Line integral and differentiability of $\Vert f \Vert^{p}$ where $f(x) = A(x)$ 
Problem. Let $f: \mathbb{R}^{n} \to \mathbb{R}^{n}$, $n \geq 2$, the function defined by $f(x) = Ax$ where $A$ is a matrix $n \times n$. For each of the statements below, show if is true or give a counterexample if is false.
(a) If $g(x) = \Vert Ax \Vert^{p}$, with $p>1$, then $g$ is differentiable in $\mathbb{R}^{n}$ (here $\Vert\;\Vert$ denote the euclidean norm in $\mathbb{R}^{n}$).
(b) If $\det A = 1$, then $\int_{\gamma}f\mathrm{d} \gamma$ not depends of the path $\gamma$ that connects the points $x_{0}$ e $x_{1}$ of $\mathbb{R}^{n}$.

(b) In my book, $\displaystyle \int_{\gamma}f\mathrm{d}\gamma$ is defined over a curve $\gamma: [a,b] \to \mathbb{R}^{n}$ of class $C^{1}$ and I proved that

If $\Omega$ is connected open in $\mathbb{R}^{n}$ and $f: \Omega \to \mathbb{R}$ is a function of class $C^{1}$, then $f'$ is a conservative field, that is,
$$\int_{\gamma_{1}}f\mathrm{d}\gamma = \int_{\gamma_{2}}f\mathrm{d}\gamma \tag{*}\label{*}$$
for any differentiable curves that connect $x$ to $y$ for every $x,y \in \Omega$.

So, since $f$ is linear, $f$ satisfies the hypothesis of (*). Moreover, $f' \equiv f$, then $f$ is conservative.
But, I don't know if in this question, the definition of $\int_{\gamma}f\mathrm{d}\gamma$ is equal of my book. Moreover, if I'm right, the hypothesis $\det A = 1$ seems unnecessary. It made me think I'm totally wrong and try another way. So I tried to apply a linear change in the variables, but I think I couldn't use it correctly

(a) I know that

If $f:\mathbb{R}^{n} \to \mathbb{R}$ is given by $f(x) = \Vert x \Vert_{p}^{p}$ where $\Vert\;\Vert_{p} = (|x_{1}|^{p} + ... + |x_{n}|^{p})^{\frac{1}{p}}$, then is differentiable

and

If $g:\mathbb{R}^{n} \to \mathbb{R}$ is differentiable, then $F(x) = g(Ax)$ (where $A$ is a $n \times n$ matrix) is differentiable and $F'(x) = A^{T}g'(Ax)$.

These two results make me think that (a) is true, because (a) would be like a generalization of them. But I couldn't prove.

Can someone help me?
 A: (a) I'll use $|\,|$ for vector norms and absolute values, and $\|\,\|$ for matrix norms.
Observe that the map $x\to |A(x)|^2$ is differentiable on $\mathbb R^n.$ It is, after all, just a polynomial.
Case 1: Assume $A(x_0)\ne 0.$ Then we can choose $r>0$ such that $|A(x)|>0$ for $x$ in $B(x_0,r).$ For such $x,$ $x\to |A(x)|^2$ maps into $(0,\infty).$ The function $t\to t^{p/2}$ is differentiable on $(0,\infty).$ It follows that $|A(x)|^p = (|A(x)|^2)^{p/2}$ is differentiable in $B(x_0,r),$ since it is the composition of differentiable maps.
Case 2: Assume $A(x_0)= 0.$ Then
$$|A(x_0+h)|^p-|A(x_0)|^p = |A(h)|^p\le (\|A\||h|)^p=\|A\|^p|h|^p.$$
Since $p>1,$ this is $o(h),$ showing the derivative of our function is $0$ at $x_0.$
(b) I don't think this is true. Let
$$A =\left (\begin{matrix}1\,\,\,1\\ 0\,\,\,1\\\end{matrix}\right ). $$
Then $\det A = 1.$ Consider the two paths $\alpha, \beta :[0,1]\to \mathbb R^2$ given by $\alpha (t) = (t,t),$ $\beta (t) = (t,t^2).$ Both of these paths connect $(0,0)$ with $(1,1).$ I'm getting
$$\int_0^1 A(\alpha(t))\cdot \alpha'(t)\, dt = 3/2,\,\, \int_0^1 A(\beta(t))\cdot \beta'(t)\, dt = 4/3.$$
