Chain rule verification

Question

Let $A\in GL_n(\mathbb R)$, $f\in C^2(\mathbb R^n)$. Let $g(x)=f(Ax)$ and $A=(a_{ij})$. Is $$\frac {\partial g(x)}{\partial x_j}=\sum_{i=1}^{n}a_{ij}f_i(Ax)$$ where $f_i$ denotes the $i$-th partial derivative of $f$ ? Thanks and regards.

Answer 1

Call

$$ y_i = (A x)_i = \sum_k a_{ik} x_k $$

and use the chain rule,

$$ \frac{\partial g}{\partial x_j} = \sum_i\frac{\partial f}{\partial y_i} \frac{\partial y_i}{\partial x_j} = \sum_i f_i(Ax) \frac{\partial}{\partial_j}\left(\sum_k a_{ik}x_k \right) = \sum_{i,k}a_{ij}f_i(Ax) \underbrace{\frac{\partial x_k}{\partial x_j}}_{\delta_{kj}} = \sum_i a_{ij}f_i(Ax) $$