I am just wondering why this is correct,
$$ \frac{d}{dR} \delta[f(t')] = -\frac{1}{c}\frac{d}{df}\delta[f(t')], $$
where $t' = t - \frac{R}{c}$.
Is this simply due to chain rule? Or something like
$$ \frac{d}{dR} = \frac{d}{df}\frac{df}{dR}. $$
Thanks!
Yes, it is the chain rule.
How do you represent the derivative of $g(h(x))$?
Lets write that as $\frac{d}{dx} g(h(x)) = g'(h(x))\cdot h'(x)$, which is just the chain rule.
For your example, we are given that $t' = t - \frac{R}{c}$.
We want to find:
$\displaystyle \frac{d}{dR} (\delta[f(t')]) = \frac{d}{dR} (\delta\left[f\left(t - \frac{R}{c}\right)\right]) = -\frac{f'(t-\frac{R}{c}) \delta'(f(t-\frac{R}{c}))}{c} = -\frac{f'(t') \delta'(f(t'))}{c} = -\frac{1}{c}\frac{d}{df}\delta[f(t')]$
