Yes.
The keystone is:
Lemma. Let $f\colon [a,b]\to\mathbb R$ be continuous and assume that $f'_+(x)$ exists and is $>0$ for all $x\in [a,b)$. Then $f$ is strictly increasing.
Assume otherwise, i.e. $f(a)\ge f(b)$.
We recursively define a map $g\colon \operatorname{Ord}\to [a,b)$ such that $g$ and $f\circ g$ are strictly inreasing. Since the class $\operatorname{Ord}$ of ordinals is a proper class and $g$ is injective, we arrive at a contradiction, thus showing the claim.
- Let $g(0)=a$.
- For a successor $\alpha=\beta+1$ assume we have already defined $g(\beta)$. For sufficently small positive $h$ we have that $g(\beta)<g(\beta)+h<b$ and $\frac{f(g(\beta)+h)-f(g(\beta))}{h}\approx f_+(g(\beta))>0$. Pick one such $h$ and let $g(\alpha)=g(\beta)+h$.
- If $\alpha$ is a limit ordinal, assume $g(\beta)$ is defined for all $\beta<\alpha$. Let $x=\sup_{\beta<\alpha} g(\beta)$. A priori only $x\le b$, but we need $x<b$. Because $f$ is continuous and $f\circ g$ is strictly increasing, we conclude that $f(x)=\sup_{\beta<\alpha} f(g(\beta))\ge f(g(1))>f(g(0))=f(a)=f(b)$. Therefore $x<b$ as desired and we can let $g(\alpha)=x$.
$\square$
Corollary 1. (something like a one-sided Rolle theorem) Let $f\colon [a,b]\to\mathbb R$ be continuous with $f(a)=f(b)$. Assume $f_+$ exists and is continuos in $[a,b)$. Then $f'_+(x)=0$ for some $x\in[a,b)$.
Proof. Assume otherwise. Then either $f_+(x)>0$ for all $x$ or $f_+(x)<0$ for all $x$. In the first case the lemma applies and gives us a contradiction to $f(a)=f(b)$; in the other case, we consider $-f$ instead of $f$. $\square$
Corollary 2. (something like a one-sided IVT) Let $f\colon [a,b]\to\mathbb R$ be continuous. Assume $f_+$ exists and is continuos in $[a,b)$. Then $f'_+(x)=\frac{f(b)-f(a)}{b-a}$ for some $x\in[a,b)$.
Proof. Apply the previous corollary to $f(x)-\frac{f(b)-f(a)}{b-a}x$. $\square$
By symmetry, we have
Corollary 3. Let $f\colon [a,b]\to\mathbb R$ be continuous. Assume $f_-$ exists and is continuos in $(a,b]$. Then $f'_-(x)=\frac{f(b)-f(a)}{b-a}$ for some $x\in(a,b]$. $\square$
Theorem. Let $f\in C(\mathbb R)$ be a function with $f'_-$ continuous on $\mathbb R$.
Then $f\in C^1(\mathbb R)$.
Proof.
Consider aribtrary $a\in \mathbb R$.
Let $\epsilon>0$ be given.
Then by continuity of $f'_-$, for some $\delta>0$ we have $|f'_-(x)-f'_-(a)|<\epsilon$ for all $x\in(a,a+\delta)$.
Thus for $0<h<\delta$ we have $\left|\frac{f(a+h)-f(a)}{h}-f'_-(a)\right|<\epsilon$ by corollary 3. We conclude that $f'_+(a)=f'_-(a)$, i.e. $f$ is differentiable at $a$. $\square$
Best Answer
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If $f(x)=|e^x+(1+i)e^{-x}|$ then $$F(x)=f(x)^2=(e^x+e^{-x})^2+e^{-2x}=e^{2x}+2+2e^{-2x}.$$ Then $F'(x)=2f(x)f'(x)$ and so $$f'(x)=\frac{F'(x)}{2f(x)}=\frac{e^{2x}-2e^{-2x}}{f(x)}.$$