Let $u$ , $v$ be differentiable functions of $x$ , $y$ and $x$ , $y$ be differentiable functions of $s$ , $t$ . Then the connection

(1)

Proof. Starting from the right hand side of (1), where one can multiply the determinants similarly as the corresponding matrices, we have $$\left|\begin{matrix} u_x & u_y \\ v_x & v_y \\ \end{matrix}\right| \cdot \left|\begin{matrix} x_s & x_t \\ y_s & y_t \\ \end{matrix}\right| \;=\; \left|\begin{matrix} u_xx_s+u_yy_s & u_xx_t+u_yy_t \\ v_xx_s+v_yy_s & v_xx_t+v_yy_t \\ \end{matrix}\right| \;=\; \left|\begin{matrix} u_s & u_t \\ v_s & v_t \\ \end{matrix}\right|. $$ Here, the last stage has been written according to the general chain rule. But thus we have arrived at the left hand side of the equation (1), which hereby has been proved.

Remark. The rule (1) is only a visualisation of the more general one concerning the case of functions of $n$ variables.