In this error approximation, why can we "kill" the $\Delta x\cdot \Delta \theta$ term, but not the term with $\Delta \theta$ alone?

Question

While trying to solve a problem I've come to this equation -- where $\Delta y$ is measurement error of $y$:

$$\Delta y = \Delta h+\Delta x \cdot \tan(\theta)+x \cdot \sec^2(\theta) \cdot \Delta\theta +\Delta x \cdot \Delta\theta\sec^2(\theta)$$

The course I'm taking gets "approximate equality" by "killing" the last term because it is quadratic.

$$\Delta y \approx \Delta h+\Delta x \cdot \tan(\theta)+x \cdot \sec^2(\theta) \cdot \Delta\theta$$

My question is:

What makes the last term so different from the second-to-last that it is the one being killed, and not the former or both?

Answer 1

In this contest, $x$ and $\sec^2\theta$ are medium-sized quantities, but $\Delta h$ and $\Delta x$ and $\Delta \theta$ are incredibly small quantities. Everything in the equation has one factor of $\Delta$-something, except that the last term has two such factors. Since the product of two incredibly small quantities is incredibly smaller than just one such quantity, it can be ignored in this context. (A formal argument would use limits.)

Answer 2

The idea here is that $\Delta x$ and $\Delta \theta$ are small. If this is true, then all the terms before the "quadratic" term are small but the quadratic term is extremely small. For example, what if $\Delta x$, $\Delta \theta$, and $\Delta h$ are on the order of $10^{-3}$. This means $\Delta x \Delta \theta$ is on the order of $10^{-6}$ and would thus not contribute all that much to $\Delta y$.