Why differentials can cancel in some versions of chain rule but not others? This makes perfect sense to me: $$\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}$$
This does not: $$\frac{\partial z}{\partial u}=\frac{\partial z}{\partial x}\cdot\frac{\partial x}{\partial u}+\frac{\partial z}{\partial y}\cdot\frac{\partial y}{\partial u}$$
Why can the differentials cancel in one but not the other? Is there any way to conceptualize the difference? Thanks!
 A: I am going to take a few liberties with language in hopes of conveying the spirit of the concepts:
In the first case, $y$ depends on $x$, which depends on $t$.
In the second, $z$ depends on $x$ and $y$, both of which depend on $u$.
In other words, the change in $z$ because $u$ changes is due to both $x$ and $y$ changing.  $$\frac{\partial z}{\partial x}\cdot \frac{\partial x}{\partial u}$$ is not the entirety of the change in $z$ because of a change in $u$.  It's only a piece of it.  Partial derivatives explicitly say that.  Only in the single variable function of a single variable case is the connection complete enough to "cancel" the intermediate step.  Because there is only one way to get from $y$ to $t$, through $x$, and so nothing is lost if you skip the intermediate step.
But for functions of more than one variable, there are multiple ways things can affect the output variable.  If you just say $\frac{\partial z}{\partial u}$, you are including all of the ways to get from $u$ to $z$.  But in $$\frac{\partial z}{\partial x}\cdot \frac{\partial x}{\partial u}$$ you are specifying which "path" you are taking, and that means you haven't taken the other path, and so you haven't accounted for everything.  That's what the partial symbol is supposed to indicate.
