Meaning of second derivative as a linear transformation (jacobian matrix) My book has an exercise that says:

Let $U\subset \mathbb{R}^m$ be open, $a\in U$ and $f:U\to\mathbb{R}^n$
  be a map of class $C^2$. The second derivative of $f$, is, by
  definition, the bilinear map
  $f''(x):\mathbb{R}^m\times\mathbb{R}^m\to\mathbb{R}^n$, given by
  $f'(x)\cdot u\cdot v = \frac{\partial }{\partial v}\frac{\partial
 f}{\partial u}(a)$. Prove that $f''(a)\cdot u \cdot v = f''(a)\cdot v
\cdot u$

All I know about derivatives, is that it is a linear transformation $f'(x)$ that can be applied to vectors. Normally, a second derivative would be a derivative of a derivative, but in this case I cannot understand this definition neither how to prove what my book asks. Could somebody explain me better? I guess it has something to do with Schwarz's theorem for partial derivatives.
 A: You are right, if $f\colon U\subseteq\mathbb{R}^m\rightarrow\mathbb{R}^n$ is smooth, then $\mathbb{d}f\colon\mathbb{R}^m\rightarrow\textrm{End}(\mathbb{R}^m,\mathbb{R}^n)$ so that one gets:
$${\mathrm{d}^2}f\colon\mathbb{R}^m\rightarrow\mathrm{End}(\mathbb{R}^m,\mathrm{End}(\mathbb{R}^m,\mathbb{R}^n)).$$
Therefore, for $x\in\mathbb{R}^m$, $\mathrm{d}^2f$ is not a bilinear map but a linear map from $\mathbb{R}^m$ to $\textrm{End}(\mathbb{R}^m,\mathbb{R}^n)$. However, one has the following:

Proposition. Let $V$ and $W$ be vector spaces over the same field, then $$\textrm{End}(V,\textrm{End}(V,W))\cong \textrm{Bil}(V,W)$$
  through the evaluation map $b\mapsto (x\mapsto b(x,\cdot))$.

Proof. The evaluation map is clearly linear and its inverse is given by $\varphi\mapsto ((x,y)\mapsto\varphi(x)(y))$. $\Box$
Finally, for all $x\in\mathbb{R}^m$, ${\mathrm{d}^2}_xf$ can be seen as a bilinear map, actually:
$${\mathrm{d}^2}_xf\cdot(u,v)={}^\intercal u\textrm{Hess}_xfv.$$
The result you are trying to prove is equivalent to $\textrm{Hess}_xf$ being symmetric which is a consequence of Schwarz's equality, as you mentioned.
