=== Necessary Condition === Suppose that $f$ is continuously differentiable on $\closedint a b$. We need to show that: :$f$ is continuously differentiable on $\openint a b$ and: :$\displaystyle \lim_{x \mathop \to a^+} \map {f'} x$ and $\displaystyle \lim_{x \mathop \to b^-} \map {f'} x$ exist. $f$ is continuously differentiable on $\openint a b$ because: :$f$ is continuously differentiable on $\closedint a b$ :$\openint a b$ is a subset of $\closedint a b$.  What remains to show is that $\displaystyle \lim_{x \mathop \to a^+} \map {f'} x$ and $\displaystyle \lim_{x \mathop \to b^-} \map {f'} x$ exist. We shall only make a proof of the existence of $\displaystyle \lim_{x \mathop \to a^+} \map {f'} x$ as the proof of the existence of $\displaystyle \lim_{x \mathop \to b^-} \map {f'} x$ is similar. Since $f$ is differentiable at $a$, $\map {f'} a$ exists. Since $f'$ is continuous at $a$, the following equation is true: :$\displaystyle \map {f'} a = \lim_{x \mathop \to a^+} \map {f'} x$ Therefore the {{RHS}} of this equation exists. {{qed|lemma}} === Sufficient Condition === Suppose that: :$f$ is continuously differentiable on $\openint a b$ and: :$\displaystyle \lim_{x \mathop \to a^+} \map {f'} x$ and $\displaystyle \lim_{x \mathop \to b^-} \map {f'} x$ exist. We need to show that $f$ is continuously differentiable on $\closedint a b$. We know that $f$ is continuously differentiable on $\openint a b$ so what remains to show is the following: :$\map {f'} a$ exists :$f'$ is continuous at $a$ :$\map {f'} b$ exists :$f'$ is continuous at $b$. We shall limit ourselves to constructing a proof of the first two of these statements as the proof of the last two statements is similar. Since $a$ is the left-most end point of the domain of $f$, $\map {f'} a$ is the right-hand derivative of $f$ at $a$, denoted by $\map {f'_+} a$. We have per definition: :$\displaystyle \map {f'_+} a = \lim_{\delta \mathop \to 0^+} \frac {\map f {a + \delta} - \map f a} \delta$ Since $f$ is continuous on $\closedint a b$ and differentiable on $\openint a b$, the Mean Value Theorem gives that a point $c$ in $\openint a  {a + \delta}$ exists such that :$\map f {a + \delta} - \map f a = \map {f'} c \delta$ We use this equation in the expression for $\map {f'_+} a$: {{begin-eqn}} {{eqn | l = \map {f'_+} a       | r = \lim_{\delta \mathop \to 0^+} \frac {\map f {a + \delta} - \map f a} \delta }} {{eqn | r = \lim_{\delta \mathop \to 0^+} \frac {\map {f'} c \, \delta} \delta }} {{eqn | r = \lim_{\delta \mathop \to 0^+} \map {f'} c }} {{end-eqn}} Since $c$ is in $\openint a {a + \delta}$, we have $\size {c - a} < \delta$. Therefore, $c$ approaches $a$ when $\delta$ approaches 0. Moreover, since $c > a$, $c$ approaches $a$ from above. Since $\displaystyle \lim_{x \mathop \to a^+} \map {f'} x$ exists, so does $\displaystyle \lim_{\delta \mathop \to 0^+} \map {f'} c$ because $c \to a^+$ when $\delta \to 0^+$. Furthermore, they are equal. Hence, since we proved above that $\displaystyle \lim_{\delta \mathop \to 0^+} \map {f'} c$ equals $\map {f'_+} a$ :$\displaystyle \map {f'_+} a = \lim_{x \mathop \to a^+} \map {f'} x$ Since $\displaystyle \lim_{x \mathop \to a^+} \map {f'} x$ exists per presupposition, we conclude that $\map {f'_+} a$ exists, and that their values are equal. Hence $\map {f'} a$ exists and equals $\displaystyle \lim_{x \mathop \to a^+} \map {f'} x$. Thus, the first of the two statements we needed to prove is proven. Now we shall prove that $f'$ is continuous at $a$. Since $a$ is the left-most end point of the domain of $f$, $f'$ being continuous at $a$ means that $\displaystyle \map {f'} a = \lim_{x \mathop \to a^+} \map {f'} x$. Since we just proved this, we conclude that $f'$ is continuous at $a$. {{qed}} Category:Differential Calculus iclshe4fmbqwbgj02hgacv3sdfkeiq5