The correct option is (a) 4
To explain: By Data \(u=\frac{yz}{x}, v=\frac{zx}{y}, w=\frac{xy}{z}\)
\(J = \frac{∂(u,v,w)}{∂(x,y,z)} = \begin{vmatrix}
\frac{∂u}{∂x} & \frac{∂u}{∂y} &\frac{∂u}{∂z}\\
\frac{∂v}{∂x} &\frac{∂v}{∂y} &\frac{∂v}{∂z}\\
\frac{∂w}{∂x} &\frac{∂w}{∂y} &\frac{∂w}{∂z}\\
\end{vmatrix} = \begin{vmatrix}
\frac{-yz}{x^2} & \frac{z}{x} &\frac{y}{x}\\
\frac{z}{y} &\frac{-zx}{y^2} &\frac{x}{y}\\
\frac{y}{z} &\frac{x}{z} &\frac{-xy}{z^2}\\
\end{vmatrix}\)
\(=\frac{-yz}{x^2}\Big\{(\frac{-zx}{y^2})(\frac{-xy}{y^2}) – (\frac{x}{z})(\frac{x}{y})\Big\} – (\frac{z}{x})\Big\{(\frac{z}{y})(\frac{-xy}{z^2})-(\frac{y}{z})(\frac{x}{y})\Big\} \\
+ \frac{y}{x}\Big\{(\frac{z}{y})(\frac{x}{z})-(\frac{y}{z})(\frac{-zx}{y^2})\Big\}\)
\(=\frac{-yz}{x^2} \Big\{\frac{x^2}{yz} – \frac{x^2}{yz}\Big\} – \frac{z}{x}\Big\{\frac{-x}{z} – \frac{x}{z}\Big\} + \frac{y}{x}\Big\{\frac{x}{y} + \frac{x}{y}\Big\} = 0 + 1 + 1 + 1 = 4\)
Therefore \(\frac{∂(u,v,w)}{∂(x,y,z)} = 4.\)