$ exists, i.e. the derivative exists, then they are both defined and equal (assuming of course that the function is defined on some open interval around $x$). In particular, $$\begin{align}\frac{f(x+h/2) - f(x-h/2)}{h}&=\frac{f(x+h/2)-f(x)}{h}-\frac{f(x-h/2)-f(x)}{h}\\ &=\frac{1}{2} \left(\frac{f(x+h/2)-f(x)}{h/2}-\frac{f(x-h/2)-f(x)}{h/2}\right)\end{align}$$ so let $h \to 0$ to give the required result.

A counterexample would be $f(x) = |x|$ at 0. Then, the derivative doesn't exist, since the limit from either direction is different, so limit is undefined. But, the 'central' derivative does exist, and is zero.