Define $f: \mathbb{R}^N \to \mathbb{R}^m$.
Let $a$ be a regular value of f (i.e. $\forall p \in f^{-1}(a), \text{D}_pf$ is surjective), $X = f^{-1}(a)$ and choose $p \in X$. Then $\text{T}_pX= \text{Ker}\text{D}_pf$, where $\text{T}_pX$ is the tangent plane of $X$ in $p$.
We defined also: $\text{T}_pX = \{ v= c'(0) \text{ st } c:]-\epsilon, \epsilon[ \to \mathbb{R}^N \text{ derivable}, c(0) = p \text{ and } \text{Im}c \subset X \}$.
I want to understand the proof of $\text{T}_pX= \text{Ker}\text{D}_pf$ which starts like this in our lecture notes:
$\text{dim}\text{KerD}_pf = N-m$ since $\text{D}_pf$ surjective. It suffices to show $\text{T}_pX \subset \text{Ker}\text{D}_pf$. Why do we only have to show this inclusion?
If you have a finite-dimensional vector space, and two subspaces of the same dimension such that one is contained in the other, they're necessarily equal. The dimensions of $T_pM$ and $\ker Df_p$ are equal, so it suffices to show one of the inclusions. And $T_pM \subseteq \ker Df_p$ is the easiest one, because if $f(c(t)) = a$ for all $t$, what happens when you take the derivative of both sides with respect to $t$?