For transverse maps, $T_x(f^{-1}(Z)=df^{-1}_x(T_{f(x)}(Z))$

For transverse maps, $T_x(f^{-1}(Z)=df^{-1}_x(T_{f(x)}(Z))$

Let $f: X \to Y$ be a map transversal to a submanifold $Z$ in $Y$. Then $W
= f^{-1}(Z)$ is a submanifold of $X$. Prove that $T_x(W)$ is the preimage
of $T_{f(x)}(Z)$ under the linear map $df_x:T_x(X) \to T_{f(x)}(Y)$. the
linear map $df_x:T_x(X)¨T_{f(x)}(Y)$.
My approach to the problem was to create parametrizations, $\phi$ and
$\psi$ of neighborhoods of $f(x)$ and $x$ (choosing all of Euclidean space
as the domain. We then take $y\in T_x(f^{-1}(Z)$ and note that there is
$v$ such that $y=d\psi_x(v)$. Now let $\pi$ as either the submersion or
embedding of the two Euclidean spaces we are mapping from. Then
$df^{-1}_x(d\phi_{f(x)}(d\pi(v))=d\psi_x(v)=y$ by the chain rule so $y\in
df^{-1}_x(T_{f(x)}(Z))$.
I do the other side mutatis mutandis.
Does this actually work, or have I made a critical error? (I'm personally
leaning toward the second)
If anyone could give some tips, I would appreciate it.