Answer
$$\dfrac{\sqrt 3}{6}-\dfrac{1}{4}$$
Work Step by Step
Our aim is to determine the directional derivative. In order to find it we have to use the expression:
$D_uf(x,y)=f_x(x,y)m+f_y(x,y)n$
Given: $f(x,y)=\sqrt{(2x+3y)}$
$D_uf(x,y)=\dfrac{1}{\sqrt {2x+3y}} \times \cos (-\pi/6)+\dfrac{3}{2\sqrt {2x+3y}} \times \sin (-\pi/6)$ This implies
From the given data, we have : $(x,y)=$ $(3,1)$
$D_uf(0,1)=(\dfrac{1}{3})(\dfrac{\sqrt 3}{2})+(\dfrac{1}{2})(\dfrac{-1}{2}) =\dfrac{\sqrt 3}{6}-\dfrac{1}{4}$
