Why $\Delta(c^2)=2c(\Delta c)$?

I read a solution of find relative error for $c^2 = a^2 + b^2 -2ab\cdot\cos(\alpha)$ and it written there the equation $\Delta(c^2)=2c(\Delta c)$ .can someone explain how to developing this equation ?

Edit : definition of $\Delta: \Delta(x) = |(x-x^*)|$ when $x^*$ is is the $x$ with the error .

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What is $\Delta(c^2)$ in this context? More general, how does $\Delta$ act on $c$? –  Pedro Tamaroff Sep 8 '12 at 19:20
@PeterTamaroff - I think the OP means the error estimate. –  nbubis Sep 8 '12 at 19:22
@nbubis Oh, right. $d(c^2)=2c dc$, maybe? –  Pedro Tamaroff Sep 8 '12 at 19:27
@PeterTamaroff : I edited my post –  URL87 Sep 8 '12 at 19:28

In general, the error can be computed by using the derivative as a first order approximation: $$\Delta f(x) = \frac{\Delta f(x)}{\Delta x}\Delta x \simeq \frac{df(x)}{dx}\Delta x$$ In your case, because $d(c^2) / dc = 2c$: $$\Delta(c^2) = 2c \Delta c$$

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Let $f : \mathbb{R} \to \mathbb{R}$ be defined by $f (x) = x^2$. Then, we have that

$$f (x + \Delta x ) = (x + \Delta x)^2 = x^2 + 2 x \Delta x + (\Delta x)^2$$

Think of $f$ as a black-box that takes values of $x$ and spits out $f (x)$. For a given $x$, we obtain $f (x)$. What happens if we perturb the input? If the input is $x + \Delta x$ then the output will be $f (x + \Delta x )$. The perturbation in the output is thus

$$f (x + \Delta x ) - f (x) = 2 x \Delta x + (\Delta x)^2$$

Note that the magnitude of the perturbation in the output depends on the input value $x$. If $\Delta x$ is "small enough", then the perturbation in the output can be given by its first-order approximation

$$f (x + \Delta x ) - f (x) \approx 2 x \Delta x$$

However, if $\Delta x$ is not "small enough", the $(\Delta x)^2$ term will have to be included.

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+1 for mentioning whether $(\Delta x)^2$ is "small enough" –  Hurkyl Sep 8 '12 at 21:25