Show that $\displaystyle x\frac{d}{dx}(\delta(x))=-\delta(x)$.
My attempt:
Using the product rule of differentiation,
$\displaystyle\frac{d}{dx}(x\delta(x))=x\frac{d}{dx}(\delta(x))+\delta(x)\cdot 1$
$\displaystyle\implies x\frac{d}{dx}(\delta(x))=-\delta(x)+\frac{d}{dx}(x\delta(x))$
How do I get rid of the $\displaystyle\frac{d}{dx}(x\delta(x))$ term?
On
Recall that a distribution is a continuous linear functional on the Schwartz space $\mathcal{S}(\mathbb{R})$. There are two important constructions to obtain new distributions from a given distribution:
1.) We can multiply with a smooth function. Namely, let $T\in \mathcal{S}(\mathbb{R})^*$ be a distribution and $f\in C^\infty(\mathbb{R})$, then we define for $\phi\in \mathcal{S}(\mathbb{R})$ $$ (f\cdot T)(\phi) := T(f\cdot \phi).$$
2.) We can also differentiate distributions. This we do the following way $$ \left(\frac{d}{dx} T\right)(\phi) := -T\left(\frac{d}{dx}\phi \right).$$
Now we recall that the dirac delta is by definition the evaluation at $0$, ie $$ \delta(\phi):= \phi(0) $$ We want to show the equality $$ f \left(\frac{d}{dx} \delta \right) = - \delta$$ where $f(x)=x$. As this are linear functionals, this means, we need to show that for all $\phi \in \mathcal{S}(\mathbb{R})$ holds $$ \left(f \left(\frac{d}{dx} \delta \right)\right)(\phi) = - \delta (\phi) $$ Now let's go through the definitions to unravel this mess. $$ \left(f \left(\frac{d}{dx} \delta \right)\right)(\phi) = \left(\frac{d}{dx} \delta \right)(f\phi) = -\delta\left( \frac{d}{dx} \left( f \cdot \phi \right) \right) = -\frac{d}{dx} \left( f \cdot \phi \right)(0) = -\left(\frac{d}{dx} f \right)(0) \cdot \phi(0) - f(0) \cdot \left(\frac{d}{dx} \phi \right)(0)$$ Now we recall that $f(x)=x$ and hence $f(0)=0$ and $\frac{d}{dx} f(0)=1$. Hence, we get $$ \left(f \left(\frac{d}{dx} \delta \right)\right)(\phi) = - \phi(0) = - \delta(\phi). $$
Added: Let me quickly explain why we define these things in exactly that way. In the dual space of the Schwartz space there are also linear functionals that can be represented by a function $g\in C^\infty(\mathbb{R})$. We usually denote these guys by $T_g$ $$ \phi \mapsto \int g(x) \phi(x) dx =: T_g(\phi). $$ Now we have $$ T_{hg} (\phi) = \int h(x) g(x) \phi(x) dx = T_{g} (h\phi) $$ Thus, our definition above is a generalization of this formula in the case when $T_g$ is a general distribution and not necessarily represented by a function. On the other hand we have $$ T_{\frac{d}{dx} g} (\phi) = \int \left(\frac{d}{dx} g(x)\right) \phi(x) dx = - \int g(x) \frac{d}{dx} \phi(x) dx = - T_g \left(\frac{d}{dx} \phi(x)\right)$$ where we used integration by parts and the properties Schwartz function to kill the boundary terms. Again our formula above is a generalization of this computation for distributions represented by a function.
Here is an elementary proof of
$$x \delta'(x)=-\delta(x)\tag{0}$$
that assumes you know the important identity, valid for any continuous function $f$ :
$$f(x)\delta(x)=f(0)\delta(x).\tag{1}$$
(See identities page 4 of the interesting document : https://www.reed.edu/physics/faculty/wheeler/documents/Miscellaneous%20Math/Delta%20Functions/Simplified%20Dirac%20Delta.pdf).
In our case, with $f(x)=x$, we have thus
$$x\delta(x)=0\tag{2}$$
Taking the derivative of (2) (formula $(uv)'=u'v+uv'$), we get
$$1\delta(x)+x\delta'(x)=0,$$
which is another way to write formula (0).
Remarks
1) See the very interesting solutions to this question Differential of Dirac Delta Function making sense of the following distributional identity, not trivially equivalent to (0) :
$$\dfrac{1}{x}\delta(x)=-\delta'(x)$$
2) (following a remark by @Severin Schraven) : Why can we still apply differentiation rule $(uv)'=u'v+uv'$ when $v$ is a "true" distribution (not associated with a function) ? Answer :
Let us define $C^{\infty}$ (rapidly decreasing) gaussian functions $\delta_n(x):=\sqrt{\tfrac{n}{2 \pi}}\exp(- \tfrac{n^2x^2}{2})$, that are known (see for example http://hitoshi.berkeley.edu/221A/delta.pdf) to converge towards distribution $\delta$, then apply the differentiation rule for a product to $x\delta_n(x)$, then apply the fact that derivation is a continuous operation for the topology in Schwartz space.
3) A practical remark : Notation $\delta(x)$ although practical, can be sometimes very misleading. If you have some problems with it, drop the "$(x)$"...