proof of $d(xy) = x(dy) + y(dx)$ I was trying to prove
$$d(xy) = x(dy) + y(dx)$$
earlier this morning and I used this post to help me understand the task. 
I understood the entirety of the post for my calculus class, apart from one step. 
Considering an area of a rectangle with dimensions $x$ and $y$ $xy$ makes sense. 
Likewise, the area of a rectangle with dimensions $(x+\Delta x)(y+\Delta y)$ giving  
$$A_1=(x+\Delta x)(y+\Delta y)=xy+x \Delta y+y\Delta x+\Delta x \Delta y$$
made sense too. 
I was confused however about the subtraction of the two areas that gives
$$x \Delta y+y\Delta x+\Delta x \Delta y$$
but the small approximation of $\Delta x $ and $\Delta y$ very small, ensuring that  $\Delta x \Delta y$ is negligible afterwards made sense.
Why is there the need to subtract the areas as part of the proof for this differentiation property?
 A: By $d(xy)$ we mean change in xy,
Remember when a differential was defined, it was "Infinitesimal Change".


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*Wikipedia says:

What exactly is change? Roughly speaking Final - Initial.
The initial here is $xy$.
We wish to express final in terms of change in the individual variables $x$ and $y$.
Hence we do $$d(xy) = (x+dx)(y+dy) - xy \ \ \ [Final \ - \ Initial]$$
A: Whenever dealing with derivatives change is an inevitable part. Now , what the person was trying to explain is let $A=x.y$ and now he was trying to find the CHANGE In Area $A$ for some infinitesimal small change in an arbitrary variable z. Mathematically that means $\frac{dA}{dz}$ which is similar to $\frac{\Delta A}{\Delta z}$ when the change in $\Delta z$ is not small and as we know $\Delta A $ means change in area of triangle i.e $A_0 -A_1$ which is what is done in the post .
Now , $\frac{dA}{dz}=\frac{d(x.y)}{dz}=
x.\frac{dy}{dz}+y.\frac{d x}{dz}$(product rule)
Taking ,$dz$ common from both sides and eliminating gives
$d(x.y)=x.dy+y.dx$
