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 .  
 A: 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 $d(c^2) / dc = 2c$:
$$\Delta(c^2) = 2c \Delta c$$
A: 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.
