Why does $\frac{d}{dx}\delta(x-a)=-\frac{d}{da}\delta(x-a)$? I am reading Shankars principles of Quantum Mechanics and in his discussion of the derivative of the delta function, he simply writes that
$$\frac{d}{dx}\delta(x-a)=-\frac{d}{da}\delta(x-a)$$
I can't seem to figure out how we get the RHS from the LHS?
Any help on this issue would be most appreciated!
                                     **EDIT**

I am aware that we need to apply the chain rule to solve this as follows: Let $u=x-a$, then
\begin{eqnarray}
\frac{d}{da}\delta(x-a) &=& \frac{d}{du}\delta(u)|_{u=x-a} \frac{du}{da} \\
\Rightarrow &=&\frac{d}{du}\delta(u)|_{u=x-a} \frac{d}{da}(x-a) \\
\Rightarrow &=& -\frac{d}{du}\delta(u)|_{u=x-a}
\end{eqnarray}
The derivative in the last line is with respect to $u$ not $x$. In order to evaluate this derivative, we must derive $δ(u)$ with respect to $u$ and then evaluate the result at $u=x−a$. But how do we know that this is equal to simply deriving $\delta(x-a)$ with respect to $x$? For an arbitrary function $f(x)$, we have that in general $\frac{df(g(x))}{dx}\neq \frac{df(x)}{dx}|_{x=g(x)}$. So why do we infer in the case of the delta function that $\frac{d\delta(x-a)}{dx}=\frac{d\delta(u)}{du}|_{u=x-a}$ ?
 A: Let $f$ be a differentiable function. Then, by chain rule we have
$$
\frac{\partial}{\partial x}f(x-y)
= f'(x-y) \cdot \frac{\partial (x-y)}{\partial x}
= f'(x-y) \cdot 1
= f'(x-y)
$$
and
$$
\frac{\partial}{\partial y}f(x-y)
= f'(x-y) \cdot \frac{\partial (x-y)}{\partial y}
= f'(x-y) \cdot (-1)
= -f'(x-y).
$$
Thus,
$$
\frac{\partial}{\partial x}f(x-y)
= -\frac{\partial}{\partial y}f(x-y).
$$
This generalizes to distributions like $\delta$ by taking limits of nascent $\delta$ functions.
A: These are actually partial derivatives.
If we let $u=x-t$, that is, $u$ is a function of $x$ and $t$, then we get $\frac{\partial u}{\partial x}=1$ and $\frac{\partial u}{\partial t}=-1$. Therefore, the chain rule says
$$
\frac{\partial}{\partial x}f(x-t)=\frac{\partial f\circ u}{\partial x}\!\!\!\!\stackrel{\large\overset{\text{chain rule}}{\downarrow\\\phantom{}}}=\!\!\!\!(f'\circ u)\frac{\partial u}{\partial x}=f'\circ u=f'(x-t)
$$
and
$$
\frac{\partial}{\partial t}f(x-t)=\frac{\partial f\circ u}{\partial t}\!\!\!\!\stackrel{\large\overset{\text{chain rule}}{\downarrow\\\phantom{}}}=\!\!\!\!(f'\circ u)\frac{\partial u}{\partial t}=-f'\circ u=-f'(x-t)
$$
Thus,
$$
\frac{\partial}{\partial t}f(x-t)=-f'(x-t)=-\frac{\partial}{\partial x}f(x-t)
$$
A: $$\frac{d}{da}\delta(x-a) = \frac{d}{dx}\delta(x-a)\frac{d}{da}(x-a)
= \frac{d}{dx}\delta(x-a)\cdot(-1)=-\frac{d}{dx}\delta(x-a)$$
