Prove $\operatorname{div}(A \circ f \circ A^{-1}) (p) = \operatorname{div}(f)(A^{-1}p)$ with $A \in \operatorname{GL}_3(\mathbb{R})$ Let $A \in \operatorname{GL}_3(\mathbb{R})$ and $f: \mathbb{R}^3 \to \mathbb{R}^3$ a continuously differentiable vector field. We define the vector field $f_A := A \circ f \circ A^{-1}$
Prove that:


*

*$\operatorname{div}(f_A) (p) = \operatorname{div}(f)(A^{-1}p)$ for all $p \in \mathbb{R^3}$

*$\operatorname{rot}(f_A) (p) = \operatorname{rot}(f)(A^{-1}p)$ for all $p \in \mathbb{R^3}$, if $A\in \operatorname{SO}_3(\mathbb{R})$


How I tried solving it
by brute force:
$\operatorname{div}(f_A) (p)$ = ($\partial_1 A_{1,}(f_1(A_{1,}^{-1})) + \partial_2 A_{2,}(f_2(A_{2,}^{-1})) + \partial_3 A_{3,}(f_3(A_{3,}^{-1})))(p)$
However I do not really know how one continues from here.
By using the definition
Our textbook defines $\operatorname{div}(f)(p) = \lim_{h\to 0} \frac{1}{h^3} 
 \int_{\partial(p+[0,h]^3)} f \cdot d\vec{n}$
Similary  $\operatorname{rot}(f)(p) = \lim_{r\to 0} \frac{1}{\pi r^2} 
 \int_{\partial B_r(p)} f \cdot d\vec{s}$
These definitions seem very technical so I am not sure how they should be used.
Your help is greatly appreciated since I don't even know where to start.
 A: For the divergence.
Applying the chain rule, the Fréchet dérivative of $f_A$ at $p$ is
$$f_A^\prime(p)=A^\prime[(f \circ A^{-1})(p)] \circ f^\prime(A^{-1}(p)) \circ \left(A^{-1}\right)^\prime(p).$$
As $A$ and $A^\prime$ are linear applications, their Fréchet derivatives at any point are themselves. Hence $A^\prime[(f \circ A^{-1})(p)]=A$ and $\left(A^{-1}\right)^\prime(p)= A^{-1}$ which implies
$$f_A^\prime(p)=A \circ f^\prime(A^{-1}(p)) \circ A^{-1}.$$
As for any $h: \mathbb{R}^3 \to \mathbb{R}^3$ continuously differentiable, $\operatorname{div}(h)(p) = \operatorname{Tr}(h^\prime(p))$ you get
$$\begin{align}
\operatorname{div(f_A)}(p) &= \operatorname{Tr}(A \circ f^\prime(A^{-1}(p)) \circ A^{-1}) = \operatorname{Tr}(A \circ A^{-1} \circ f^\prime(A^{-1}(p)))\\
&=\operatorname{Tr}(f^\prime(A^{-1}(p))) = \operatorname{div}(f)(A^{-1}(p))
\end{align}$$
For the curl equality
Use the equality
$$\operatorname{div}(X_0 \wedge F) = - \operatorname{rot}F \cdot X_0$$ where $X_0$ is any constant vector field and the identity above for $\operatorname{div}$. 
