Why $J_{g\circ f}(x)=0$ for all $x\in \mathbb R^3$. Let $f:\mathbb R^3\longrightarrow \mathbb R$ and $g:\mathbb R\longrightarrow \mathbb R^3$ two derivables function. 
In a True/False question, I had the following questions :
1) $\det J_{f\circ g}(t)=0$ for all $t\in \mathbb R$.
2) $\det J_{g\circ f}(x)=0$ for all $x\in\mathbb R^3$.
I recall that $J_h$ denote the jacobian matrix of a function $h$.
In the solution it's writtent that 1) is false and 2) is true without any explanation. I really don't understand those answer. Could someone give me an intuition behind these result ?
 A: You can consider $f\circ g\colon \mathbb R\to\mathbb R$. Therefore $J_{f\circ g}(t) = \frac{d}{dt}(f\circ g)(t) = (f\circ g)'(t)$ is just the derivative of $f\circ g$. The determinant of a scalar is the identity ($\det(t) = t$). Since $f\circ g$ could be anything there is no need for its derivative to vanish.
Now we can consider $g\circ f\colon \mathbb R^3\to\mathbb R^3$. Intuitively speaking, $f$ gives and $g$ takes a scalar, so two of three informations are lost. Therefore the Jacobian cannot be regular/have full rank. ($\det = 0$ means the matrix is not regular, which means the matrix can not have full rank.)
We can also just calculate the jacobian by using the chain rule and the notation $J_h = h'$:
$$ J_{g\circ f}(x) = (g\circ f)'(x) = g'(f(x))\cdot f'(x) $$
where $g'(y)\in\mathbb R^{1\times 3}$ and $f'(x)\in\mathbb R^{3\times 1}$. This means $J_{g\circ g}$ is the dyadic product of two vectors and thus has to have rank 1. Since it does not have full rank it has to have vanishing determinant.
A: First of all remember the formula for the Jacobian of function composition:
$$ J_{f \circ g} = \left( J_f \circ g  \right) \cdot J_g$$
So if $ f:\mathbb R^3\longrightarrow \mathbb R $ and $ g:\mathbb R\longrightarrow \mathbb R^3 $ then $J_f$ is a $1 \times 3 $ matrix and $J_g$ is a  $3 \times 1$ matrix. So both have at most a rank of $1$ and so is their product.
Concerning 2), $J_{g \circ f}$ is a $3 \times 3 $ matrix with rank at most $1$. So it is not invertible and his determinant is $0$ everywhere.
Concerning 1), $J_{f \circ g}$ is a $1 \times 1 $ matrix with rank at most $1$. So it is a scalar and we simply need to find a simple counter example where it is non zero somewhere. 
So let's take $ g:t \longrightarrow (t,0,0) $ and  $ f :x \longrightarrow x \cdot (1,0,0) $. Then $ (f \circ g)(t) = f\left((t,0,0)\right) = (t,0,0)\cdot (1,0,0) = t $ and the Jacobian is $1$ everywhere.
