Derivatives: simplifying "d" of a number without being over "dx" I understand why $\dfrac {\mathrm d(x^2)}{\mathrm dx} = 2x$ since we're taking the derivative of $x^2$ with respect to $x$. Or $\dfrac {\mathrm dx^2}{\mathrm dx^2} = 1$ since we're taking the derivative of $x^2$ with respect to $x^2$ as the base variable.
From the textbook:
$$\begin{align}
x^2 &= u^3-1\tag1\\[5pt]
\frac {\mathrm dx^2}{\mathrm du} &= 3u^2\tag2\\[5pt]
dx^2&=3u^2\mathrm du\tag3\\[5pt]
2x\mathrm dx&=3u^2\mathrm du\tag4
\end{align}$$
How did the textbook go from step 3 to step 4? Specifically how does $\mathrm dx^2 = 2x\mathrm dx$?
 A: Thanks for the all hints, I figured it out.
Continuing from $(3)$:
$$\begin{align}
\mathrm dx^2&=3u^2\mathrm du\tag3\\[5pt]
\frac{\mathrm dx}{\mathrm dx}\mathrm dx^2&=3u^2\mathrm du\tag{3a}\\[5pt]
\mathrm dx\frac{\mathrm dx^2}{\mathrm dx}&=3u^2\mathrm du\tag{3b}\\[5pt]
\mathrm dx \cdot 2x&=3u^2\mathrm du\tag{3c}\\[5pt]
2x\mathrm dx&=3u^2\mathrm du\tag4
\end{align}$$
A: Differentials satisfy all of the familiar derivative laws. e.g. (where $x,y$ are variables and $n$ is a constant)
$$ \mathrm{d}n = 0 $$
$$ \mathrm{d}(x+y) = \mathrm{d}x + \mathrm{d}y $$
$$ \mathrm{d}(x-y) = \mathrm{d}x - \mathrm{d}y $$
$$ \mathrm{d}(xy) = x \mathrm{d}y + y \mathrm{d}x $$
$$ \mathrm{d}\left( \frac{x}{y} \right) = \frac{y \mathrm{d}x - x \mathrm{d}y}{y^2} $$
$$ \mathrm{d}(x^n) = n x^{n-1} \mathrm{d}x$$
$$ \mathrm{d}f(x) = f'(x) \mathrm{d}x $$
Since you can apply $\mathrm{d}$ to an identity to get a new identity, the equation you were aiming for could be derived in just one step, by applying the derivative laws
$$ \mathrm{d}(x^2) = 2x \mathrm{d}x $$
$$ \mathrm{d}(u^3 - 1) = \mathrm{d}(u^3) - \mathrm{d}1 = 3 u^2 \mathrm{d}u$$
A: This is a total derivative and you appear to be using it in the form taught as implicit differentiation.  The set of variables is $\{x,u\}$.  (Note that when we write $x^2 = u^3 - 1$, there is no clear division between dependent and independent variables.)  Then 
\begin{align}
d(x^2) &= \frac{\partial x^2}{\partial x} \mathrm{d} x + \frac{\partial x^2}{\partial u} \mathrm{d}u   \\
    &= (2x) \mathrm{d}x + (0) \mathrm{d}u  \\
    &= 2x \mathrm{d}x  \text{.}
\end{align}
It probably would have been better if your book had demonstrated the use of the chain rule here:  \begin{align}
\frac{\mathrm{d}(x^2)}{\mathrm{d}u} &= \frac{\mathrm{d}(x^2)}{\mathrm{d}x}\frac{\mathrm{d}x}{\mathrm{d}u}  \\
    &= 2x \frac{\mathrm{d}x}{\mathrm{d}u}  \text{,}
\end{align}
and then continue on by "multiplying" both sides by $\mathrm{d}u$.  (Warning:  there is a very good chance your "algebra brain" has just drawn an incorrect conclusion about how the products of derivatives work.)
There are a number of horrible ideas present in what you claim is in your book:  $\frac{\mathrm{d}(x^2)}{\mathrm{d}u}$ is not a fraction and you cannot go around multiplying things by plain "$\mathrm{d}u$"s safely without knowing what you are doing.  In particular, if taken at face value, the step of multiplying both sides of your equation by $\mathrm{d}u$ is equivalent to destroying all information in your equation by multiplying both sides by zero.  What is actually meant is that a limit on the left and a limit on the right of the equals sign approach zero in some particularly helpful way that is indicated by everything else in the equation.  However, this formal manipulation can be handy for solving certain types of equations.  But like any formal manipulation, it can only suggest a solution; you must still verify that the suggestion is a solution.
A: Really, $\frac{d(x^2)}{dx} = 2x$ and $d(x^2) = 2xdx$ are just slightly different notations for the same statement. There is no multiplication going on, it just kinda looks like it.

This is a better chain of reasoning:
$x^2+c = \int{2xdx}$.
By differentiating both sides, you get
$d(x^2)+d(c) = d(\int{2xdx})$.
$d(c)$ is $0$, and $d(\int{()})$ cancels out, so
$d(x^2) = 2xdx$

And you can (hopefully) obviously extend that reasoning to other functions
A: Just as you said in the first line, $\frac{d(x^2)}{dx}=2x$. If you multiply both sides by $dx$, you get number 4.
