Proof of Derivative of a Scaled Exponential Function

First, $\mathbb{R}$ is an interval, and every $t\in\mathbb{R}$ is an interior point of it, since $t-1<t<t+1$ with $t-1,t+1\in\mathbb{R}$. Hence Derivative at an Interior Point applies at every point of $\mathbb{R}$, and every increment $h\ne0$ is admissible.

Case $c=0$. By item 1 of Basic Properties of the Exponential Function, $\exp(0)=1$, so $E_0$ is the constant function $1$. By the constant clauses of claims 2 and 4 of the one-dimensional rules lemma, $E_0$ is differentiable at every $t$ with derivative $0=c\,\exp(ct)$ and continuous at every $t$.

Case $c\ne0$. Fix $t\in\mathbb{R}$ and let $\varepsilon>0$. By item 3 of Basic Properties of the Exponential Function, $\exp$ is differentiable at the point $ct$ with derivative $\exp(ct)$, the derivative being the one-dimensional one of Derivative at an Interior Point. Applying that condition at $ct$ with tolerance $\varepsilon/|c|$, there is $\delta_0>0$ such that

$$\Bigl|\frac{\exp(ct+k)-\exp(ct)}{k}-\exp(ct)\Bigr|<\frac{\varepsilon}{|c|}\qquad(0<|k|<\delta_0).$$

Put $\delta=\delta_0/|c|$ and let $0<|h|<\delta$. With $k=ch$ we have $0<|k|<\delta_0$ and $E_c(t+h)=\exp(ct+k)$, so

$$\frac{E_c(t+h)-E_c(t)}{h}=c\,\frac{\exp(ct+k)-\exp(ct)}{k},$$

and therefore

$$\Bigl|\frac{E_c(t+h)-E_c(t)}{h}-c\,\exp(ct)\Bigr|=|c|\,\Bigl|\frac{\exp(ct+k)-\exp(ct)}{k}-\exp(ct)\Bigr|<|c|\cdot\frac{\varepsilon}{|c|}=\varepsilon.$$

Hence $E_c$ is differentiable at $t$ with $E_c'(t)=c\,\exp(ct)$. Continuity at every $t$ then follows from claim 1 of the one-dimensional rules lemma. $\blacksquare$