Error propagation for quadratic I have a very simple question I am struggling with. Lets say I want to propagate the error for some expression $$ y = x^2$$
Lets say I known that $x = 0 \pm 100$. Using standard error propagation I get that $$\sigma_y = 2x\sigma_x$$
This means that the value I get $$y = 0 \pm 0$$
Which I find very counter-intuitive. I know $x$ incredibly imprecisely, yet I know $y$ with perfect precision? What am I missing here? Or is this really the true result?
 A: Your intuition is correct, it certainly can't be the true result. It seems to me that you're using a formula that isn't quite adapted to the problem. The "standard" formula for the propagation of error to $y$ is usually derived in the following way:
\begin{equation*}
\begin{aligned}
y &= x^2 \\ \implies \log{y} &= 2 \log{x}\\
\implies \frac{\Delta y}{y} &= 2 \frac{\Delta x}{x}
\end{aligned}
\end{equation*}
We arrive at the formula you quote by then multiplying by $y$, and identifying $\Delta y \equiv \sigma_y$ and $\Delta x \equiv \sigma_x$.
Of course, the relative error is not defined when $y=0=x$! Thus, naively using this formula isn't a good idea, since the relative error isn't nicely defined when the "true" value is zero. There is an interesting discussion of this on the Stats StackExchange as well as the Math StackExchange.
Happily, such situations don't arise very often (in introductory labs at least). In an actual experiment I suspect that quick workarounds could be arranged.

EDIT:
I understand that your question is more related to the "theory" but practically, it seems to me that if you actually had such a situation in a laboratory, it's an indication that you are not using the correct apparatus to measure the quantity in question. The value being zero when the uncertainty is so large is equivalent to trying to measure the mass of a single hair from your head using a kitchen weighing balance.
A: The method of differentiating to find error works only when the error is much smaller compared to the measured value.
ie: if $x>>\Delta x$.
In general, if $\Delta x$ is the error associated with $x$, then the maximum error associated with $y (=x^2)$ is:
$(x+\Delta x)^2$
$= x^2 + 2x\Delta x + {\Delta x}^2$
So we see the deviation from $x^2$ is $2x\Delta x + (\Delta x) ^2$.
usually error of the instrument would be small compared to measured value and hence ${\Delta x}^2$ can be neglected.
Here that's not the case. Now, we get a large deviation as our intuition says.
A: You made a mistake in your error propagation equation. You want to see how an error $\Delta x$ propagates and gives rise to a $\Delta y$. It should be something like $\Delta y = 2 x \Delta x$ which can be rewritten as $ \Delta y / y = 2 \Delta x / x$. So as usual with multiplication you're adding the relative errors in first order.
A: If $x$ is a random variable from the normal distribution $N(\mu_x, \sigma_x)$, then $y=x^2$ is a random variable from the (scaled) $\chi^2_\nu$ distribution with $\nu=k=1$.

(copied from wiki)
The image shows the $\chi^2_\nu$ distributions for the scaled variable $\chi =Z^2$, where $Z = (x - \mu_x)/\sigma_x$ is $N(0,1)$ distributed. Hence, the image shows that even if $\mu_x = 0$ the mean of $y$ is non-zero,  as you expected. Doing the calculation one can show that the expectation value of $\chi$ is given by $E[\chi] = E[Z^2] = \nu$. Since one can always scale the variable $x\to Z$ such that the mean value becomes zero, we will consider this special case, but keep $\sigma_z$ explicitly. Hence, we consider the case $x \sim N(0,\sigma_x)$.
The formal way to tackle your problem is to use the concept of transforming the random variable, which leads to the result described above. A "simpler method", which is also directly connected to your question, is to calculate the variance of y,
$$
\sigma^2_y = %Var[y] = 
\left(
\frac{\partial y}{\partial z}
\right)^2 \sigma^2_z
= (2Z)^2 \sigma^2_z
$$
which is mainly the formula you already had. However, in "my" relation the random variable $Z$ does not appear linear, but squared. Taking the expectation value of this expression, we obtain
$$
\sigma^2_y = (2 \sigma_z)^2 \, E[Z^2] 
=  (2 \sigma_z)^2 \nu = (2\sigma_z)^2
$$
where we used $\nu=1$. Now taking the square root we obtain the result
\begin{align}
 E[y] &= \nu \\
\sigma_y &= 2\sigma_z
\end{align}
or in your notation $y = \nu \pm 2\sigma_z$.
