Question

Given \(u=\frac{yz}{x}, v=\frac{zx}{y}, w=\frac{xy}{z}\) then the value of \(\frac{∂(u,v,w)}{∂(x,y,z)}\) is ________

(a) 4

(b) -4

(c) 0

(d) 1

This question was posed to me in an interview for job.

My enquiry is from Jacobians in section Partial Differentiation of Engineering Mathematics

Answer 1

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  \(\frac{∂(u,v,w)}{∂(x,y,z)} = 4.\)