I don't follow one step in the first proof of Theorem $7$ of Appendix $C$ of Partial Differential Equations by Lawrence Evans. This subsection goes through properties of mollifiers. The step I don't understand is highlighted in red text below.
Theorem: $$f^{\epsilon}\in C^{\infty}(U_{\epsilon}).$$
Proof: Fix $x\in U_{\epsilon}$, $i\in\{1,2,\ldots,n\}$, and $h$ so small that $x+he_i\in U_{\epsilon}$. Then \begin{align} \frac{f^{\epsilon}(x+he_i) - f^{\epsilon}(x)}{h} &= \frac{1}{\epsilon^n}\int_U \frac{1}{h}\left[\eta\left(\frac{x+he_i-y}{\epsilon}\right) - \eta\left(\frac{x-y}{\epsilon}\right)\right]f(y)\,dy \\&= \frac{1}{\epsilon^n}\int_V \frac{1}{h}\left[\eta\left(\frac{x+he_i-y}{\epsilon}\right) - \eta\left(\frac{x-y}{\epsilon}\right)\right]f(y)\,dy \end{align} for some open set $V \subset\subset U$. As $$\frac{1}{h}\left[\eta\left(\frac{x+he_i-y}{\epsilon}\right) - \eta\left(\frac{x-y}{\epsilon}\right)\right] \to \color{red}{\frac{1}{\epsilon}}\eta_{x_i}\left(\frac{x-y}{\epsilon}\right)$$ uniformly on $V$, the partial derivative $f_{x_i}^{\epsilon}(x)$ exists and $$f_{x_i}^{\epsilon}(x)=\int_U \eta_{\epsilon,x_i}(x-y)f(y)\,dy.$$ A similar argument shows that $D^{\alpha}f^{\epsilon}(x)$ exists, and $$D^{\alpha}f^{\epsilon}(x)=\int_U D^{\alpha}\eta_{\epsilon}(x-y)f(y)\,dy \quad (x\in U_{\epsilon}),$$ for each multiindex $\alpha.$
Question: The uniform convergence follows from a similar argument as this question. Where does the constant $\color{red}{\frac{1}{\epsilon}}$ come from? I think this should be $$\frac{1}{h}\left[\eta\left(\frac{x+he_i-y}{\epsilon}\right) - \eta\left(\frac{x-y}{\epsilon}\right)\right] \to \color{blue}{\epsilon}\cdot\eta_{x_i}\left(\frac{x-y}{\epsilon}\right)$$ as $$\frac{d}{dx}\left(\frac{x-y}{\epsilon}\right)=\frac{1}{\epsilon}$$ which would multiply (and not divide) the above quantity by $\color{blue}{\epsilon}.$ I am probably forgetting something about the definition of the directional derivative in multivariate calculus.
Background Notation (as discussed by Evans in the appendix).
If $U\subset \mathbb R^n$ is open and $\epsilon > 0,$ we write $$U_{\epsilon}:=\{x\in U ~|~ \text{dist}(x,\partial U) > \epsilon \},$$ where dist = distance between $x$ and the boundary of $U$.
We define $\eta\in C^{\infty}(\mathbb R^n)$ by $$\eta(x):= \begin{cases} C~\text{exp}\left(\frac{1}{|x|^2-1}\right) & \text{if $|x| < 1$} \\ 0 & \text{if $|x| \ge 1$}, \end{cases} $$ where the constant $C$ is selected so that $\int_{\mathbb R^n} \eta \,dx = 1$.
For every $\epsilon > 0 $, set $$\eta_{\epsilon}(x):=\frac{1}{\epsilon^n}\eta\left(\frac{x}{\epsilon}\right).$$ We call $\eta$ the standard mollifier. The functions $\eta_{\epsilon}$ are $C^{\infty}$ and satisfy $$\int_{\mathbb R^n} \eta \,dx = 1,\quad \text{spt}(\eta_{\epsilon})\subset B(0,\epsilon).$$
If $f:U\to\mathbb R$ is locally integrable, we define its mollification by $$f^{\epsilon}:= \eta_{\epsilon}\ast f\quad \text{in $U_{\epsilon}$.}$$ That is, $$f^{\epsilon}(x)=\int_U \eta_{\epsilon}(x-y)f(y)\,dy=\int_{B(0,\epsilon)}\eta_{\epsilon}(y)f(x-y)\,dy$$ for all $x\in U_{\epsilon}.$