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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?

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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.

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    $\begingroup$ Ohhh, that makes a lot of sense by adding and subtracting $f(x)$, thank you! $\endgroup$ – EmZ Apr 18 '18 at 22:03
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    $\begingroup$ No worries! I think the $|x|$ example is easy to visualise as a counterexample to this also! The 'symmetric' derivative you've given is actually still very useful though - numerically it converges faster for differentiable functions, and so some calculators use it for numeric differentiation. An old calculator of mine supposedly did numeric differentiation, and it evaluated the derivative of $|x|$ at 0 to be 0. $\endgroup$ – B. Mehta Apr 18 '18 at 22:06

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