Show that for smooth manifolds $X$ and $Y$, $T[X \times Y]_{(x, y)} = TX_x \times TY_y$
Suppose $X \subseteq \mathbb{R}^k$ is a smooth manifold of dimension $n$ and $Y \subseteq \mathbb{R}^l$ is a smooth manifold of dimension $m$. Then $X \times Y \subseteq \mathbb{R}^{k+l}$ is a smooth manifold of $\text{dim} = k+l$
Choose parameterizations $\phi : U \to \phi[U]$ and $\psi : V \to \psi[V]$, of neighbourhoods $\phi[U]$ of $x \in X$ and $\psi[V]$ of $y \in Y$, and suppose $\phi(u) = x$ and $\psi(v) = y$.
Then $\phi \times \psi : U \times V \to \phi[U] \times \psi[V]$ defined by $\phi \times \psi(u, v) = (\psi(u), \psi(v))$ is a local parameterization of $X \times Y$ around $(x, y)$.
Now $$T[X \times Y]_{(x, y)} = d( \phi \times \psi)_{(u, v)}\left[\mathbb{R}^{k+l}\right] = d\left((u, v) \mapsto (\phi(u), \psi(v))\right)_{(u, v)}[\mathbb{R}^{k+l}]$$
The next step would be to look at the Jacobian of the RHS above, but the thing is that that the tangent space is the image $d( \phi \times \psi)_{(u, v)}\left[\mathbb{R}^{k+l}\right]$
$$T[X \times Y]_{(x, y)} = \text{Im}\left( \begin{bmatrix} d\phi_u & 0 \\ 0 & d\psi_u \end{bmatrix} \right) $$
From the above how could I complete the proof?
