## Thomas' Calculus 13th Edition

$12$
Since, we have $\dfrac{\partial w}{\partial r}=\dfrac{\partial w}{\partial x}\dfrac{\partial x}{\partial r}+\dfrac{\partial w}{\partial y}\dfrac{\partial y}{\partial r}+\dfrac{\partial w}{\partial z}\dfrac{\partial z}{\partial r}$ and, $2(x+y+z) +2(x+y+z)[-\sin (r+s) ]+2(x+y+z))[\cos (r+s) ]=2[r-s+\cos (r+s)+\sin (r+s)][1-\sin (r+s)+\cos (r+s)]$ Now, for the point $r=1,s=-1$: Thus, $\dfrac{\partial w}{\partial r}=2[r-s+\cos (r+s)+\sin (r+s)][1-\sin (r+s)+\cos (r+s)]=2[1-(-1)+\cos (1-1)+\sin (1-1)][1-\sin (1-1)+\cos (1-1)]=12$