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!

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

## Best Answer

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.