Second derivative expression I have $f:\mathbb R^n\to \mathbb R$ and $\gamma:\mathbb R \to \mathbb R^n$, which are both $\mathrm C^2$. Considering $g=f\circ \gamma$, how could I express $g''$, second derivative of $g$ in terms of partial derivatives of $f$ and $\gamma$?. The first I know is that 
$$ \mathrm Dg(a) = \mathrm Df(\gamma(a)) \circ D\gamma(a), $$
so I consider $Dg:x\mapsto Dg(x)$, but this gets complicated. Thanks in advance.
Edit: I know what is the Hessian matrix, I would like to apply it here.
Edit II: Well, it is clear that
$$ g'(a) h = \langle \nabla  g(a), h\rangle = \langle \nabla f(\gamma(a)), \langle \nabla \gamma(a),h\rangle\rangle $$
so, what now?

Since you want the derivative of the scalar function $$g(t):=f\bigl(\gamma(t)\bigr)\tag{1}$$ in terms of partial derivatives I'd argue as follows: Differentiating $(1)$ with respect to $t$ using the chain rule gives
$$g'(t)=\sum_{i=1}^n f_{.i}\bigl(\gamma(t)\bigr)\>\gamma_i'(t)\ ,$$
so $g'$ is a finite sum of products of functions of $t$. For the second derivative we have to use the product rule, and the chain rule again:
$$g''(t)=\sum_{i,\> k\>=1}^n f_{.ik}\bigl(\gamma(t)\bigr)\>\gamma_k'(t)\gamma_i'(t)+\sum_{i=1}^n f_{.i}\bigl(\gamma(t)\bigr)\>\gamma_i''(t)\ .$$
