Is this the right way of solving $frac{d}{dx}(sin(x)cdot x^2)$

I don’t know much about theorems related to limits and I’m currently learning calculus from 3Blue1Brown’s online series : Essence of Calculus.

This example’s generalization is what I’ll use to derive the product rule.

Now, if we have a rectangle with length and breadth equal to $sin(x)$ and $x^2$ for some values of $x$, then it’s area will be : $sin(x)cdot x^2$. Now, if we "nudge" the value of $x$ by some little amount, say, $dx$, then there will be corresponding changes in the values of $sin(x)$ and $x^2$. Let the little change in $sin(x)$ be $d(sin(x))$ and the little change in $x^2$ be $dx^2$.

Now, the change in the value of $(sin(x)cdot x^2)$ i.e. $d(sin(x)cdot x^2)$ will be the sum of the three new strips of area.
$$therefore~ d(sin(x)cdot x^2) = sin(x)cdot dx^2+x^2cdot d(sin(x))+dx^2cdot d(sin(x))$$.
Now, $dx^2 = 2xcdot dx$, $d(sin(x)) = cos(x)cdot dx$.
$$therefore~ dfrac{d(sin(x)cdot x^2)}{dx} = dfrac{sin(x)cdot dx^2+x^2cdot d(sin(x))+dx^2cdot d(sin(x))}{dx}$$
$$ = dfrac{sin(x)cdot 2xcdot dx + x^2cdotcos(x)cdot dx + 2xcdot dxcdot cos(x)cdot dx}{dx}$$
$$ = sin(x)cdot 2x + x^2cdot cos(x) + 2xcdot cos(x)cdot dx$$
Now, as $dx rightarrow 0$, $dfrac{d(sin(x)cdot x^2)}{dx} rightarrow sin(x)cdot 2x + x^2cdotcos(x)$ because anything in the form of $p(dx)^n$, where $p in Bbb R$ and $n in Bbb Z^+$ will approach $0$ as well.

So, we can say that $dfrac{d(sin(x)cdot x^2)}{dx} = sin(x)cdot 2x + x^2cdotcos(x)$ as
$$dfrac{d}{dx} f(x) = lim_{Delta x rightarrow 0}dfrac{f(x+Delta x) – f(x)}{Delta x}$$

I want to know if I’ve done all of this correctly, without any conceptual mistakes.

Thanks!