Prove $ dy_1 \, dy_2=|J|\,dx_1 \, dx_2$, where $|J|$ is the determinant of the Jacobian. Suppose,$y_1=y_1(x_1,x_2)$ and $y_2=y_2(x_1,x_2)$,
such that,
$$dy_1=\frac{\partial y_1}{\partial x_1}dx_1+\frac{\partial y_1}{\partial x_2} \, dx_2$$
$$dy_2=\frac{\partial y_2}{\partial x_1}dx_1+\frac{\partial y_2}{\partial x_1} \, dx_2$$
Then,I've taken the product of the above two,but unable to reach to the result.
 A: You need not the usual product, but the exterior product of the two (see Arnol'd's Mathematical Methods of Classical Mechanics):
$$
dy_{1} \wedge dy_{2} = \left({\partial y_{1} \over \partial x_{1}} {\partial y_{2} \over \partial x_{2}}  - 
{\partial y_{1} \over \partial x_{2}} {\partial y_{2} \over \partial x_{1}} 
\right) dx_{1} \wedge dx_{2}.
$$
The expression in ()'s is your $J$.
A: Let $y^1 = y^1(x^1,x^2)$ and $y^2 = y^2(x^1,x^2)$. Suppose also that;
$$\\$$
$$dy^1=\frac{\partial y^1}{\partial x^1}dx^1+\frac{\partial y_1}{\partial x^2}dx^2$$
$$dy^2=\frac{\partial y^2}{\partial x^1}dx^1+\frac{\partial y^2}{\partial x^1}dx^2 $$
Then we have;
$$dy^1dy^2 = dy^1 \wedge dy^2$ =\left( \frac{\partial y^1}{\partial x^1}dx^1+\frac{\partial y^1}{\partial x^2}dx^2\right) \wedge \left(\frac{\partial y^2}{\partial x^1}dx^1+\frac{\partial y^2}{\partial x^1}dx^2\right)$$
The above gives the following;
$$\\$$
$$ \left(\frac{\partial y^1}{\partial x^1} \frac{\partial y^2}{\partial x^1}\right) dx^1\wedge dx^1 +  \left(\frac{\partial y^1}{\partial x^1} \frac{\partial y^2}{\partial x^1}\right) dx^1\wedge dx^2 +  \left(\frac{\partial y^1}{\partial x^2} \frac{\partial y^2}{\partial x^1}\right) dx^2\wedge dx^1 + \left(\frac{\partial y^1}{\partial x^2} \frac{\partial y^2}{\partial x^1}\right) dx^2\wedge dx^2  $$
$$\\$$
Now recall that $dx^i \wedge dx^i = 0, dx^j \wedge dx^i = -dx^i \wedge dx^j$ and so you have;
$$\\$$
$$\left(\frac{\partial y^1}{\partial x^1} \frac{\partial y^2}{\partial x^1}\right) dx^1\wedge dx^2 +  \left(\frac{\partial y^1}{\partial x^2} \frac{\partial y^2}{\partial x^1}\right) dx^2\wedge dx^1 = \underbrace{ \left(\frac{\partial y^1}{\partial x^1} \frac{\partial y^2}{\partial x^1}-\frac{\partial y^1}{\partial x^2} \frac{\partial y^2}{\partial x^1} \right)}_{\textbf{Jacobian}} \ dx^1\wedge dx^2  $$
