I'll give you the whole context: In solving the heat equation $u_t = ku_xx$ with bounds $u(x,0)=0, u(0,t)=0, u(l,t)=f(t)$, let $v(x,t)$ be the solution for the special case $f(t)=1$.
Use the Laplace transform to obtain Duhamel's formula: $u(x,t)=\frac{\partial}{\partial t}\int_0^{t} f(s) v(x,t-s) \,ds$.
So I take the Laplace transform with respect to t och end up with (using the derivative transform + z/z trick + convolution transform) to get
$u(x,t) = \int_0^{t} f'(s) v(x,t-s) \,ds$.
My question: is this the answer I am looking for? Can I change the "order" of convolution and bring out the derivative?
You could just as easily write
$$u(x,t) = \frac{\partial}{\partial t} \int_0^t ds \, f(t-s) v(x,s) $$
which is equal to
$$u(x,t) = f(0) v(x,t) - f(t) v(x,0) + \int_0^t ds \, f'(t-s) v(x,s) $$
I assume you enforce $f(0)=0$ for continuity and that $v(x,0)=0$. Therefore, you can write
$$u(x,t) = \int_0^t ds \, f'(s) v(x,t-s) $$
Note that we did not assert changing the order of differentiation and integration, but simply used the Fundamental theorem of calculus.