If the right derivative is positive at each point, can I conclude that $f$ is increasing at right of each point? Let $f:\mathbb R\to \mathbb R$ a derivable function at right  of each point, i.e. $$\lim_{h\to 0^+}\frac{f(x+h)-f(x)}{h}=f_d'(x)$$
exist for all $x$. We suppose that $f_d'(x)\geq 0$ for all $x$. Can we conclude that $f$ is increasing at right of each point ? i.e for all $x$, there is $\delta =\delta_x $ s.t. $f(x)\leq f(y)$ for all $y\in [x, x+\delta ]$ ?
I tried as follow :
$$f(x+h)=f(x)+f'_d(x)h+h\varepsilon (h)$$
for $h>0$ where $\varepsilon (h)\to 0$ when $h\to 0^+$. In particular, $$f(x+h)\geq f(x)+h\varepsilon (h).$$
Now, I don't see why this would prove the statement since $\varepsilon (h)$ could be negative. Any idea ?
 A: Here is a counterexample. Define 
$$
f(x) = \begin{cases}
0 & (x \leq 0)\\
-\frac{1}{2^n} & \left(\frac{1}{\sqrt{2^n}} \leq x < \frac{1}{\sqrt{2^{n-1}}};\  n > 1 \right)\\
-\frac{1}{2} & \left(x\geq \frac{1}{\sqrt 2}\right)
\end{cases}
$$
Clearly 
$$f'_+(x) = \lim_{h\rightarrow 0^+}\frac{f(x+h)-f(x)}{h}=0$$
for all $x \neq 0$.
To show that the right derivative in the origin is $0$, note that for $x>0$
$$ -x^2 \leq f(x) \leq 0,$$
so that 
$$-x \leq \frac{f(x)}{x} \leq 0.$$
Thus
$$f'_+(0) = \lim_{h\rightarrow 0^+} \frac{f(h)}{h}=0,$$
by the squeezing rule.
Note, by the way, that $f(x)$ is continuous and differentiable in $0$.
In the Figure below you see a plot of the function. The green line corresponds to the function $y=-x^2$ and the red line to the function $y=-\frac{1}{2}x^2$.
So $f(x)$ satisfies the requirements, i.e. $f'_+(x) \geq 0$, for all $x\in \Bbb R$, but for any $\delta > 0$, $x\in (0,\delta) \Rightarrow f(x) < 0$. 
In conclusion, if $f(x)$ is not continuous your statement is clearly false.

EDIT 
Following the same approach you can even force the right derivative at each point (except $0$) to be strictly positive. Consider, e.g., the function in the Figure below, where red lines are graphs of $y=-x|x|$ and $y=-\frac{1}{2}x|x|$. Then $f(x)$ is defined as follows.
$$f(x) = \begin{cases}\frac{\sqrt{2^{-k}}-\sqrt{2^{-k+1}}}{\sqrt[4]{2^{-k+1}}-\sqrt[4]{2^{-k}}}(x+\sqrt[4]{2^{-k+1}})- \sqrt{2^{-k+1}} & \left(-\sqrt[4]{2^{-k+1}}\leq x<-\sqrt[4]{2^{-k}}; \ k\in \Bbb Z\right)\\ 0 & (x=0)\\ \frac{\sqrt{2^{-k}}-\sqrt{2^{-k+1}}}{\sqrt[4]{2^{-k+1}}-\sqrt[4]{2^{-k}}}(x-\sqrt[4]{2^{-k+1}})+ \sqrt{2^{-k+1}}& \left(\sqrt[4]{2^{-k}}\leq x<\sqrt[4]{2^{-k+1}}; \ k\in \Bbb Z\right) .\end{cases}$$
Again the squeezing rule guarantees continuity and differentiability in $0$, with $f'(0) = 0$.

