Integration is a bounded operator on $C[0,1]$; differentiation, under the sup norm, is not

Let $C^1[0,1]=\{f:[0,1]\to\mathbb{R}:f\text{ is differentiable on }[0,1] \text{ and }f'\text{ is continuous}\}$ and $C[0,1]=\{f:[0,1]\to\mathbb{R}:f\text{ is continuous on }[0,1]\text{ with } \|f\|=\sup\{|f(t)|:t\in[0,1]\}\}$. Let $D:C^1[0,1]\to C[0,1]$ be defined by $Df=f'$ and $I:C[0,1]\to C[0,1]$ be defined as $I(f)(x)=\int_0^xf(t)dt$. Then: