Product of Jacobians and chain Rule I've got a question on how to achieve this proof correctly (2nd year vector calculus) : 
Assuming that u = u(r, s) ,  v = v(r, s) , r = r(x, y) and s = s(x, y) are $C^1$ functions, prove that $$\frac{\partial(u, v)}{\partial(r, s)}\frac{\partial(r, s)}{\partial(x, y)} = \frac{\partial(u, v)}{\partial(x, y)}$$.
Hint: Use the chain rule and the identity $det(AB) = det(A)det(B)$
So far, I've done the following: 
$$\frac{\partial(u, v)}{\partial(r, s)} = \begin{vmatrix}\frac{\partial u}{\partial r} & \frac{\partial u}{\partial s} \\ \frac{\partial v}{\partial r} & \frac{\partial v}{\partial s}\end{vmatrix} = \frac{\partial u}{\partial r}\frac{\partial v}{\partial s} - \frac{\partial u}{\partial s}\frac{\partial v}{\partial r}$$
$$\frac{\partial(r, s)}{\partial(x, y)} = \begin{vmatrix}\frac{\partial r}{\partial x} & \frac{\partial r}{\partial y} \\ \frac{\partial s}{\partial x} & \frac{\partial s}{\partial y}\end{vmatrix} = \frac{\partial r}{\partial x}\frac{\partial s}{\partial y} - \frac{\partial s}{\partial x}\frac{\partial r}{\partial y}$$
$$\frac{\partial(u, v)}{\partial(x, y)} = \begin{vmatrix}\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\end{vmatrix} = \frac{\partial u}{\partial x}\frac{\partial v}{\partial y} - \frac{\partial v}{\partial x}\frac{\partial u}{\partial y}$$
so $$\frac{\partial(u, v)}{\partial(r, s)}\frac{\partial(r, s)}{\partial(x, y)} \\ =\Biggl[ \frac{\partial u}{\partial r}\frac{\partial v}{\partial s} - \frac{\partial u}{\partial s}\frac{\partial v}{\partial r}\Biggl ]\Biggl[ \frac{\partial r}{\partial x}\frac{\partial s}{\partial y} - \frac{\partial s}{\partial x}\frac{\partial r}{\partial y}\Biggl] \\ = \frac{\partial u}{\partial r}\frac{\partial r}{\partial x}\frac{\partial v}{\partial s}\frac{\partial s}{\partial y} - \frac{\partial u}{\partial s}\frac{\partial s}{\partial y}\frac{\partial v}{\partial r}\frac{\partial r}{\partial x} - \frac{\partial u}{\partial r}\frac{\partial r}{\partial y}\frac{\partial v}{\partial s}\frac{\partial s}{\partial x} + \frac{\partial u}{\partial s}\frac{\partial s}{\partial x}\frac{\partial v}{\partial r}\frac{\partial r}{\partial y} \\ = \frac{\partial u}{\partial x}\frac{\partial v}{\partial y} - \frac{\partial u}{\partial y}\frac{\partial v}{\partial x} - \frac{\partial u}{\partial y}\frac{\partial v}{\partial x} + \frac{\partial u}{\partial x}\frac{\partial v}{\partial y}$$ which equals $$2 \frac{\partial u}{\partial x}\frac{\partial v}{\partial y} - 2 \frac{\partial u}{\partial y}\frac{\partial v}{\partial x}$$
But I'm struggling to see what I'm doing wrong, so just wondering if anyone could direct me on the correct path please, thanks!
 A: You cannot just cancel things like $$\frac {\partial u}{\partial r} \frac {\partial r}{\partial x} \to \frac {\partial u}{\partial x},  $$ this is not the $\mathrm d/\mathrm dx $ symbol. Unlike univariate Calculus, $\partial u/\partial x$ is only a notation, which is actually not a quotient: $\partial u$ is meaningless. 
Demonstration
Let $\boldsymbol g\colon (r,s) \mapsto (u,v) , \boldsymbol f \colon (x,y)\mapsto (r,s)$, then $(\boldsymbol g \circ \boldsymbol f)\colon (x,y) \mapsto (u,v)$. The chain rule indicates that
$$
\mathrm D (\boldsymbol g \circ \boldsymbol f) = \mathrm D\boldsymbol g \mathrm D \boldsymbol f,
$$
In matrix form, it is just [$\partial_x u$ is just $\partial u/\partial x$, for brevity]
$$
\begin{bmatrix}
\partial_x u & \partial_y u \\ \partial_x v & \partial_y v 
\end{bmatrix}=
\begin{bmatrix}
\partial_r u & \partial_s u \\ \partial_r v & \partial_s v 
\end{bmatrix}
\begin{bmatrix}
\partial_x r & \partial_y s \\ \partial_x r & \partial_y s 
\end{bmatrix},
$$
take determinants, making use of the equation
$$
\det (\boldsymbol {AB}) = \det(\boldsymbol A)\det(\boldsymbol B)
$$
for $n \times n [n \in \mathbb N^*]$ matrices $\boldsymbol A,\boldsymbol B$
and note that
$$
\frac {\partial (u,v)}{\partial (x,y)} = \det \left( 
\begin{bmatrix}
\partial_x u & \partial_y u \\ \partial_x v & \partial_y v 
\end{bmatrix} \right), 
$$
and the rest also have similar correspondence, we have the desired equation. 
Remark
According to the chain rule, one has
$$
\frac {\partial u}{\partial x} = \frac {\partial u}{\partial r} \frac {\partial r}{\partial x} + \frac {\partial u}{\partial s} \frac {\partial s}{\partial x} \neq \frac {\partial u}{\partial r}\frac {\partial r}{\partial x}
$$
in general, so you cannot do the cancellation in your post. 
A: Derivative Matrix
If f : $\Bbb R^n \to \Bbb R^m$ is a differentiable function, then the derivative is an  m x n matrix given by 
$$\mathbf D f = \begin{bmatrix} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} & \cdots & \frac{\partial f_1}{\partial x_n}\\ \vdots & \vdots & \ddots & \vdots \\ \frac{\partial f_m}{\partial x_1} & \frac{\partial f_m}{\partial x_2} & \cdots & \frac{\partial f_m}{\partial x_n}\end{bmatrix}$$ 
where $ f(x_1, x_2, ... , x_n) = (f_1(x_1, ... x_n), f_2(x_1, ..., x_n)..., f_m(x_1,...,x_n))$
Matrix Version of Chain Rule
If f : $\Bbb R^m \to  \Bbb R^p $ and g : $\Bbb R^n \to \Bbb R^m$ are differentiable functions and the composition f $\circ$ g is defined then $$\mathbf D(f \circ g) = \mathbf Df \mathbf Dg$$  
Note: (f $\circ$ g)$(x_1, x_2,...,x_n)$ = f [g ($x_1,x_2,...,x_n)]$
