Please show your answer to 4 decimal places.
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 $(5,1)$. Give your answer as a unit vector.

Step 1 ：Given the function \(f(x, y)=4 x \sin (x y)\), we need to find the direction of maximum rate of change at the point \((5,1)\).

Step 2 ：The direction of maximum rate of change of a function at a given point is given by the gradient of the function at that point. The gradient is a vector that points in the direction of the steepest ascent of the function.

Step 3 ：To find the gradient, we need to compute the partial derivatives of the function with respect to each variable. In this case, we need to compute the partial derivatives of \(f(x, y)\) with respect to \(x\) and \(y\).

Step 4 ：The partial derivative of \(f(x, y)\) with respect to \(x\) is \(4x y \cos(x y) + 4 \sin(x y)\).

Step 5 ：The partial derivative of \(f(x, y)\) with respect to \(y\) is \(4x^2 \cos(x y)\).

Step 6 ：Evaluating these at the point \((5,1)\), we get \(4\sin(5) + 20\cos(5)\) for the derivative with respect to \(x\), and \(100\cos(5)\) for the derivative with respect to \(y\).

Step 7 ：The magnitude of the gradient vector is \(\sqrt{(4\sin(5) + 20\cos(5))^2 + 10000\cos(5)^2}\).

Step 8 ：The unit vector in the direction of the gradient is given by dividing each component of the gradient by the magnitude. This gives us \(\left(\frac{4\sin(5) + 20\cos(5)}{\sqrt{(4\sin(5) + 20\cos(5))^2 + 10000\cos(5)^2}}, \frac{100\cos(5)}{\sqrt{(4\sin(5) + 20\cos(5))^2 + 10000\cos(5)^2}}\right)\).

Step 9 ：\(\boxed{\left(\frac{4\sin(5) + 20\cos(5)}{\sqrt{(4\sin(5) + 20\cos(5))^2 + 10000\cos(5)^2}}, \frac{100\cos(5)}{\sqrt{(4\sin(5) + 20\cos(5))^2 + 10000\cos(5)^2}}\right)}\) is the direction in which the maximum rate of change occurs for the function \(f(x, y)=4 x \sin (x y)\) at the point \((5,1)\).