Given that $f,g$ are continuous real and differentiable functions on $[0,T]$ and
$$ (f*g)(t) := \int_0^tf(t-s)g(s)ds $$
Prove that:
$$ \frac{d}{dt}(f *g)(t) = \int_0^tg(t-s)f^{'}(s)ds \; \; + f(0)g(t) $$
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what you tried? – Masacroso Oct 15 '16 at 04:08
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1I don't mean to be nit picky, but you'll need to at least assume that $f$ is differentiable; otherwise the right hand side doesn't make sense. What have you tried here? – User8128 Oct 15 '16 at 04:11
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By commutativity of convolution we would also have $$ \frac{\mathsf d}{\mathsf dt}(f\star g)(t) = \frac{\mathsf d}{\mathsf dt}(g\star f)(t) = \int_0^t f(t-s)g'(s)\ \mathsf ds + g(0)f(t), $$ so we need to assume that $f$ and $g$ are differentiable on $(0,T)$. – Math1000 Oct 15 '16 at 04:15
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See here: http://math.stackexchange.com/questions/177239/derivative-of-convolution – Math1000 Oct 15 '16 at 04:19
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@User8128, Ah yes, sorry I forgot about the differentiability, I'll fix that now. – amazonprime Oct 15 '16 at 17:09
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@Masacroso I looked around in books and online and found the Leibniz integral rule. I tried playing around with that, but it didn't get me the answer. – amazonprime Oct 15 '16 at 17:11
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You might want to consider the convolution theorem and the relation between the derivative and its Fourier transform. – Oct 15 '16 at 17:39