With taylor series, $f(x + h) = f(x) + \frac{df}{dx}h + O(h^2)$,
we know the derivative of $f(x)$ is the coefficient of the first order term $\frac{df}{dx}h$.
Using this definition of derivative, we can easily understand the linearity of derivative:
$\frac{d(u+v)}{dx} = \frac{du}{dx} + \frac{dv}{dx}$ as
$u(x + h) = u(x) + \frac{du}{dx}h + O(h^2)$
$v(x + h) = v(x) + \frac{dv}{dx}h + O(h^2)$
Thus: $(u+v)(x+h) = u(x+h) + v(x+h)$
$= u(x) + \frac{du}{dx}h + O(h^2) + v(x) + \frac{dv}{dx}h + O(h^2)$
$= u(x) + v(x) + (\frac{du}{dx} + \frac{dv}{dx})h + O(h^2)$
from the coefficient of the first order term, we know the derivative of $u+v$ is $\frac{du}{dx} + \frac{dv}{dx}$
The question is how to use the same reasoning process to verify reciprocal rule, i.e.
$\frac{d(1/v)}{dx} = -\frac{1}{v^2}\frac{dv}{dx}$
I was starting with:
$\frac{1}{v}(x + h) = \frac{1}{v(x + h)} = \frac{1}{v(x) + \frac{dv}{dx}h + O(h^2)}$
And I stuck here, could you please give some directions on how to continue?
Assuming things are well defined etc...
$$\frac{1}{v+\frac{dv}{dx}h + O(h^2)} = \frac{1}{v}\cdot\frac{1}{1+\frac{1}{v}\frac{dv}{dx}h + O(h^2)}$$
$$ = \frac{1}{v}\left(1 - \frac{1}{v}\frac{dv}{dx}h + O(h^2)\right) = \frac{1}{v} - \frac{1}{v^2}\frac{dv}{dx}h + O(h^2)$$
by geometric series.