Spivak Calculus on Manifolds: Problem 2-13 I'm going through Spivak's Calculus on Manifolds, and I'm currently working on Problem 2-13 part (b). The problem statement is

If $f,g: \mathbb{R} \rightarrow \mathbb{R}^{n}$ are differentiable and $h: \mathbb{R} \rightarrow \mathbb{R}$ is defined by $h(t) = \langle f(t),g(t)\rangle$, show that
  $h'(a) = \langle f'(a)^T,g(a) \rangle + \langle f(a),g'(a)^T \rangle.$ 

What's confusing me is that $f'(a)$ is a vector in $\mathbb{R}^{n}$ and taking its transpose yields a vector in the dual space to $\mathbb{R}^{n}$, so taking the inner product with $g(a)$, which is a vector in $\mathbb{R}^{n}$ yields a scalar in $\mathbb{R}$, corresponding to the left term in the equation. Whereas the right term in the equation is an outer product, so it sends a vector and a dual vector to a linear map in Hom($\mathbb{R}^{n}$), since the vector $f(a)$ is $n\times 1$ and the dual vector $g'(a)^T$ is $1\times n$ so their product is $n\times n$. How can these two terms be added to yield a scalar in $\mathbb{R}$?
 A: Note that $f'(a):\mathbb{R} \to \mathbb{R}^n$, $g'(a):\mathbb{R} \to \mathbb{R}^n$, and $\langle \cdot , \cdot \rangle: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$.
Alternatively, consider the expansion:
Suppose $f(t+\delta) = f(t) + f'(t) \delta + o_f(\delta)$, and similarly for $g$.
\begin{eqnarray}
h(t+\delta)-h(t) &=& \langle f(t+\delta), g(t+\delta) \rangle - \langle f(t), g(t) \rangle \\
&=& \langle f'(t)\delta, g(t) \rangle+\langle f(t), g'(t)\delta \rangle +
\langle f'(t) \delta , g'(t) \delta \rangle +
\langle o_f(\delta), g(t) + g'(t) \delta \rangle +
\langle f(t) + f'(t) \delta , o_g(\delta) \rangle + 
\langle o_f(\delta) , o_g(\delta) \rangle \\
&=&  (\langle f'(t), g(t) \rangle+\langle f(t), g'(t) \rangle)\delta +o_h(\delta)
\end{eqnarray}
Hence $h'(t) = \langle f'(t), g(t) \rangle+\langle f(t), g'(t) \rangle$.
A: Spivak writes the following in parantheses after stating the problem quoted in the question details.

Note that $f'(a)$ is an $n \times 1$ matrix; its transpose $f'(a)^T$ is a $1 \times n$ matrix which we consider as a member of $\mathbb{R}^n$.

My understanding is that Spivak just wants to view the matrix $f'(a)$ as a vector in $\mathbb{R}^n$, and so takes its transpose and identifies $\mathbb{R}^n$ with $M_{1,n}(\mathbb{R})$. This allows him to take the inner product of this vector with $g(a)$.
So, I believe you are overthinking it when you say that $f'(a)$ is a member of the dual space of $\mathbb{R}^n$.
