I was digging deeply in the fundamentals of calculus, I found that in the famous '3Blue1Brown' channel, when demonstrating the process of finding the derivative of the product of two functions , (dx) ² is ignored because it becomes so tiny when dx is going closer and closer to zero.
I think that our calculus can be more precise if we don't ignore it, actually I think we shouldn't ignore it as this is math and not physics or such, the question is how our calculus is still valid and reliable if we give up even a tiny bit of precision.
The following image shows the details of the demonstration:
With all of the terms in $$df=\sin(x)d(x^2)+x^2d\sin(x)+d(x^2)d\sin(x)$$ to move forward, you will divide by $dx$: $$\frac{df}{dx}=\sin(x)\frac{d(x^2)}{dx}+x^2\frac{d\sin(x)}{dx}+d(x^2)\frac{d\sin(x)}{dx}$$ (I'm using $d$ as a positive quantity prior to taking the limit.)
Now take the limit as $dx\to0$ (which compels $d(x^2)\to0$).
$$\frac{df}{dx}=\sin(x)2x+x^2\cos(x)+0\cdot\cos(x)$$
So after taking the limit, this additional term certainly contributes $0$. (Note that I could have paired the last denominator with $d(x^2)$ instead, but then $d\sin(x)\to0$ and the end result is the same.)