Given gradient and hessian of $\phi$ and the first and second derivatives of $h$ I would like to find the gradient and hessian matrix of $f$
$f(x)=h(\phi(x))$
Where $f:R^n \longrightarrow R$
$\phi:R^n \longrightarrow R$
$h:R \longrightarrow R$
According to the chain rule: $\nabla f = h'(\phi(x))\nabla\phi(x)$ Is it correct and trivial and doesn't need any other proof? And how can I find the hessian of $f$?
Since
$$f(x_1,x_2,...,x_n)=h(\phi(x_1,x_2,...,x_n))$$
by chain rule we have that
$$\frac{\partial f}{\partial x_i}=h'(\phi(\vec x))\frac{\partial \phi}{\partial x_i}\implies \nabla f(\vec x) = h'(\phi(\vec x))\nabla\phi(\vec x)$$
For the Hessian note that
$$\frac{\partial^2 f}{\partial x_i^2}=h''(\phi(\vec x))\left(\frac{\partial \phi}{\partial x_i}\right)^2+h'(\phi(\vec x))\frac{\partial^2 \phi}{\partial x_i^2}$$
$$\frac{\partial^2 f}{\partial x_ix_j}=h''(\phi(\vec x))\frac{\partial \phi}{\partial x_i}\frac{\partial \phi}{\partial x_j}+h'(\phi(\vec x))\frac{\partial^2 \phi}{\partial x_ix_j}$$