Find the gradient and hessian of $f(Ax+b)$ for real value $f$ and matrix $A$ 
Let $A\in\mathbb{R}^{m\times n}$, $b\in \mathbb{R}^m$. For
  $x\in\mathbb{R}^n$, we define $q(x) = f(Ax+b)$ with
  $f:\mathbb{R}^m\to\mathbb{R}$. Find the gradient and hessian of the
  function $q$.

This question is kinda strange. If I were to take the jacobian, I would just compose the jacobian of the outer funciton with the jacobian of the inner function. Now, how do I take the partial derivative of $q$?
$$\frac{\partial f(Ax+b)}{\partial x_1} = \lim_{h\to 0}\frac{f(A(x_1+h,x_2,\cdots,x_n) + b) - f(Ax+b)}{h}$$
I don't think it helps in thinking this way. I have no means of finding this limit without using some chain rule or so. Maybe I can apply the chain rule to $q$, but how? 
UPDATE:
By the hint given below,
$$q(x_1,\dots, x_n)=f(f_1,\cdots,f_n) = f\left(\sum_{i=1}^n a_{1i}x_i+b_1,\dots,\sum_{i=1}^n a_{mi}x_i+b_m\right)$$
I think the multivariable chain rule can be applied:
$$\frac{\partial f}{\partial x_1} = \frac{\partial f}{\partial f_1}\frac{\partial f_1}{\partial x_1} + \cdots + \frac{\partial f}{\partial f_n}\frac{\partial f_n}{\partial x_1}$$
And see that 
$$\frac{\partial f_1}{\partial x_1} = a_{11}\\\cdots\\\frac{\partial f_n}{\partial x_1} = a_{m1}$$
So we get
$$\frac{\partial f}{\partial x_1} = \frac{\partial f}{\partial f_1}a_{11} + \cdots + \frac{\partial f}{\partial f_n}a_{m1}$$
In general:
$$\frac{\partial f}{\partial x_j} = \frac{\partial f}{\partial f_1}a_{1j} + \cdots + \frac{\partial f}{\partial f_n}a_{mj}$$
I think there's still a lot of work to do. 
 A: Hint
We have that
$$q(x_1,\dots, x_n)=f\left(\sum_{i=1}^n a_{1i}x_i+b_1,\dots,\sum_{i=1}^n a_{mi}x_i+b_m\right). $$
Thus we have
$$\dfrac{\partial q}{\partial x_i}=a_{1i}\dfrac{\partial f}{\partial u_1}+\dots +a_{mi}\dfrac{\partial f}{\partial u_m}.$$ That is
$$(\nabla q(x))^T=A(\nabla f q(x))^T.$$
Edit to get the Hessian
\begin{align}
\dfrac{\partial^2 q}{\partial x_j\partial x_i} &=\sum_{k=1}^m a_{ki}\dfrac{\partial}{\partial x_j}\left( \dfrac{\partial f}{\partial u_k}\right)
\\&= \sum_{k=1}^m a_{ki}\sum_{l=1}^ma_{lj}\dfrac{\partial^2 f}{\partial u_l\partial u_k}
\\&= \sum_{k,l=1}^ma_{ki}a_{lj} \dfrac{\partial^2 f}{\partial u_l\partial u_k}.
\end{align}
We have used that
$$\dfrac{\partial}{\partial x_j}\dfrac{\partial f}{\partial u_k}=\dfrac{\partial}{\partial u_1}\left(\dfrac{\partial f}{\partial u_k}\right)\dfrac{\partial u_1}{\partial x_j}+\dots +\dfrac{\partial}{\partial u_m}\left(\dfrac{\partial f}{\partial u_k}\right)\dfrac{\partial u_m}{\partial x_j}$$
Thus, whe have that
$$\nabla^2 q (x)=A^T (\nabla^2 f(u)) A.$$
A: Background knowledge: if $F:\mathbb R^n \to \mathbb R^m$ is differentiable at $x$, then $F'(x)$ is an $m \times n$ matrix.

Let $g(x) = f(Ax + b)$. By the chain rule,
$$
g'(x) = f'(Ax + b)A.
$$
If we use the convention that the gradient is a column vector, then
$$
\nabla g(x) = g'(x)^T = A^T \nabla f(Ax + b).
$$
The Hessian of $g$ is the derivative of the function $x \mapsto \nabla g(x)$. By the chain rule,
$$
\nabla^2 g(x) = A^T \nabla^2 f(Ax + b) A.
$$
A: Let $\phi = x \mapsto Ax + b$  such that $q = f \circ \phi$.
Denote by $J_g$ the Jacobian matrix of any function $g$.
Applying the chain rule leads to $J_q (x) =  J_f(\phi(x)) J_{\phi}(x) $
Since $J_{\phi}(x) = A$, and $\nabla g = (J_g)^T$ for any scalar function $g$, this boils down to $$(\nabla q (x))^T = (\nabla f (Ax+b) )^TA.$$
Finally, we find:  $$\nabla q (x) = A^T \nabla f (Ax+b).$$
If you want a more gradient-specific formula, you can directly state that $\nabla q (x)= J_{\phi}(x)^T \nabla f (\phi(x))$.
