Is this notation inconsistent? If not, can some explain why not? Im working through a textbook section on particle kinematics. An example given is relating vertical velocity to horizontal velocity and states:
$y$ has a constant velocity of $10 \ \rm [m/s]$
$y=(0.001x^2)\ \rm [m]$
when $y=100\ \rm [m]$ then $x=316.2\ \rm [m]$ and $t=10 \ \rm [s]$
Here comes my problem. The book states exactly this:
$v_y=y'=\frac{d}{dt}(0.001x^2)=(0.002x)x'=0.002xv_x$
My problem is that $x'$ seems to have either appeared from nowhere or been omitted in the first stages.
$\frac{d}{dt}(0.001x^2)=0.002x$ not $(0.002x)x'$
To me this seems sloppy or inconsistent. I keep running into little problems like this throughout this book and then waste time trying to figure out what's going on. It also erodes my trust in what I'm reading .
Can anyone offer some insight into this?
 A: I think you are getting confused with derivatives. You write:
$$\frac{d}{d\color{red}t} (0.001 x^2) = 0.002 x\,, \tag{1}$$
but this is wrong. It would be true if the derivative was with respect to $x$, i.e.:
$$\frac{d}{{d\color{red}x}} (0.001 x^2) = 0.002 x\,. \tag{2}$$
Here you have to take a derivative with respect to the time $t$, and for that you have to use the chain rule, which states:
$$\frac{d}{dt} f(g(t)) = f'(g(t)) g'(t)\,, \tag{3}$$
where the notation $f'$ means derivative with respect to the argument of the function, or more concretely:
$$f'(g(t)) = \frac{d}{dg(t)} f(g(t))\,, \qquad g'(t) = \frac{d}{dt} g(t)\,. \tag{3}$$
In your example $x'$ is a common shorthand notation that means:
$$x' := \frac{d}{dt} x(t)\,, \tag{4}$$
and the derivative of the position $x$ with respect to the time $t$ is the velocity $v$ (note that the notation $\dot{x}$ is also often used for denoting derivatives with respect to time). It is a sensible operation to do, as this derivative means the change in position as time passes by, which is a function you are usually eager to know in physics.
In your case the functions $f$ and $g$ presented in the chain rule above would be:
$$f(x) = x^2\,, \qquad g(t) = x(t)\,, \tag{5}$$
which result the following derivatives:
$$f'(x) = \frac{d}{dx} x^2 = 2 x\,, \qquad g'(t) = \frac{d}{dt} x(t) = x'(t)\,. \tag{6}$$
And you should find:
$$\begin{align} y' &= \frac{d}{dt} (x(t))^2 \\ &= \frac{d}{d (x(t))} (x(t))^2 \frac{d}{dt} x(t) \\ &= 2 x(t) x'(t)\,. \tag{7} \end{align}$$
(I have omitted the $0.001$ prefactor for clarity. I have also written $x(t)$ everywhere to emphasise the fact that $x$ is really a function of time, but if it bothers you you can remove the $(t)$ as long as you don't forget that this dependence is there!)
Hope this helps! Otherwise here is a good source for reviewing the chain rule.
A: This comes from the chain rule.
Your $y$ expression is expressed as a function of $x$. However, both $x$ and $y$ depend on time. Therefore, it can be written that:
$$y(t)=\left( x(t) \right)^2$$
I omitted your constant for readability.
Then, by the chain rule,
$$\dfrac{\rm d}{\mathrm dt}y(t)=\dfrac{\mathrm d}{\mathrm dt}\left( x(t) \right)^2 = 2x(t)\dfrac{\mathrm d}{\mathrm dt} x(t) =2xx'$$
