Answer
We show that
${\rm{Jac}}\left( {{G_2} \circ {G_1}} \right) = {\rm{Jac}}\left( {{G_2}} \right){\rm{Jac}}\left( {{G_1}} \right)$
Work Step by Step
Write the mappings ${G_1}$ and ${G_2}$:
${G_1}:\left( {s,t} \right) \to \left( {u,v} \right)$, ${\ \ \ \ \ }$ where $\left( {s,t} \right) \in {{\cal D}_1}$, $\left( {u,v} \right) \in {{\cal D}_2}$
${G_2}:\left( {u,v} \right) \to \left( {x,y} \right)$, ${\ \ \ \ \ }$ where $\left( {x,y} \right) \in {{\cal D}_3}$
So, ${G_2} \circ {G_1}:\left( {s,t} \right) \to \left( {x,y} \right)$.
Thus, $x$ and $y$ are composite functions: $x\left( {u\left( {s,t} \right),v\left( {s,t} \right)} \right)$ and $y\left( {u\left( {s,t} \right),v\left( {s,t} \right)} \right)$.
By the Multivariable Chain Rule (Theorem 1 in Section 15.6)
$\frac{{\partial x}}{{\partial s}} = \frac{{\partial x}}{{\partial u}}\frac{{\partial u}}{{\partial s}} + \frac{{\partial x}}{{\partial v}}\frac{{\partial v}}{{\partial s}}$, ${\ \ \ \ }$ $\frac{{\partial x}}{{\partial t}} = \frac{{\partial x}}{{\partial u}}\frac{{\partial u}}{{\partial t}} + \frac{{\partial x}}{{\partial v}}\frac{{\partial v}}{{\partial t}}$
$\frac{{\partial y}}{{\partial s}} = \frac{{\partial y}}{{\partial u}}\frac{{\partial u}}{{\partial s}} + \frac{{\partial y}}{{\partial v}}\frac{{\partial v}}{{\partial s}}$, ${\ \ \ \ }$ $\frac{{\partial y}}{{\partial t}} = \frac{{\partial y}}{{\partial u}}\frac{{\partial u}}{{\partial t}} + \frac{{\partial y}}{{\partial v}}\frac{{\partial v}}{{\partial t}}$
We can write these as dot products:
$\frac{{\partial x}}{{\partial s}} = \left( {\frac{{\partial x}}{{\partial u}},\frac{{\partial x}}{{\partial v}}} \right)\cdot\left( {\frac{{\partial u}}{{\partial s}},\frac{{\partial v}}{{\partial s}}} \right)$, ${\ \ \ }$ $\frac{{\partial x}}{{\partial t}} = \left( {\frac{{\partial x}}{{\partial u}},\frac{{\partial x}}{{\partial v}}} \right)\cdot\left( {\frac{{\partial u}}{{\partial t}},\frac{{\partial v}}{{\partial t}}} \right)$
$\frac{{\partial y}}{{\partial s}} = \left( {\frac{{\partial y}}{{\partial u}},\frac{{\partial y}}{{\partial v}}} \right)\cdot\left( {\frac{{\partial u}}{{\partial s}},\frac{{\partial v}}{{\partial s}}} \right)$, ${\ \ \ }$ $\frac{{\partial y}}{{\partial t}} = \left( {\frac{{\partial y}}{{\partial u}},\frac{{\partial y}}{{\partial v}}} \right)\cdot\left( {\frac{{\partial u}}{{\partial t}},\frac{{\partial v}}{{\partial t}}} \right)$
Recall that for two $2 \times 2$ matrices $A$ and $B$, the $\left( {i,j} \right)$-entry of $AB$ is the dot product of the $i$th row of $A$ and the $j$th column of $B$. Thus, using these results, we can write the Jacobian matrix
$\left( {\begin{array}{*{20}{c}}
{\frac{{\partial x}}{{\partial s}}}&{\frac{{\partial x}}{{\partial t}}}\\
{\frac{{\partial y}}{{\partial s}}}&{\frac{{\partial y}}{{\partial t}}}
\end{array}} \right)$
as
$\left( {\begin{array}{*{20}{c}}
{\frac{{\partial x}}{{\partial s}}}&{\frac{{\partial x}}{{\partial t}}}\\
{\frac{{\partial y}}{{\partial s}}}&{\frac{{\partial y}}{{\partial t}}}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\
{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{\frac{{\partial u}}{{\partial s}}}&{\frac{{\partial u}}{{\partial t}}}\\
{\frac{{\partial v}}{{\partial s}}}&{\frac{{\partial v}}{{\partial t}}}
\end{array}} \right)$
Now, we evaluate the Jacobian of ${G_2} \circ {G_1}$, which is the determinant of the Jacobian matrix:
${\rm{Jac}}\left( {{G_2} \circ {G_1}} \right) = \left| {\begin{array}{*{20}{c}}
{\frac{{\partial x}}{{\partial s}}}&{\frac{{\partial x}}{{\partial t}}}\\
{\frac{{\partial y}}{{\partial s}}}&{\frac{{\partial y}}{{\partial t}}}
\end{array}} \right| = \left| {\left( {\begin{array}{*{20}{c}}
{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\
{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{\frac{{\partial u}}{{\partial s}}}&{\frac{{\partial u}}{{\partial t}}}\\
{\frac{{\partial v}}{{\partial s}}}&{\frac{{\partial v}}{{\partial t}}}
\end{array}} \right)} \right|$
Using the result in Exercise 49 for two matrices $A$ and $B$ that
$\det \left( {AB} \right) = \det \left( A \right)\det \left( B \right)$,
we obtain
${\rm{Jac}}\left( {{G_2} \circ {G_1}} \right) = \left| {\begin{array}{*{20}{c}}
{\frac{{\partial x}}{{\partial s}}}&{\frac{{\partial x}}{{\partial t}}}\\
{\frac{{\partial y}}{{\partial s}}}&{\frac{{\partial y}}{{\partial t}}}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\
{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}}
\end{array}} \right|\left| {\begin{array}{*{20}{c}}
{\frac{{\partial u}}{{\partial s}}}&{\frac{{\partial u}}{{\partial t}}}\\
{\frac{{\partial v}}{{\partial s}}}&{\frac{{\partial v}}{{\partial t}}}
\end{array}} \right|$
But ${\rm{Jac}}\left( {{G_2}} \right) = \left| {\begin{array}{*{20}{c}}
{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\
{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}}
\end{array}} \right|$ and ${\rm{Jac}}\left( {{G_1}} \right) = \left| {\begin{array}{*{20}{c}}
{\frac{{\partial u}}{{\partial s}}}&{\frac{{\partial u}}{{\partial t}}}\\
{\frac{{\partial v}}{{\partial s}}}&{\frac{{\partial v}}{{\partial t}}}
\end{array}} \right|$. Therefore,
${\rm{Jac}}\left( {{G_2} \circ {G_1}} \right) = {\rm{Jac}}\left( {{G_2}} \right){\rm{Jac}}\left( {{G_1}} \right)$
