## Calculus 8th Edition

$2uvw$
Since, $Jacobian =\begin{vmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}&\dfrac{\partial x}{\partial w}\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}&\dfrac{\partial y}{\partial w}\\\dfrac{\partial z}{\partial u}&\dfrac{\partial z}{\partial v}&\dfrac{\partial z}{\partial w}\end{vmatrix}$ Given: $x=uv$ Here, we have $\dfrac{\partial x}{\partial u}=v; \dfrac{\partial x}{\partial v}=u$ and $\dfrac{\partial x}{\partial w}=0$ Given: $y=vw$ Also, $\dfrac{\partial y}{\partial u}=0; \dfrac{\partial y}{\partial v}=w$ and $\dfrac{\partial y}{\partial w}=v$ Given: $z=wu$ Also, $\dfrac{\partial z}{\partial u}=w; \dfrac{\partial z}{\partial v}=0$ and $\dfrac{\partial z}{\partial w}=u$ Now, $Jacobian =\begin{vmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}&\dfrac{\partial x}{\partial w}\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}&\dfrac{\partial y}{\partial w}\\\dfrac{\partial z}{\partial u}&\dfrac{\partial z}{\partial v}&\dfrac{\partial z}{\partial w}\end{vmatrix}=\begin{vmatrix} v&u&0\\0&w&v\\w&0&u\end{vmatrix}$ or, $=v(wu-0)-u(0-vw) +0$ Hence, $Jacobian=2uvw$