Example taken from here. Suppose we want to calculate derivative of $f(x) = 3x + 2$ at the point $x = 4$. We can use dual numbers; $x = 4 + 1\varepsilon$ because the derivative of $x$ is 1; then substitute it in to get $(3 + 0\varepsilon) \cdot (4 + 1\varepsilon) + (2 + 0\varepsilon) = 14 + 3\varepsilon$. The derivative is then $3$.
However, I'm a bit confused on why this is so. Why does the coefficient of the dual component represent the derivative? I know that dual numbers are defined as $\varepsilon^2 = 0$ so there is some sort of connection between infinitesimally small numbers, which suggests the link, but I don't know the concrete details.
Nvm, I found it out. What is being done in that blog post is just taylor expansion (plug in $x + y\varepsilon$ into Taylor expansion and it will be clear). I was trying to work with traditional limit definition of derivative.