## Calculus 8th Edition

$1+2uvw-u^2-v^2-w^2$
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: $x=u+vw$ Here, we have $\dfrac{\partial x}{\partial u}=1; \dfrac{\partial x}{\partial v}=w$ and $\dfrac{\partial x}{\partial w}=v$ Given that $y=v+wu$ Thus, $\dfrac{\partial y}{\partial u}=w; \dfrac{\partial y}{\partial v}=1$ and $\dfrac{\partial y}{\partial w}=u$ Given: $z=w+uv$ Thus, $\dfrac{\partial z}{\partial u}=v; \dfrac{\partial z}{\partial v}=u$ and $\dfrac{\partial z}{\partial w}=1$ Now, $Jacobian =\begin{vmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}&\dfrac{\partial x}{\partial w}\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}&\dfrac{\partial y}{\partial w}\\\dfrac{\partial z}{\partial u}&\dfrac{\partial z}{\partial v}&\dfrac{\partial z}{\partial w}\end{vmatrix}=\begin{vmatrix} 1&w&v\\w&1&u\\v&u&1\end{vmatrix}=1(1-u^2)-w(w-uv)+v(uw-v)=1+2uvw-u^2-v^2-w^2$