Distributional Derivative of Heaviside Step Function times Sine

Theorem
Let $H$ be the Heaviside step function.

Let $\delta$ be the Dirac delta distribution.

Then in the distributional sense:


 * $T'_{\map H x \sin x} = T_{\map H x \cos x} $

Proof
$x \stackrel f {\longrightarrow} \map H x \sin x$ is a continuously differentiable real function on $\R \setminus \set 0$ and possibly has a discontinuity at $x = 0$.

By Differentiable Function as Distribution we have that $T'_f = T_{f'}$.

Moreover:


 * $x < 0 \implies \paren {{\map H x} \map \sin x}' = 0$.


 * $x > 0 \implies \paren {{\map H x} \map \sin x}' = \cos x$.

Altogether:


 * $\forall x \in \R \setminus \set 0 : \paren {{\map H x} \map \sin x}' = \map H x \cos x$

Furthermore:


 * $\ds \map f {0^+} - \map f {0^-} = \map H {0^+} \map \sin {0^+} - \map H {0^-} \map \sin {0^-} = 0$

The result follows from the Jump Rule.