Can we say : $f(x)$ is differentiable at $x=a$ $\Leftrightarrow $ $f'(x)$ is continuous at $x=a$ ? Why NOT? '$\Leftrightarrow$' Is very much important in this question . Actually, it seems very obvious to me.
We say a function is differentiable at $x=a$ iff 
$\lim_{ h\rightarrow 0 }{ \frac { f(a+h)-f(a) }{ h }  } = lim_{ h\rightarrow 0 }{ \frac { f(a-h)-f(a) }{ -h }  }$
Now, let
$f'(x)=g(x)$
If $f(x)$ is differentiable at $x=a$ , it means
$f'(a^+)=f'(a^-)$
Which is nothing else
$g(a^+)=g(a^-)$
i.e. $g(x)$ is continuous at $x=a$ $[$since $g(a)=f(a)=f'(a^-)=f'(a^+)=g(a^+)=g(a^-)]$ 
For those who agree with me, here comes a so called EXCEPTION function, 
$f(x) = \begin{cases}
{ x }^{ 2 }\cos { \frac { 1 }{ x }  } & \text{if} ~~~~ x\neq 0\\
0 & \text{if} ~~~~ x=0 \end{cases}$
here
$f'(0^+)=\lim_{ h\rightarrow 0 }{ \frac { { h }^{ 2 }\cos { \frac { 1 }{ h }  }-0 }{ h }  } = 0$
and
$f'(0^-)=\lim_{ h\rightarrow 0 }{ \frac { { -h }^{ 2 }\cos { \frac { 1 }{ -h }  }-0 }{ -h }  } = 0$
Since $f'(0^+)=f'(0^-)=0$ , $f(x)$ is well differentiable at $x=0$ 
Now take a look at the following 
$f'(x) = g(x) = sin(\frac 1x)+2xcos(\frac 1x)$
Whoa! , here $\lim_{ x\rightarrow 0 }{g(x)} = \lim_{ x\rightarrow 0 }{sin(\frac 1h)}$ = Oscillatory Value!
This suggests that $g(x)$ is not continuous at $x=0$ 
i.e. $f(x)$ is not differentiable at $x=0$
Why So ?
Any help will be appreciated. 
 A: The part
If $f(x)$ is differentiable at $x=a$ it means
$$f′(a^+)=f′(a^−)$$
is incorrect.
The statement 
$$f′(a^+)=f′(a^−)$$
means
$$\lim_{t \to 0^+} f'(a+t)=\lim_{t \to 0^-} f'(a+t)$$
which is not the same as the limit you wrote, it is actually the double limit:
$$\lim_{t \to 0^+} \lim_{h \to 0} \frac{f(a+t+h)-f(a+t)}{h}=\lim_{t \to 0^-} \lim_{h \to 0} \frac{f(a+t+h)-f(a+t)}{h}$$
P.S. The strongest relation between the two concepts which I am aware of is the following result, which is a trivial application of the Mean Value Theorem.
Lemma If $f$ is differentiable on $(a- \delta, a) \cup (a, a+\delta)$ and 
$$\lim_{x \to a} f'(x)$$
exists, then $f$ is differentiable at $x=a$ and $f'$ is continuous at $x=a$.
The proof of this lemma also emphasizes a subtle issue which happens here:
If $f$ is differentiable on $(a- \delta, a) \cup (a, a+\delta)$, then there exists a non-constant sequence $x_n \to a$ such that 
$$\lim_{n \to \infty} f'(x_n)$$
exists. But to prove continuity, you need to get this limit to exists when $x \to 0$ (or equivalently for all such sequences), which is not always the case..
A: $f^\prime(x)$ may not even exist for $x \neq 0$.
Consider the function defined by
$$f(x)=x^2 \cdot 1_{\mathbb Q}(x) = \begin{cases}
0 & \text{ for } x \in \mathbb{Q}\\
x & \text{ for } x \notin \mathbb{Q}
\end{cases}$$
$f$ is differentiable at $0$ (with $f^\prime(0)=0$) but $f^\prime$ doesn't exist for other real values.
A: In addition to the previous answer, let me try to convince you that your $f$ is differentiable at $x = 0$.
I claim that $f'(0) = 0$. To prove this, I have to demonstrate that, for any $\epsilon > 0$ that you give me, I can find a $\delta > 0$ such that
$$ | x| < \delta \implies | \  \frac{f(x) - f(0)}{x - 0} - 0  \ | < \epsilon \ \ \ \ \ (\star)$$
Now, your function $f$ satisfies the nice property that $|f(x) - f(0)| < x^2$ for any $x \in \mathbb R$. So if I pick my $\delta$ to be equal to your $\epsilon$, the inequality $(\star)$ is satisfied. Hence $f'(0) = 0$.
