## Thomas' Calculus 13th Edition

$-8$
Consider $\dfrac{\partial w}{\partial v}=\dfrac{\partial w}{\partial x}\dfrac{\partial x}{\partial v}+\dfrac{\partial w}{\partial y}\dfrac{\partial y}{\partial v}+\dfrac{\partial w}{\partial z}\dfrac{\partial z}{\partial v}$ and, $(y)(\dfrac{2v}{u})+(x)(1)+(\dfrac{1}{z})(0)=(\dfrac{2v}{u})(u+v)+(\dfrac{v^2}{u})$ For the point $u=-1,v=2$: Thus, $\dfrac{\partial w}{\partial v}=(\dfrac{4}{-1}) \times (-1+2)+\dfrac{4}{(-1)}=-8$