Before this gets downvoted into oblivion, let me explain what I'm asking.
There's a well known theorem that states the following
If $f : \mathbb{R}^n \to \mathbb{R}^m$ is a linear transformation, then $f' = f$, where $f'$ is the $m \times n $ matrix of partial derivatives of $f$ (i.e. $f'$ is the derivative of $f$)
But we know that the derivative itself is a linear transformation $f' : \mathbb{R}^n \to \mathbb{R}^m$, so if we let $g = f'$, then $g' = f'$, and since $g'$ is the second derivative of $f$ (that being $f''$), does that not imply that all second derivatives (or any $n^{th}$ derivative) of any function $f: \mathbb{R}^n \to \mathbb{R}^m$ equals the first derivative $f'$?
This obviously cannot be the case, so there must be some error in my interpretation of this theorem. But I'm not sure what my error is. Also I apologize for asking such a silly question such as this.
You need to be careful. As DonAntonio pointed out in his comment, you should not confuse the derivative and the differential. Let $a\in \mathbb{R}^n$ and $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ be twice differentiable. Then you have
$$ df: \mathbb{R}^n \rightarrow L(\mathbb{R}^n, \mathbb{R}^m), \ a \mapsto (x\mapsto df_a x) $$
where $L(\mathbb{R}^n, \mathbb{R}^m)$ is the space of all linear maps from $\mathbb{R}^n$ to $\mathbb{R}^m$. This map is in general not linear. Take for example $n=m=1$ and $f(x)= x^3$, then you have
$$ df: \mathbb{R} \rightarrow L(\mathbb{R}, \mathbb{R}), \ a \mapsto (x\mapsto 3a^2 x).$$
Thus, $d(df)\neq df$ in general (you differentiate with respect to $a$, whereas $df_a(x)$ is only linear with respect to $x$).
On the other hand, you have that for every $a$ that $df_a$ is indeed linear and therefore $d(df_a)=df_a$.