Derivative of limited convolution

Question

If I take infinite convolution, I know that its derivative is the convolution between one term and other term's derivative.

However, I am interested on derivative of

$$\dfrac {d}{dt}\int_0^t f (t) g (t-a) da$$

i.e. in terms of limited convolution.

Many thanks in advance!

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EDIT: Please review comments on answer

Answer 1

I think @tossimmar 's answer gives some preliminary remarks, but I think this answer will provide the asker with more of what he/she is looking for. First of all, Leibniz's integral rule, in most general form, is $$\frac{\mathrm{d}}{\mathrm{d}t}\left(\int_{a(t)}^{b(t)}f(t,s)\mathrm{d}s\right)=f\left(t,b(t)\right)b'(t)-f\left(t,a(t)\right)a'(t)+\int_{a(t)}^{b(t)}\frac{\partial}{\partial t}f(t,s)\mathrm{d}s$$ In our case, this gives us $$\frac{\mathrm{d}}{\mathrm{d}t}\left(\int_{0}^{t}f(t)g(t-s)\mathrm{d}s\right)=f(t)g(0)+\int_0^tf'(t)g(t-s)\mathrm{d}s+\int_0^tf(t)g'(t-s)\mathrm{d}s$$ Of course, we could write this in terms of this partial convolution which I'll denote with an tilded asterisk: $$\mathrm{D}(f~\tilde{*}~g)=g(0)\cdot f+f'~\tilde{*}~g+f~\tilde{*}~g'$$

Answer 2

Apply the special case of Liebniz's Rule, $$\frac{d}{dx}\bigg(\int_{a}^{x}h(x,t)dt\bigg) = h(x,x) + \int_{a}^{x}\frac{\partial}{\partial x}h(x,t)dt.$$