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Ok, so I was doing a substitution problem and I realized that $dy = u\ dx + x\ du$ and not $\frac{dy}{dx} = u\ dx + x\ du$ and I was wondering what the difference was between those two. My first guess would be that $\frac{dy}{dx}$ means differential of $y$ with respect to $x$, and $dy$ would be differential of $y$, but I don't know what that implies exactly.
Can you provide examples so I can wrap my head around that concept?
If $\frac{dy}{dx}$ = integral of $3x$, what would $y$ be?
Both $dx$ and $dy$ are just symbols which mean nothing out of context, but probably there is a way to explain. $dx$ is increase of $x$ and $dy$ is increase of $y$. That means that if you add "infinitesimal" $dx$ to $x$, $y$ will grow by $dy$. In other words, $$dy=y(x+dx)-y(x).$$ If you write $\frac{dy}{dx}$, you mean $$ \frac{dy}{dx} = \lim_{\Delta x\rightarrow 0}\frac{y(x+\Delta x) - y(x)}{\Delta x}, $$ since $dx$ was assumed infinitesimal.
When you write $dy$, you always assume there is some $dx$. For example, for $y(x)=\sin(x)$ $$dy = \cos(x)\,dx;$$ for $y(x)=e^x$ $$dy = e^xdx = y(x) dx.$$
So, you can't just write something like $dy = 5$. May seem to have no sense, but the logic of infinitesimals may simplify proofs sometimes (and it was the only way to derivate before Cauchy!)
In contrast, $\frac{dy}{dx}$ is always a number (we are still talking about real functions, right?), which can depend on $x$. For example, for $y(x) = \sin(x)$ $$\frac{dy}{dx} = \cos(x).$$
In your example $y(x, u) = x\cdot u$, and $$dy=y(x + dx, u + du) - y(x, u) = u\,dx + x\,du + dx\cdot du,$$ and $dx \cdot du$ is omitted because it is much less than other summands and thus "insignificant". There can't be calculated $\frac{dy}{dx}$ anymore because we would have to divide infinitesimal $du$ by infinitesimal $dx$, but there is $$\frac{\partial y}{\partial x} = \lim_{\Delta x\rightarrow 0}\frac{y(x+\Delta x, u) - y(x, u)}{\Delta x} = u,$$ the partial derivative, which allows us to write equation for $dy$ in form $$dy = \frac{\partial y}{\partial x}dx + \frac{\partial y}{\partial u}du.$$
In conclusion, $dy$ is the full increase of $y$ depending on increases of all variables ($dx$, $du$, etc.) and $\frac{dy}{dx}$ or $\frac{\partial y}{\partial x}$ is the growth coefficient determining how much greater will grow $y$ with growth of only one variable $x$.
Differentials are closely related to the linear approximation of a function at a point. Take some continuous function $y(x)$, then it can be approximated well by $y(x)\approx y(x_0)+y'(x_0)(x-x_0)$, that is $y(x)= y(x_0)+y'(x_0)(x-x_0)+\epsilon(x_0)(x-x_0)$, where $\epsilon$ is a the error in the approximation. That equation basically says that a function is approximately equal to its tangent at $x_0$. Note that this approximation is good only for $x \approx x_0$.
As $x$ approaches $x_0$, the error term shrinks to $0$ and $y(x)=y(x_0)+y'(x_0)(x-x_0)$. To signify that $x-x_0$ must be very small, call it $dx$, so $y(x)=y(x_0)+y'(x_0)dx$. A differential of a function is $dy=y'(x_0)dx$: the change in $y$ co-ordinate of the tangent with an incremental increase in $x$, $dx$. Divide by $dx$ and take the limit as $dx \rightarrow 0$ to find the the increase in $y$ co-ordinate ($dy$) per incremental increase in $x$ co-ordinate (i.e. the derivative at that point, $\frac{dy}{dx}$, sort of).
For a concrete example, consider $y(x)=u(x)v(x)$, then, using the product rule, $y'(x_0)=u'(x_0)v(x_0)+u(x_0)v'(x_0)$, so $dy=u'(x_0)v(x_0)dx+u(x_0)v'(x_0)dx$, but note that $dv=v'(x_0)dx$ and $du=u'(x_0)dx$, so substituting we obtain $dy=v(x_0)du+u(x_0)dv$.
Note that a similar thing exists, $\Delta y(x)=y'(x_0)(x-x_0)+\epsilon(x_0)(x-x_0)$, that is the change in $y$ co-ordinate of the curve (not the tangent) with a change $x$ in $x$ co-ordinate, often written $y'(x_0)\Delta x+\epsilon(x_0) \Delta x$
