I have already seen some posts about a different way to write a Taylor series of a smooth function $f : \mathbb{R} \longrightarrow \mathbb{R} $, but all of them are written using the lagrange notation for derivative, $f'(x)$, and I'm trying to develop the same formula in the leibniz notation $\frac{d}{dx} f(x)$. Here's my attempt:
Suppose $f$ is a one-variable function smooth at some point $x \in \mathbb{R}$. Then its Taylor series around $x$ is given by
$$ f(y) = f(x) + (y-x)\frac{d}{dy}[f(y)] \Bigg|_{y=x} + \frac{1}{2!}(y-x)^2 \frac{d^2}{dy^2} [f(y)] \Bigg|_{y=x} + \cdots$$
If $\epsilon := y-x \implies f(y) = f(x + (y-x)) = f(x+\epsilon)$, such that
$$ f(x+\epsilon) = f(x)+ \epsilon\frac{d}{dy}[f(y)]+\Bigg|_{y=x} + \frac{1}{2!}\epsilon^2 \frac{d^2}{dy^2} [f(y)] \Bigg|_{y=x} + \cdots $$
If I state that $\frac{d^n}{dy^n} [f(y)] \Bigg|_{y=x} \equiv \frac{d^n}{dx^n} [f(x)]$, the taylor series is
$$ f(x+\epsilon) = f(x)+ \epsilon\frac{d}{dx}[f(x)]+ \frac{1}{2!}\epsilon^2 \frac{d^2}{dx^2} [f(x)] + \cdots$$
Is this notation for changing the variable in the derivatives from $y$ to $x$ confusing?
Note: I know how to proceed using Lagrange notation for derivatives, my problem is due exclusively to Leibniz notation.
