$(Given)\ x=uv \cdots(1) $

• $Hence,\ x_u = \dfrac{\delta x}{\delta u} = v$

$x_v = \dfrac{\delta x}{\delta v} = u$

• $(Given)\ y=\dfrac {u}v \cdots(2) $

$ y_u=\dfrac {\delta y}{\delta u} = \dfrac {1}v$

$ y_v=\dfrac {\delta y}{\delta v} = u.\dfrac {-1}{v^2}$

$J = \dfrac{\delta(x,y)}{\delta(u,v)}$

$= \begin{vmatrix} x_u & x_v \\[0.3em] y_u & y_v \end{vmatrix}$

$= x_u .y_v - x_v. y_u $

$= v.\dfrac {-u}{v^2}- u.\dfrac {1}v$

$= \dfrac {-2u}{v}$

$= - 2y \cdots (3)$

$From\ 2, \ u= v.y \cdots(4)$

• $Substituting\ 'u'\ in\ (1)\ we\ get,$

$x = (v y)v$

$ \dfrac {x}y = v^2$

$v = \dfrac {\sqrt x} {\sqrt y}$

$= x^{1/2}.y^{-1/2} \cdots (5)$

$v_x=y^{-1/2}.\dfrac {1} 2 x^{-1/2}\ and$

$v_y=x^{1/2}.\dfrac {-1} 2 y^{-3/2}$

• $Hence\ from\ (4)\ and\ (5)\ we\ get,$

$u = (x^{1/2}.y^{-1/2})y$

$=x^{1/2}.y^{1/2}$

$u_x=y^{1/2}.\dfrac {1} 2 x^{-1/2}\ and$

$u_y=x^{1/2}.\dfrac {1} 2 y^{-1/2}$

$J' = \dfrac{\delta(u,v)}{\delta(x,y)}$

$= \begin{vmatrix} u_x & u_y \\[0.3em] v_x & v_y \end{vmatrix}$

$= u_x .v_y - u_y. v_x$

$=\bigg (y^{1/2}.\dfrac {1} 2 x^{-1/2} \bigg) \times\ \bigg (x^{1/2}.\dfrac {-1} 2 y^{-3/2} \bigg) - \bigg (x^{1/2}.\dfrac {1} 2 y^{-1/2}\bigg) \times \bigg(y^{-1/2}.\dfrac {1} 2 x^{-1/2} \bigg)$

$ = -\dfrac {1} 4 .\dfrac {1} y -\dfrac {1} 4 .\dfrac {1} y$

$= -\dfrac {1} {2y} \cdots(6)$

• $We\ equate\ (3)\ and\ (6)\ we\ get,$

$J. J' = -2y .\dfrac {-1} {2y} $

$J. J' = 1\ \cdots Hence\ Proved.$