Say we have a rectangle where the sides lengths are given by $w(t)$ and $l(t)$. Then the area of the rectangle is $A(t) = w(t) l(t)$.
Say that we calculate the area at a given time $t_0$. We will call $w_0 = w(t_0)$ and the same for $l_0$ and $A_0$. Then $A_0 = w_0 l_0$.
After some time $\Delta t$ we will be at $t_1 = t_0 + \Delta t$, and both the width and length would have variate $\Delta w$ and $\Delta h$, respectively. Then the area at this point would be:
$A_1 = (w_0 + \Delta w)(h_0 + \Delta h) = w_0 h_0 + w_0 \Delta h + h_0 \Delta w + \Delta w \Delta h$.
We can calculate how the area has changed in this time, i.e.,
$\Delta A = A_1 - A_0 = w_0 \Delta h + h_0 \Delta w + \Delta w \Delta h$.
Finally, we want to calculate the "slope" or "velocity" of the variation,
$s = \frac{\Delta A}{\Delta t}$.
And to find the derivative in $t_0$, we want to see what happens when $\Delta t$ is close to zero.
$\frac{dA}{dt} = \lim_{\Delta t \to 0} \frac{\Delta A}{\Delta t},$
$\frac{dA}{dt} = \lim_{\Delta t \to 0} \frac{w_0 \Delta h + h_0 \Delta w + \Delta w \Delta h}{\Delta t} = \lim_{\Delta t \to 0} \frac{w_0 \Delta h}{\Delta t} + \lim_{\Delta t \to 0} \frac{h_0 \Delta w}{\Delta t} + \lim_{\Delta t \to 0} \frac{\Delta w \Delta h}{\Delta t}$,
$\frac{dA}{dt} = w_0 \lim_{\Delta t \to 0} \frac{\Delta h}{\Delta t} + h_0 \lim_{\Delta t \to 0} \frac{\Delta w}{\Delta t} + \lim_{\Delta t \to 0} \frac{\Delta w \Delta h}{\Delta t}$,
$\frac{dA}{dt} = w_0 \frac{dh}{dt} + h_0 \frac{dw}{dt} + \lim_{\Delta t \to 0} \frac{\Delta w \Delta h}{\Delta t}$.
Now, in order for things to work, the last term must be zero. I guess that since you have two operands both tending to zero being multiplied, the numerator will tend to zero "quicker" than the denominator, but this is vague idea.
Going back to the rectangle example, I find it counter-intuitive that the last term is zero, see this image: rectangle area
For the last term we have indeed that
$$\lim_{\Delta t \to 0} \frac{\Delta w \Delta h}{\Delta t}=\lim_{\Delta t \to 0} \frac{\Delta w}{\Delta t}\frac{\Delta h}{\Delta t}\Delta t=0$$