How can I express $\frac{dy}{dx}$ if I make the change of variable $x-1 = t$? If $y$ is a function of x and I make the change of variable $x-1=t$. Now what would $\frac{dy}{dx}$ equal in terms of $\frac{dy}{dt}$.
 A: I am writing this answer in a slightly different notation. Although this notation may look cumbersome at the first sight but after a while you will appreciate it as this is really the true way. Use the chain rule.
$$D\Big[f \circ g\Big]=\Big[Df\circ g\Big] \cdot Dg \tag{1}$$
In your example
\begin{align*}
g:&\mathbb{R}\to\mathbb{R} \\
&\verb!#! \to \verb!#!+1
\tag{2}
\end{align*}
and so the derivative of $g$ will be
\begin{align*}
Dg:&\mathbb{R}\to\mathbb{R} \\
&\verb!#! \to 1
\tag{3}
\end{align*}
evaluating the functional equation $(1)$ at a point $x$, we obtain
\begin{align*}
D\Big[f \circ g\Big](x) &= \Big[Df\circ g\Big](x) \cdot Dg(x) \\
&= \Big[Df\circ g\Big](x) \cdot 1 \\
&= \Big[Df\circ g\Big](x)
\tag{4}
\end{align*}
so the final result is that
$$D\Big[f \circ g\Big]=Df\circ g \tag{5}$$

The final result in the other two answers is $\frac{dy}{dx}=\frac{dy}{dt}$. The method that these answers follow is the common way that is usually taught in a calculus course. However, I believe that such a notation is totally misleading as  it does not make any difference between functions and their inputs and outputs! Also, the compositions are not expressed precisely.

A: As $x - 1= t$, hence $\frac{dt}{dx} = 1$
$$\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = \frac{dy}{dt} \cdot (1)$$
So you get $\frac{dy}{dx} = \frac{dy}{dt}$.
A: As: $x-1=t$ then $x(t)=t+1$ so $\frac{dx}{dt}=1$
Now: $$y(x(t))'=\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}=\frac{dy}{dx}\times1=\frac{dy}{dx}$$
Therefore: $$\frac{dy}{dx}=\frac{dy}{dt}$$
