Derivative of the Heaviside step function

Question

The Heaviside step function is defined as $$H(x) = \begin{cases} 0 & \text{if }x<0 \\ 1 & \text{if }x\geq0\end{cases}$$ Set also $K(x)=H(2x)$ for all $x\in\mathbb R$.

Now it is well known (and can be easily proven) that the derivative of $H$ in the sense of distributions is the Dirac delta $\delta_0\,$: $$H' = \delta_0 \;.$$ Using standard calculus rules I would then expect $$K' = 2\,H' = 2\,\delta_0 $$ but this is of course not true since $K=H$.

I'd like to understand why I cannot use standard calculus rules and which rules fail (and which do not) when dealing with distributional derivatives.

Answer 1

Using the chain rule gives you $$ K'(x) = \frac{d}{dx} H(2x) = 2 H'(2x) = 2\delta(2x) = 2\cdot\frac12\delta(x) = \delta(x) $$ since $\delta(ax) = \frac{1}{|a|}\delta(x).$