Answer
$\frac{1}{\sqrt{2s+3t}}i + \frac{3}{2\sqrt{2s+3t}}j$
Work Step by Step
Derive $f(s, t) $with respect to $s$, which is:
$f_{s}(s, t) = (\frac{1}{2})\cdot(2s + 3t)^\frac{-1}{2}\cdot 2 = \frac{1}{\sqrt{2s+3t}}$
Derive $f(s, t) $with respect to $s$, which is:
$f_{t}(s, t) = (\frac{1}{2})\cdot(2s + 3t)^\frac{-1}{2}\cdot 3 = \frac{3}{2\sqrt{2s+3t}}$
Thus the gradient of $f$ is calculated:
$\nabla f(s,t) = f_{s}(s, t)i + f_{t}(s, t)j$
$ = \frac{1}{\sqrt{2s+3t}}i + \frac{3}{2\sqrt{2s+3t}}j$
