Why is $\frac{d}{dx}(x^3)$ not $3x^2 + 3x$? In 3Blue1Brown's 'The Essence of Calculus' chapter 3 he shows a geometric analogy of why the derivative of $f(x)=x^3$ is $3x^2$.

I understand why we can ignore the tiny cube in the corner. But why do we also ignore the three lines along the edges of the cube? Each has a volume of $x$ as $dx$ approaches zero.
 A: Let's replace the infinitesimal $\mathrm dx$ with a finite $h\in\Bbb R^+$, so$$f(x+h)-f(x)=3x^2h+3xh^2+h^3.$$(One could adopt axioms about powers of $\mathrm dx$ that let me work with it instead of $h$, but right now it's not worth the hassle.) We define$$f^\prime(x):=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\left(3x^2+3xh+h^2\right)=3x^2.$$The $h\to0$ limit doesn't just delete the numerator's fastest-shrinking term $h^3$, i.e. the one with the highest power of $h$. It deletes every term that shrinks faster than the denominator $h$.
The geometric insight is that, for small $h$, the strips along three of the cube's edges have much less volume than the squares covering three of its faces, even if the volume at one vertex is much smaller still. When taking the first derivative, the highest-order correction (i.e. largest piece added) wins.
A: $$d(x^3)=(x+d\!x)^3-x^3=3x^2d\!x+3xd\!x^2+d\!x^3$$
where $\!x$ is one symbol name.
Therefore
$$\frac{d}{d\!x}(x^3)=3x^2+3xd\!x+d\!x^2$$
and further
$$\frac{d}{d\!x}(x^3)=3x^2$$
if we ignore the two infinitesimal summands.
A: Take a cube of $1$ meter size. Table another one of $1.01$ meter. The new volume is
$$(1.01)^3=1.030301000$$ So
$$\frac{(1.01)^3 -(1.00)^3} {0.01 }=3.0301$$ Now doing it for a cube of $1.001$
$$\frac{(1.001)^3 -(1.000)^3} {0.001 }=3.003001$$
