How can I express $\frac{dy}{dx}$ if I make the change of variable $x-1 = t$?

Question

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 $\frac{dy}{dt}$.

Answer 1

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: $$\frac{dy}{dx}=\frac{dy}{dt}$$

Answer 2

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.

Answer 3

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 $\frac{dy}{dx} = \frac{dy}{dt}$.