Chapter 13 - Partial Derivatives - 13.3 Partial Derivatives - Exercises Set 13.3 - Page 937: 40

$${w_x}\left( {\frac{1}{2},\pi } \right) = - \frac{\pi }{4},\,\,\,{\text{and }}\,\,\,{w_y}\left( {\frac{1}{2},\pi } \right) = - \frac{1}{8}$$

$$\eqalign{ & w = {x^2}\cos xy;\,\,\,\,\,\,\,\,\,\frac{{\,\partial w}}{{\partial x}}\left( {\frac{1}{2},\pi } \right),\,\,\,\,\,\frac{{\,\partial w}}{{\partial y}}\left( {\frac{1}{2},\pi } \right) \cr & \cr & {\text{Calculate }}{w_x}\left( {x,y} \right){\text{ differentiating with respect to }}x;{\text{ treat }}y{\text{ as a constant}} \cr & {\text{and use the product rule}} \cr & {w_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left( {{x^2}\cos xy} \right) \cr & {w_x}\left( {x,y} \right) = \cos xy\frac{\partial }{{\partial x}}\left( {{x^2}} \right) + {x^2}\frac{\partial }{{\partial x}}\left( {\cos xy} \right) \cr & {w_x}\left( {x,y} \right) = \cos xy\left( {2x} \right) + {x^2}\left( { - \sin xy} \right)\left( y \right) \cr & {w_x}\left( {x,y} \right) = 2x\cos xy - {x^2}y\sin xy \cr & \cr & {\text{Evaluating }}{w_x}\left( {\frac{1}{2},\pi } \right) \cr & {w_x}\left( {\frac{1}{2},\pi } \right) = 2\left( {\frac{1}{2}} \right)\cos \left( {\frac{1}{2}\pi } \right) - {\left( {\frac{1}{2}} \right)^2}\left( \pi \right)\sin \left( {\frac{1}{2}\pi } \right) \cr & {w_x}\left( {\frac{1}{2},\pi } \right) = 2\left( {\frac{1}{2}} \right)\left( 0 \right) - {\left( {\frac{1}{2}} \right)^2}\left( \pi \right)\left( 1 \right) \cr & {w_x}\left( {\frac{1}{2},\pi } \right) = - \frac{\pi }{4} \cr & \cr & {\text{Calculate }}{w_y}\left( {x,y} \right){\text{ differentiating with respect to }}y;{\text{ treat }}x{\text{ as a constant}} \cr & {w_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left( {{x^2}\cos xy} \right) \cr & {w_y}\left( {x,y} \right) = {x^2}\frac{\partial }{{\partial y}}\left( {\cos xy} \right) \cr & {w_y}\left( {x,y} \right) = {x^2}\left( { - \sin xy} \right)\left( x \right) \cr & {w_y}\left( {x,y} \right) = - {x^3}\sin xy \cr & \cr & {\text{Evaluating }}{w_y}\left( {\frac{1}{2},\pi } \right) \cr & {w_y}\left( {\frac{1}{2},\pi } \right) = - {\left( {\frac{1}{2}} \right)^3}\sin \left( {\frac{1}{2}\pi } \right) \cr & {w_y}\left( {\frac{1}{2},\pi } \right) = - \frac{1}{8} \cr} $$