Proof of Derivative and Continuity of the Scaled Exponential Function

Throughout, $|\cdot|$ is the absolute value on $\mathbb{R}$, so that $d_{\mathbb{R}}(s,t)=|s-t|$ for all real $s,t$ by The Absolute Value Metric on the Real Line.

Claim 1. Fix $t\in\mathbb{R}$.

Case $c=0$. By claim 1 of Basic Properties of the Exponential Function, $\exp(0)=1$, so $E_0(u)=1$ for every $u\in\mathbb{R}$ and every difference quotient of $E_0$ is $0$. Since also $c\exp(ct)=0$, for every real $\varepsilon>0$ the choice $\delta=1$ witnesses that $E_0$ is differentiable at $t$ with derivative $0=c\exp(ct)$.

Case $c\ne0$. Then $|c|>0$. By claim 3 of Basic Properties of the Exponential Function, $\exp$ is differentiable at $0$ with $\exp'(0)=\exp(0)=1$, the last equality by claim 1 there. By claim 2 of Basic Properties of the Exponential Function, $\exp(ct)>0$.

Let $\varepsilon\in\mathbb{R}$ with $\varepsilon>0$. Then

is a positive real, so by the differentiability of $\exp$ at $0$ there is a real $\delta'>0$ such that every real $k$ with $0<|k|<\delta'$ satisfies

the difference quotient of $\exp$ at $0$ being $(\exp(k)-\exp(0))/k=(\exp(k)-1)/k$.

Put $\delta=\delta'/|c|$, a positive real, and let $h$ be real with $0<|h|<\delta$. Set $k=ch$; then $0<|k|=|c|\,|h|<|c|\,\delta=\delta'$. By claim 1 of Basic Properties of the Exponential Function,

and since $h=k/c$,

Therefore

As $\varepsilon>0$ was arbitrary, $E_c$ is differentiable at $t$ with $E_c'(t)=c\exp(ct)$.

Claim 2. By claim 1, $E_c$ is differentiable at every $t\in\mathbb{R}$, and every such $t$ is an interior point of the interval $\mathbb{R}$, as recorded in the statement. Hence Differentiability at an Interior Point Implies Continuity There, applied with $I=\mathbb{R}$ at each point $t$, shows that $E_c$ is continuous at $t$ relative to $\mathbb{R}$ for every $t$, that is, continuous on $\mathbb{R}$. $\blacksquare$