Geometric proof for derivative of $f(x)=\sqrt{x}$ Using a square of  area $x$,
we obtain
$$dx = (2\sqrt{x})d\sqrt{x}+(d\sqrt{x})^2$$
$$\frac{d\sqrt{x}}{dx} = \frac{1}{2 \sqrt{x}} - \frac{(d\sqrt{x})^2}{2\sqrt{x}dx}$$
How do I show that $\frac{(d\sqrt{x})^2}{2\sqrt{x}dx} \rightarrow 0$ as $dx\rightarrow0$?

By your geometric construction, and also by your first line, when $dx \rightarrow 0$, $d\sqrt x \rightarrow 0$.
$\frac {dx} {d\sqrt x} = 2\sqrt x + d\sqrt x$
So when $dx \rightarrow 0, \frac {dx} {d\sqrt x} \rightarrow 2\sqrt x$
So when $x\ne 0, \frac {(d\sqrt x)^2} {2\sqrt x dx}$ is equivalent when $dx \rightarrow 0$ to $\frac {d\sqrt x} {4x}$.
When $dx\rightarrow 0$, this $\rightarrow 0$.
