Show that the map $B(f) = f^{\prime}(0)$ defines a linear functional on the normed vector space $(C^{(1)}[0,1], \| \cdot \|)$ that is not bounded. 
Let $C^{(1)}[0,1]$ be the space of all continuous functions on $[0,1]$
  that have continuous derivative. (This space is called the space of
  continuously differentiable functions.)   Define a norm $\|\cdot \|$
  on $C^{(1)}[0,1]$ by $\| f \|=\max _{t \in T}|f(t)|.$   Show that the
  map   $$B(f) = f^{\prime}(0)$$  defines a linear functional on the
  normed vector space $(C^{(1)}[0,1], \| \cdot \|)$ that is not bounded.

My attempt: 
For any $f,g \in C^{(1)}[0,1],$ we have $B(f+g) = (f + g)^{\prime}(0) = f^{\prime}(0) + g^{\prime}(0) = B(f) + B(g).$
    Therefore, $B$ is a linear functional. 
To show that $B$ is not bounded, we wish to show that for any $M > 0$, there exists $f \in C^{(1)}[0,1]$ such that $\| f \| \leq 1$ but $|B(f)| > M.$
Let $M > 0 $ be given. I have trouble constructing a function with maximum $\leq1$ and has derivative at $x = 0$ greater than $M.$
I plan to construct $f$ which looks similar to $\sin(x)$. For any given $M > 0$, I can draw $\sin(x)$ such that it is above the line $y = Mx$. However, I do not know how to formulate this phenomena using mathematical symbols. 

Define $ f_n(x) = \sin (nx)   $ then $\sup|f_n(x)| =1$ however 
$f_n'(x) = n\cos nx$ hence $$|f'_n(0)| =n.$$
