# Proving the existence of right derivative

Suppose we have a continuous function $f:[a,b]\to\mathbb{R}$. We know that $f$ is differentiable on $(a,b]$ and that there is a finite limit:

$$\lim_{x\to a^{+}}f^{\prime}(x)$$

Can we prove that $f$ has a right derivative in point $a$?

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For every $x>a$ there exists $\xi \in (a,x)$ such that $$f(x)-f(a)=f'(\xi)(x-a).$$ If $x \to a$, then $\xi \to a$, and ...

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Hm... Well, we can write the mean value theorem as: $$f^{\prime}(\xi)=\frac{f(x)-f(a)}{x-a}$$ So, as you said, if $x\to{a}^{+}$, then $\xi\to{a}^{+}$ as well. Therefore, we can take limits from both sides: – Johnny Westerling Sep 11 '12 at 12:12
...and we have: $$\lim_{\xi\to{a}^{+}}f^{\prime}(\xi)=\lim_{x\to{a}^{+}}\frac{f(x)-f(a)}{x-a}$$ we know that the limit on the left side exists, and the limit on the right side is the right derivative in $a$... is that correct? – Johnny Westerling Sep 11 '12 at 12:14
Yes. If you want to use kill a fly with a cannon, you can also use De l'Hospital. But it is too much, in this case. – Siminore Sep 11 '12 at 12:27
I'm not sure if de l'Hospital would work, because $f^{\prime}(a)$ does not exist... We have only one-sided derivative in $a$. – Johnny Westerling Sep 11 '12 at 12:35

$$f'_+(a)=\lim_{x \rightarrow a^+} \frac{f(x)-f(a)}{x-a}.$$ By continuity of $f$, we have here indefinite symbol $\frac{[0]}{[0]}$. Using de L'Hospital rule we obtain

$$f'_+(a)=\lim_{x \rightarrow a^+} f'(x).$$ (The last limit exists by assumption).

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Oh yes, now I understand what Siminore meant (I got confused by the fact that $f^{\prime}(a)$ doesn't exist, but since $f(a)$ is a constant it doesn't bother us in this case). Thanks. – Johnny Westerling Sep 11 '12 at 13:05