#math/calculus
Near $x_{0}$, a differentiable function is approximated by its tangent line:
$
f(x)\approx L(x)=f(x_{0})+f'(x_{0})(x-x_{0}).
$
Writing $x=x_{0}+\Delta x$ gives the equivalent change formula $\Delta y\approx f'(x_{0})\Delta x$. More precisely, the approximation error is small compared with $\Delta x$ as $\Delta x\to 0$.
**Example.** For the area $A(r)=r^{2}$ of a square, linearize at $r_{0}=2$. Since $A(2)=4$ and $A'(2)=4$,
$
A(2.01)\approx 4+4(0.01)=4.04.
$
The exact area is $4.0401$; the missed corner has area $(0.01)^{2}=0.0001$.
```tikz
\begin{document}
\begin{tikzpicture}[scale=1.0,line cap=round]
\fill[blue!10] (0,0) rectangle (3,3);
\fill[teal!25] (3,0) rectangle (3.45,3);
\fill[teal!25] (0,3) rectangle (3,3.45);
\fill[red!35] (3,3) rectangle (3.45,3.45);
\draw[very thick] (0,0) rectangle (3.45,3.45);
\draw[thick] (3,0) -- (3,3.45);
\draw[thick] (0,3) -- (3.45,3);
\node at (1.5,1.5) {$2^{2}=4$};
\node[rotate=90] at (3.23,1.5) {$2\Delta r$};
\node at (1.5,3.23) {$2\Delta r$};
\node[right,red!70!black] at (3.45,3.23) {$(\Delta r)^{2}$};
\end{tikzpicture}
\end{document}
```
> [!Practice]
> Marsden & Weinstein, pp. 94–97, exercises 45, 46, 48, 50, and 86–91.