What does my calculus textbook imply that differentials can't be manipulated algebraically? The textbook defines differentials like this.
Let $y=f(x)$ be a differentiable function of $x$. The differential of $x$ (denoted by $dx$) is any nonzero real number. The differential of $y$ (denoted by $dy$) is equal to $f'(x)dx$.
It goes on to say that the derivative rules can be written in differential form using Leibniz notation. For example, it says the chain rule in differential form is
$$\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$$
The book says it appears to be true because the $du$'s would divide out, and although the reasoning is incorrect, it helps you remember the chain rule.
Why is the reasoning incorrect? Given those definitions of differentials, what's stopping you from manipulating them algebraically?
 A: First, $\mathrm{d}x$ is not defined.  The limit of the difference quotient is what is defined:  $\frac{\mathrm{d}y}{\mathrm{d}x} = \lim_{h \rightarrow 0} \frac{y(x+h) - y(x)}{(x+h)-x}$.  Pretending that the parts of the expression $\frac{\mathrm{d}y}{\mathrm{d}x}$ are individually defined makes as much sense as treating the "$\mathrm{i}$" in "$\lim$" as if it were a separable thing.  Pretending that you can cancel parts of these expressions makes as much sense as the "derivation":  $\frac{\sin x}{n} = \frac{\mathrm{si}\not\mathrm{n} x}{\not n} = \mathrm{six} = 6$.
Note that in the definition, the "$y$" appearing in the numerator is treated as a function of the "$x$" appearing in the denominator and the "$x$" is treated as an independent variable.  So in $\frac{\mathrm{d}y}{\mathrm{d}u}$, $u$ is an independent variable, but in $\frac{\mathrm{d}u}{\mathrm{d}x}$, $u$ depends on $x$.  These are not the same $u$, in spite of the fact that we write the same thing for them.  So even if we were to somehow change all the definitions so that the sequences of symbols $\mathrm{d}y$, $\mathrm{d}x$, and $\mathrm{d}u$ had independent existence, one would still need a big theorem to justify cancelling the $\mathrm{d}u$s.  The other way to correct this is to realize that $y(u)$ is really $y(u(x))$ and apply the chain rule, yielding $y'(u(x)) u'(x)$, which should look familiar.  (It's the same expression as $\frac{\mathrm{d}y}{\mathrm{d}u} \frac{\mathrm{d}u}{\mathrm{d}x}$, but in other notation.)  Here it's more clear that it's crazy to "cancel the $u$s".
