Solution
= Solution
At $A=(1/2,1/4)$, the <gradient> is $(-3/64,-9/32)^T$. The position <vector> has length $\sqrt5/4$, so its normalized direction is $(2,1)^T/\sqrt5$. Hence the <directional derivative> is
$$
\boxed{D_{\hat{\mathbf a}}f(A)
=\frac{(2,1)}{\sqrt5}\cdot\left(-\frac3{64},-\frac9{32}\right)
=-\frac{3\sqrt5}{40}}.
$$