Find the direction in which the maximum rate of change occurs for the function $f(x, y)=4 x \sin (x y)$ at the point $(4,5)$. Give your answer as a unit vector.
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Step 1 ：Define the function \(f(x, y) = 4x \sin(xy)\)

Step 2 ：Compute the partial derivative of \(f(x, y)\) with respect to x, \(f_x(x, y) = 4x \sin(xy) + 4y \cos(xy)\)

Step 3 ：Compute the partial derivative of \(f(x, y)\) with respect to y, \(f_y(x, y) = 4x^2 \cos(xy)\)

Step 4 ：Evaluate the partial derivatives at the point (4,5) to get the gradient vector at that point, \(\nabla f(4,5) = (f_x(4,5), f_y(4,5)) = (4 \sin(20) + 80 \cos(20), 64 \cos(20))\)

Step 5 ：Compute the magnitude of the gradient vector, \(|\nabla f(4,5)| = \sqrt{(4 \sin(20) + 80 \cos(20))^2 + (64 \cos(20))^2}\)

Step 6 ：Normalize the gradient vector to get the unit vector in the direction of maximum rate of change, \(\hat{\nabla} f(4,5) = \frac{\nabla f(4,5)}{|\nabla f(4,5)|} = \left(\frac{4 \sin(20) + 80 \cos(20)}{\sqrt{(4 \sin(20) + 80 \cos(20))^2 + (64 \cos(20))^2}}, \frac{64 \cos(20)}{\sqrt{(4 \sin(20) + 80 \cos(20))^2 + (64 \cos(20))^2}}\right)\)

Step 7 ：\(\boxed{\hat{\nabla} f(4,5) = \left(\frac{4 \sin(20) + 80 \cos(20)}{\sqrt{(4 \sin(20) + 80 \cos(20))^2 + (64 \cos(20))^2}}, \frac{64 \cos(20)}{\sqrt{(4 \sin(20) + 80 \cos(20))^2 + (64 \cos(20))^2}}\right)}\)