## University Calculus: Early Transcendentals (3rd Edition)

$-8$
Since, we have $\dfrac{\partial w}{\partial v}=\dfrac{\partial w}{\partial x}\dfrac{\partial x}{\partial v}+\dfrac{\partial w}{\partial y}\dfrac{\partial y}{\partial v}+\dfrac{\partial w}{\partial z}\dfrac{\partial z}{\partial v}$ or, $=(y)(\dfrac{2v}{u})+(x)(1)+(\dfrac{1}{z})(0)$ or, $=(u+v)(\dfrac{2v}{u})+(\dfrac{v^2}{u})$ Now, at point $u=-1,v=2$, we get $\dfrac{\partial w}{\partial v}=(-1+2)(\dfrac{2(2)}{-1})+(\dfrac{2^2}{(-1)})=-8$