Let $F:N\times M \to M'$ be a smooth map, which we interpret as a "smooth" family of maps $M \to M'$, parametrized by N, so we have a map $F(y,\cdot):M \to M' \: \forall y \in N$.
Show that the family of the corresponding differentials, that is, the map:
$\tilde{F}:N \times TM \to TM', (y,v) \mapsto d(F(y,\cdot))(v)$ is also smooth.
Unfortunately my ideas didn't work out. Thanks in advance for any help.
As already pointed out by @Thorgott, The differential of a smooth map is again smooth.
Therefore, taking the differential of the map F we get $dF:T(N\times M)\to TM'$. There is a natural, linear diffeomorphism $\Phi:TN \times TM \to T(N \times M)$. So we can instead regard $dF$ as a map $dF: TN \times TM \to TM'$. Since $dF$ is linear, for tangent vectors $(p,v_1) \in TN$ and $(q,v_2)$ in $TM$ we have $dF_{(p,q)}(v_1+v_2) = d(F(\cdot,q))(v_1) + d(F(p,\cdot))(v_2)$. Which one could also understand by taking $d(F(\cdot,q))$ to be the first $m' \times n$ -block of the Jacobian and $d(F(p,\cdot))$ the latter $m' \times m$ - block. But then by restricting the map $dF:TN \times TM \to TM'$ to $N \times \{0\}$ (which is an embedded submanifold of N), we obtain the smoothness of the map $(p,(q,v_2)) \to d(F(p,\cdot))(v_2)$.