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

calculusderivativesdirac delta

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

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.