Why does the d/dx operator output a derivative, whereas the integral operator accepts a differential?

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Say I have differentiation operators $D_1$ and $D_2$, where $D_1$ is the normal gradient operator, and $D_2$ yields a differential

$$D_2 \Big{[} y = sin(x) \Big{]}$$ $$dy = cos(x) dx$$

It would seem like $D_2$ is the reverse of integration, not $D_1$, i.e.

$$\int [ \hphantom{.}dy = cos(x)dx \hphantom{.}] $$ $$ y = sin(x) + c$$

It recently occurred to me that the integration operator accepts a differential $df = {df \over dx}dx$, whereas the standard differentiation operator $D_2$ produces a derivative ${df \over dx} = {df \over dx}$.

I also noticed that chain rule is much easier when you have $y = \cos(u) du$, because you can use u-sub notation to do chain rule, i.e. $u = g(x), du = [...]dx$.

Is there a reason we don't use $D_2$? I.e. is it mathematically invalid to have an operator that takes that returns a differential $f'(x)dx$ instead of $f'(x)$. I've found one video that claims "d" is the differential operator whereas $\nabla$ is the gradient operator, but I can't find anything else online where it's recognized that there is some operator that returns differentials instead of derivatives?

I feel like many people first assume the integration operator is $\int (*) dx$ instead of $\int(*)$ since they are used to working with derivatives instead of differentials.