Let $M$, $N$, $P$ all be manifolds and $f: M \rightarrow P$, $g: N \rightarrow P$ smooth functions. We say that (or at least the exercise does) that $f$ is transversal to $g$ if $T_r P = df_p(T_pM) + dg_q(T_q N)$ were $p\in M$, $q\in N$ and $r\in P$ are such that $f(p)=g(q)=r$.
The exercise asks to prove that this is equivalent to the function $f\times g$ is transversal to the diagonal of $P$. (i.e $d(f\times g)_{(p,q)}(T_pM\times T_qN) + T_{(r,r)}\Delta = T_{(r,r)} P\times P$)
My idea is from one formula to deduce the next, but I'm not reaching anywhere. My first idea is to use the first definition and "multiply" by itself obtaining:
$T_r P \times T_r P =[df_p(T_pM) + dg_q(T_q N)]\times[df_p(T_pM) + dg_q(T_q N)]$ does this product work distributively like normal operations? Besides is $d(f\times g)_{(p,q)}(T_pM \times T_qN)=df_pT_pM\times dg_qT_qN$?