Let $\theta\left(x\right)$be the step function
$\theta\left(x\right)\equiv\begin{cases}1 & if \ x>0\\0 & if\ x<0 \end{cases}$
In the rare case where it actually matters, we define $\theta\left(0\right)=\frac{1}{2}$
Show that $\frac{d\theta}{dx}=\delta\left(x\right)$
I'm definitely a bit lost on how to start on this problem. NO ANSWERS!
How should I think about this problem?
Remember what distributional derivative means: It's exactly the thing that makes integration by parts work. If $u$ and $v$ are smooth functions, then
$$\int u' v = uv - \int uv'$$
Now if we assume $u, v$ are compactly supported so that the boundary term disappears, then
$$\int u' v = - \int uv'$$
So the distributional derivative of $u$, which is denoted $u'$, is the thing so that for all smooth functions $v$ the above works.
So what you need to establish is that for all compactly supported smooth functions $v$,
$$\int \delta(x) v(x) = - \int \theta(x) v'(x)$$