Why doesn't it work when I calculate the second order derivative? Let $y=y(x)$ be determined by the equation \begin{align*}\begin{cases} x=t-\sin{t}\\ y=1-\cos{t}.\end{cases}
\end{align*}  I understand the solution:   $$\frac{d^2y}{dx^2}=\frac{d(\frac{dy}{dx})}{dt}\frac{1}{\frac{dx}{dt}}=-\frac{1}{(1-\cos{t})^2}$$ But what's wrong with the following calculations:                       $$\frac{d^2y}{dx^2}=\frac{d^2y}{dt^2}\frac{dt^2}{dx^2}=\frac{\cos{t}dt^2}{dt^2}\frac{dt^2}{((1-\cos{t})dt)^2}=\frac{\cos{t}}{(1-\cos{t})^2}$$
 A: By the use of the chain's rule you get
$\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}$
then 
$$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}.\qquad(1)$$
Now $\frac{d}{dt}\!\!\left(\frac{dy}{dt}\right)=\frac{d}{dt}\!\!\left(\frac{dy}{dx}\frac{dx}{dt}\right)=\frac{d}{dx}\left(\frac{dy}{dx}\frac{dx}{dt}\right)\frac{dx}{dt}=\frac{d^2y}{dx^2}\left(\frac{dx}{dt}\right)^2+\frac{dy}{dx}\frac{d}{dx}\left(\frac{dx}{dt}\right)\frac{dx}{dt}$. 
Then
$$\frac{d^2y}{dt^2}=
\frac{d^2y}{dx^2}\left(\frac{dx}{dt}\right)^2+\frac{dy}{dx}\frac{d^2x}{dt^2}.\qquad(2)$$
So upon substitution of $(1)$ in $(2)$ and solving for $\frac{d^2y}{dx^2}$ we get
$$\frac{d^2y}{dx^2}=\frac{\frac{d^2y}{dt^2}\frac{dx}{dt}-\frac{d^2x}{dt^2}\frac{dy}{dt}}{\left(\frac{dx}{dt}\right)^3},$$
which can be used to solve your problem.
A: Good question! :) The thing to remember is that differentiation is an operator. It isn't a scalar variable that can be tossed around. Here is an example:
As you know, the definition of the first derivative would be:
$$\frac{dy}{dx} \approx \frac{\Delta( y)}{\Delta x}$$
However, we don't know what the change in $y$ is for a given change in $x$ directly. This is why we need to use a common variable, $t$. So the equation becomes:
$$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}} \approx\frac{\frac{\Delta y}{\Delta t}}{\frac{\Delta x}{\Delta t}}$$
If we choose exactly the same bounds in $t$ for $x$ and $y$, this convention is sort of a crude way to describe it. What you are actually doing is this:
$$\frac{dy}{dx}=\frac{\frac{d}{dt}(y)}{\frac{d}{dt}(x)}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$
When we deal with the second derivative, the approximation would be:
$$\frac{d^2y}{dx^2}\approx\frac{\Delta\left(\frac{\Delta y}{\Delta x}\right)}{\Delta x}$$
What we are approximating is the change in slope over the same interval $\Delta x$. Going back to the differentials, we get:
$$\frac{d^2y}{dx^2}=\frac{\frac{d}{dt} (\frac{dy}{dx})}{\frac{d}{dt}(x)}$$
In all these cases, the results of the differentiation are in terms of $t$, not $x$ or $y$, which is part of the issue.
