Answer:
Step-by-step explanation:
The directional derivative of f at the given point in the direction indicated by the angle θ is expressed as [tex]\nabla f(x, y)*u[/tex] where u is the unit vector in the direction θ.
Lets first calculate [tex]\nabla f(x, y)\ at\ (0, 1)[/tex]
[tex]\nabla = \frac{\delta}{\delta x} i + \frac{\delta}{\delta y} j \\\nabla f(x, y) = \frac{\delta (y cos(xy))}{\delta x} i + \frac{\delta(y cos(xy))}{\delta y} j\\\nabla f(x, y)= -y^{2} sinxy\ i + (cosxy -xysinxy) j\\[/tex]
[tex]\nabla f(x, y)\ at\ (0, 1)\\= -1^{2}sin0 \ i +(cos 0 - 0sin0) \j\\= 0i+j\\\\[/tex]
The unit vector u in the direction of θ is expressed as [tex]cos\theta \ i + sin\theta \ j[/tex]
unit vector u at θ = π/6 is cos π/6i + sin π/6 j
u= √3/2 i +1/2 j
Taking the dot product i.e [tex]\nabla f(x, y)*u[/tex]
= (0i+j)*(√3/2 i +1/2 j)
= 1/2
The directional derivative of f is 1/2
