Please explain the logic behind $d(xy) = y(dx) + x(dy)$ I've seen $d(xy) = y(dx) + x(dy)$, but I don't understand the principle behind it and memorizing it is lame.  Can anyone explain what is going on here?
For example from physics,
$$F = {{dP} \over {dt}}$$
$$F = {{d(mv)} \over {dt}}$$
$$F = {{v(dm) + m(dv)} \over {dt}}$$
Since $$dm = 0$$
$$F = {{m(dv)} \over {dt}}$$
Since $$a = {{dv} \over {dt}}$$
$$F = ma$$
 A: Hint
Suppose a rectangle of dimensions $x$ and $y$; its area is $A_0=xy$. Now change $x$ to $x+\Delta x$ and $y$ to $y+\Delta y$. The area of the new rectangle is given by $$ A_1=(x+\Delta x)(y+\Delta y)=xy+x \Delta y+y\Delta x+\Delta x \Delta y$$ So, the change of the area is $$\Delta A=A_1-A_0=x \Delta y+y\Delta x+\Delta x \Delta y$$ Now, let us make $\Delta x $ and $\Delta y$ very small; then $\Delta x \Delta y$ is negligible.
I am sure that you can take from here.
In order to illustrate, let us consider $x=10$ meters, $y=5$ meters and $\Delta x=\Delta y=1$ centimeter that is to say $0.01$ meters. So, using the last formula, $$\Delta A=10 \times 0.01+5 \times 0.01+0.01\times 0.01=0.1501$$ You see how small is the last term compared to the previous ones.
A: Here you go: https://math.stackexchange.com/a/3393408/633100
I'll actually Paste the same answer here (I tried sharing link alone but apparently answers have to be longer than 30 characters)
$d(xy)=(x+dx)(y+dy)-xy \ $     [from Definition of a "differential"]
$\ \ \ \ \ \ \ \  \ =ydx + xdy + dxdy $
Here what people normally do is, "$dxdy$ is too small and negligible". But that is SO BAD.
Instead, go back to definition of the derivative.
$\lim_{ h -> 0 }\frac{f(x+h)+f(x)}{h}$
Divide the first equation we got by $dx$
$(xy)' = y + x(y') + dy $ 
$dy -> 0$ by definition of a derivative/a differential.
Another way of looking at this would be,
Integrate both sides and all the terms will become bigger values except this $dxdy$ will alone have to be integrated twice to give it a good value.
for any quantity z, dz is defined as the infinitesimally small change in that quantity. ie approaches 0. These are generally used in combination with one or more 'differentials' and so we normally don't cancel of these to equal 0. for instance, $\frac{dv}{dt}$ is not 0/0. But with $dxdy$ if you integrate the whole equation, all the terms of the equation move from being differentials to regular terms but except this last one. There are no other differentials in the equation which makes it legal to cancel the last one to 0.
At least that's how I see it. 
