Answer
$\frac{5}{2}\sqrt {2\pi}$
Work Step by Step
Gradient of $f$= $\nabla f(x,y)=\frac{\partial f}{\partial x}\textbf{i}+\frac{\partial f}{\partial y}\textbf{j}$
$=\frac{\partial }{\partial x}(5\sin x^{2}+\cos 3y)\textbf{i}+\frac{\partial }{\partial y}(5\sin x^{2}+\cos 3y)\textbf{j}$
$=10x\cos x^{2}\textbf{i}-3\sin 3y\textbf{j}$
and so
$\nabla f(\frac{\sqrt {\pi}}{2},0)=10\times\frac{\sqrt {\pi}}{2}\times\cos (\frac{\sqrt {\pi}}{2})^{2}\textbf{i}-3\sin (3\times0)\textbf{j}$
$=5\sqrt {\pi}\times\cos (\frac{\pi}{4})\textbf{i}-3\sin 0\textbf{j}$
$=5\sqrt {\pi}\times\frac{\sqrt {2}}{2}\textbf{i}-0=\frac{5}{2}\sqrt {2\pi}$
