Fundamental theorem of calculus on a convolution, confusion

Question

I need to calculate the following, where $K(t),m(t): \mathbb{R}^+ \to \mathbb{R}$ are sufficiently smooth and integrable. $$\dfrac{\operatorname d}{\operatorname d t} (K\ast m)(t),$$ I was thinking of the following (when the conditions are met)

$$\dfrac{\operatorname d}{\operatorname d t} \int_0^tK(t-s) m(s) \operatorname ds = \int_0^t \dfrac{\operatorname d}{\operatorname ds} K(t-s)m(s) ds= K(0)m(t) - K(t) m(0).$$

But this doesn't seem to agree with the theorem $\dfrac{\operatorname d}{\operatorname d t} (K\ast m)(t) = (K' \ast m)(t)= (K\ast m')(t).$

I guess it's because of the dependence of $t$ in the integral, since when I try: $$\int_0^t \sin(t-s) \operatorname ds = 1-\cos(t)$$ while $$\dfrac{\operatorname d}{\operatorname dt} (1-\cos(t) ) = \sin(t) \not = \int_0^t\frac{\operatorname d}{\operatorname ds} \sin(t-s)\operatorname ds = \sin(0) - \sin(t) = -\sin(t)$$

What's going on here?

Answer 1

There is no mystery, when sufficient conditions are satisfied, such as $a, b, f \in \mathscr{C}_{0}^{\infty}$ $$ \frac{d}{dt} \int_{a(t)}^{b(t)} f(t, s) ds = b^\prime(t)f(t, b(t)) - a^\prime(t)f(t, a(t)) + \int_{a(t)}^{b(t)} \frac{\partial}{\partial t} (f(t, s)) ds $$

Answer 2

For some mysterious reason you use a different formula for the convolution than the correct one. Remember that

$$(f * g) (t) = \int \limits _{-\infty} ^\infty f(t-s) g(s) \ \Bbb d s .$$

Assuming that all the nice conditions are met, this allows you to write

$$(f*g)' (t) = \int \limits _{-\infty} ^\infty \frac {\Bbb d} {\Bbb d t} f(t-s) g(s) \ \Bbb d s = (f' *g) (t).$$

The change of variable $u = t-s$ and integration by parts will also show that this is equal to $(f * g') (t)$, too.