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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

$$ \frac{d}{dR} = \frac{d}{df}\frac{df}{dR}. $$


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up vote 4 down vote accepted

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:

$\displaystyle \frac{d}{dR} (\delta[f(t')]) = \frac{d}{dR} (\delta\left[f\left(t - \frac{R}{c}\right)\right]) = -\frac{f'(t-\frac{R}{c}) \delta'(f(t-\frac{R}{c}))}{c} = -\frac{f'(t') \delta'(f(t'))}{c} = -\frac{1}{c}\frac{d}{df}\delta[f(t')]$

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Great thanks, this is very reassuring! – Shinobii Mar 14 '13 at 1:52
@Shinobii: You are very welcome! Regards – Amzoti Mar 14 '13 at 2:11
Helpful, and great feedback from the OP! – amWhy Apr 21 '13 at 0:12

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