## Thomas' Calculus 13th Edition

$u_{max}= \lt 0,1 \gt$ $u_{min}= \lt 0,-1 \gt$ $D_u f_{max} = 2$ $D_u f_{min} =-2$
$f_x (1,0)= 2xy+ye^{xy} \sin y = 0$ and $f_y(1,0)= x^2+e^{xy} \cos y +x e^{xy} \sin y =(1)^2+e^{(1)(0)} \cos (0) +(1) e^{(1)(0)} \sin (0)=2$ $u_{max}=\dfrac{\nabla f (,0)}{|\nabla f (1,0) |}= \lt 0,1 \gt$ $u_{min}=- u_{max}= \lt 0,-1 \gt$ Since, $D_u f = \nabla f \cdot u$ Now, $D_u f_{max} =\nabla f (1,0) \cdot u_{max} =\lt 0,2 \gt \cdot \lt 0,1 \gt = 2$ and $D_u \ f_{min} =-D_u \ f_{max} = \ -2$