Find the direction in which the maximum rate of change occurs for the function $f(x,y)=4x\sin (xy)$ 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, $\mathrm{\nabla }f(4,5)=(f_x(4,5),f_y(4,5))=(4\sin (20)+80\cos (20),64\cos (20))$

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

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

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