Why does $\frac{d}{dx}\delta(x-a)=-\frac{d}{da}\delta(x-a)$?

Question

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 $\frac{d\delta(x-a)}{dx}=\frac{d\delta(u)}{du}|_{u=x-a}$ ?

Answer 1

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.

Answer 2

$$\frac{d}{da}\delta(x-a) = \frac{d}{dx}\delta(x-a)\frac{d}{da}(x-a) = \frac{d}{dx}\delta(x-a)\cdot(-1)=-\frac{d}{dx}\delta(x-a)$$

Answer 3

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, $$ \frac{\partial}{\partial t}f(x-t)=-f'(x-t)=-\frac{\partial}{\partial x}f(x-t) $$