Answer
$D_{u}f(4,3)=\frac{21}{2}\sqrt 2$
Work Step by Step
The directional derivative of f where f is a function of x and y is denoted by $D_{u}f(x,y)$ and is found by the formula:
$D_{u}f(x,y)=f_{x}(x,y)cosB+f_{y}(x,y)sinB$
Where $cosB$ and $sinB$ are found from the directional unit vector
$u=cosBi+sinBj$
$f(x,y)=x^3-y^3$
Our directional unit vector is given.
$v=\frac{\sqrt 2}{2}i+\frac{\sqrt 2}{2}j$
In this case, $cosB$ and $sinB=\frac{\sqrt 2}{2}$
Now we use the formula to find the directional derivative.
$D_{u}f(x,y)=3x^2\frac{\sqrt 2}{2}-3y^2\frac{\sqrt 2}{2}$
Substituting in the point $(4,3)$ we have
$D_{u}f(4,3)=3(4)^2\frac{\sqrt 2}{2}-3(3)^2\frac{\sqrt 2}{2}$
Evaluating this we get our answer
$D_{u}f(4,3)=\frac{21}{2}\sqrt 2$
