Here,$u = \dfrac{yz}{x}, v =\dfrac{zx}{y}, w = \dfrac{xy}{z}$

• Differentiating Partially,

$u_x = \dfrac{-yz}{x^2} , \space u_y = \dfrac{z}{x}, \space u_z = \dfrac{y}{x}$

$v_x = \dfrac{z}{y}, \space v_y = \dfrac{-zx}{y^2}, \space v_z = \dfrac {x}{y}$

$w_x = \dfrac{y}{z}, \space w_y =\dfrac{x}{z}, \space w_z = \dfrac{-xy}{z^2}$

• Now, is given by,

$j[\dfrac{u, v, w} {x, y, z}] = \left|\begin{matrix} \dfrac{-yz}{x^2}\space \dfrac{z}{x}\space \dfrac{y}{x} \\ \dfrac{z}{y} \space \dfrac{-zx}{y^2} \space \dfrac{x}{y} \\ \dfrac{y}{z} \space \dfrac{x}{z} \space \dfrac{-xy}{z^2} \end{matrix}\right| $

$= \dfrac{-yz}{x^2} (\dfrac{x^2yz}{y^2 z^2}-\dfrac{x^2}{yz}) -\dfrac{z}{x} ({-\dfrac{xyz}{yz^2} - \dfrac{xy}{zy})}+\dfrac{y}{x}({\dfrac{xz}{yz}+\dfrac{xyz}{y^2z}})$ 

$= -\dfrac{yz}{x^2} (\dfrac{{x^2}}{y z}-\dfrac{x^2 }{yz}) -\dfrac{z}{x} ({ -\dfrac{x}{z} - \dfrac{x}{z})}+\dfrac{y}{x}({\dfrac{x}{y}+\dfrac{x}{y}})$

$=0+1+1+1+1 =4$

∴Ans :-$j[\dfrac{u, v, w}{x, y, z}] = 4$