Module $12$: Problem $3$

$(1$ point$)$

Results for this submission

$2$ of the answers are NOT correct.

Suppose $f(x,y)=\sqrt[\phantom{2}]{tan(x)y}$ and $u$ is the unit vector in the direction of $(-3,-3).$ Then,

$($a$)$ gradf$(x,y)=($:$\frac{se{c}^2(x)}{2\sqrt[\phantom{2}]{tan(x)}},1$:$)$

$($b$)$ gradf$(1\mathrm{.}2,6)=$

$(1)$

$($c$)f_u(1\mathrm{.}2,6)=D_{u}f(1\mathrm{.}2,6)=$