Another option which isn't geometric, but which reinforces the concept of derivative as linear approximation, is as follows.

First derive (by any means) that $\frac{\textrm{d}}{\textrm{d}u} \frac{1}{u} = -u^{-2}$.

Convey that the numerical meaning of this is $\frac{1}{u + \Delta u} \approx \frac{1}{u} - \frac{\Delta u}{u^2}$.

Use this to do some back of the envelop approximations like

$$ \begin{align*} \frac{9}{19} &= \frac{9}{20-1}\\ &\approx \frac{9}{20} - \frac{-9}{20^2}\\ &= 0.45 + 0.0225\\ &=0.4725 \end{align*} $$

Compare the approximation to the true result of $0.47368...$.

The general quotient rule repeats the same sort of calculation generally.

$$ \begin{align*} \frac{f + \Delta f}{g + \Delta g} &\approx (f + \Delta f)(\frac{1}{g} - \frac{1}{g^2}\Delta g)\\ &= \frac{f}{g} + \frac{g\Delta f - f\Delta g}{g^2} - \frac{f}{g^2} \Delta f \Delta g\\ &\approx \frac{f}{g} + \frac{g\Delta f - f\Delta g}{g^2} - \frac{f}{g^2} \end{align*} $$

Another option which isn't geometric, but which reinforces the concept of derivative as linear approximation, is as follows.

First derive (by any means) that $\frac{\textrm{d}}{\textrm{d}u} \frac{1}{u} = -u^{-2}$.

Convey that the numerical meaning of this is $\frac{1}{u + \Delta u} \approx \frac{1}{u} - \frac{\Delta u}{u^2}$.

Use this to do some back of the envelop approximations like

$$ \begin{align*} \frac{9}{19} &= \frac{9}{20-1}\\ &\approx \frac{9}{20} - \frac{-9}{20^2}\\ &= 0.45 + 0.0225\\ &=0.4725 \end{align*} $$

Compare the approximation to the true result of $0.47368...$.

The general quotient rule repeats the same sort of calculation generally.

$$ \begin{align*} \frac{f + \Delta f}{g + \Delta g} &\approx (f + \Delta f)(\frac{1}{g} - \frac{1}{g^2}\Delta g)\\ &= \frac{f}{g} + \frac{g\Delta f - f\Delta g}{g^2} - \frac{f}{g^2} \Delta f \Delta g\\ &\approx \frac{f}{g} + \frac{g\Delta f - f\Delta g}{g^2} \end{align*} $$