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In my course we have stated and used the error propagation formula:

$$|y-y_0|\approx|f^\prime(x)|\cdot|x-x_0|$$

But it was presented with no proof and I wonder if you can help me understand the formula holds?

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    You don't recognize the first few terms of the Taylor expansion? :)2012-07-19
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    @J.M. OK but to me the first two terms of Taylor expansion is f(a)+f'(a)(x-a)/2. I'm working on how the two common Taylor expansions are equivalent.2012-07-19
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    @NickRosencrantz: The $2$ at the bottom shouldn't be there. The **next** term is $\frac{f''(a)}{2!}(x-a)^2$.2012-07-19
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    @AndréNicolas Yes, exactly. I was wrong from memory.2012-07-19

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Possibly the best way to understand it is via the mean value theorem: $$f(x)-f(x_0)=f'(c)\cdot(x-x_0)$$ for some $c$ between $x_0$ and $x$. If $f'$ is continuous, $f'(c)$ can be expected to be close to $f'(x)$ or $f'(x_0)$ when $\lvert x-x_0\rvert$ is small enough.

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    Thank you. It's somewhat more info for me to go on than just the Taylor expansion again.2012-07-19