How can $\frac{\mathrm{d}}{\mathrm{d}x}\left[y(u(x))\right] = \frac{\mathrm{d}y}{\mathrm{d}x}$? I just saw a video on the chain rule, and it stated that
$$\frac{\mathrm{d}}{\mathrm{d}x}\left[y(u(x))\right] = \frac{\mathrm{d}y}{\mathrm{d}x}$$
I don't understand this; if I let $y(x) = x^2$ and $u(x) = \sqrt x$ then
$$\frac{\mathrm{d}y}{\mathrm{d}x} = 2x$$
and
$$\frac{\mathrm{d}}{\mathrm{d}x}\left[y(u(x))\right] = \frac{\mathrm{d}}{\mathrm{d}x} \left[x\right] = 1$$
Clearly, I am completely misunderstanding something. What is it?
EDIT: It is my understanding right now that $y(u(x)) = (\sqrt x)^2 = x$. Is this wrong?
 A: But it is not $y(x)$,
it is $y(u(x))$.
So, if $y(v) = v^2$
and
$u(x) = \sqrt{x}$,
then
$y(u(x))
=y(\sqrt{x})
=(\sqrt{x})^2
=x
$.
If we apply the chain rule,
then,
since $y'(v) = 2v$,
$(y(u(x))'
=y'(u(x))u'(x)
=2u(x) (\sqrt{x})'
=2\sqrt{x} (\frac12 x^{-{1/2}})
=1
$.
By the way,
the proof in that video
is incomplete.
It fails to consider
the possibility that
$u'(x) = 0$.
Hardy's classic
"A Course of Pure Mathematics"
has a discussion of this
 on page 217 (section 114) in the
10th edition.
He says that this is a common error.
A: Im going to use the following notation, hoping that it will be clearer for you.
$\frac{d}{dx}(f(x)) = f'(x)$. Using this notation the chain rule says that
$\frac{d}{dx}[y(u(x))]=[y(u(x))]'=u'(x)[y'(u(x))]=\frac{du}{dx}(x)\frac{dy}{dx}(u(x))$
I used to translate this symbol pattern to english in the following way, given a composite function $y(u(x))$, I call $y(x)$ the "outside" and $u(x)$ the "inside" function. Then, the derivative of a composite function $y(u(x))$ is the derivative of the "inside" function ($u'(x)$) times the derivative of the "outside" function, $evaluated$ in the "inside" function ($y'(u(x))$). I hope it helps!
A: Your confusion is reading the abbreviation of $\frac{\mathrm d y}{\mathrm d x}$ as $\frac{\mathrm d y(x)}{\mathrm d x}$ instead of $\frac{\mathrm d y(u(x))}{\mathrm d x}$ as you ought.
$$\begin{align}
 u(x) & \mathop{:=} \surd x
\\ y(u) & \mathop{:=} u^2
\\ \therefore y(u(x)) & = y(\surd x)
\\ & = (\surd x)^2
\\ & = x
\end{align}$$
The dependent variable $y$ is defined as a function of the dependent variable $u$, which is in turn defined as a function of the independent variable $x$. Thus $y$ is a composition function when expressed with respect to $x$.
It is to be understood that when we differentiate the dependent variable $y$ with respect to $x$ we are differentiating this composition.
For clarity let's use different letters for the dependent variables and the functions by which they are defined.   Then we have an independent variable $x$, and dependent variables $u$ and $y$ are defined as functions $f$ and $g$, such that $y=f(u)$ and $u=g(x)$.   When differentiating $y$ with respect to $x$ we apply the chain rule to the composition of $f$ and $g$ (that is $f\circ g).$
$\begin{align}
u & \mathop{:=}g(x) 
\\[1ex] y & \mathop{:=} f(u) 
\\[1ex] & = [f\circ g](x) 
\\[3ex] \therefore \dfrac{\mathrm d y}{\mathrm d x}
 & = \dfrac{\mathrm d [f\circ g](x)}{\mathrm d x} 
\\[1ex] & = \dfrac{\mathrm d f(g(x))}{\mathrm d g(x)} \cdot \dfrac{\mathrm d g(x)}{\mathrm d x} 
\\[1ex] & = \frac{\mathrm d y}{\mathrm d u}\frac{\mathrm d u}{\mathrm d x}
\\[2ex] & = \frac{\mathrm d u^2}{\mathrm d u}\frac{\mathrm d \surd x}{\mathrm d x}
\\[1ex] & = 2u\cdot\frac {1}{2\surd x}
\\[1ex] & = \frac{2\surd x}{2\surd x}
\\[1ex] & = 1
\end{align}$
