To find the partial derivatives of the function $f(x, y) = 8y \sin(xy)$ with respect to $x$ and $y$, we proceed as follows:

Partial Derivative with respect to $x$, $f_x(x, y)$

Using the product rule and chain rule: $f_x(x, y) = \frac{\partial}{\partial x} \left( 8y \sin(xy) \right)$ Since $8y$ is a constant with respect to $x$, we only differentiate $\sin(xy)$ with respect to $x$, treating $y$ as a constant: $f_x(x, y) = 8y \cdot \cos(xy) \cdot y = 8y^2 \cos(xy)$ So, $f_x(x, y) = 8y^2 \cos(xy)$

Partial Derivative with respect to $y$, $f_y(x, y)$

Now, differentiating $f(x, y) = 8y \sin(xy)$ with respect to $y$: $f_y(x, y) = \frac{\partial}{\partial y} \left( 8y \sin(xy) \right)$ Using the product rule: $f_y(x, y) = 8 \sin(xy) + 8y \cdot \cos(xy) \cdot x$ Simplifying, $f_y(x, y) = 8 \sin(xy) + 8xy \cos(xy)$

Gradient Vector Field $\nabla f$

The gradient vector field $\nabla f$ of $f$ is given by: $\nabla f(x, y) = \left( f_x(x, y), f_y(x, y) \right)$ Substituting the partial derivatives we found: $\nabla f(x, y) = \left( 8y^2 \cos(xy), \, 8 \sin(xy) + 8xy \cos(xy) \right)$

Summary

$f_x(x, y) = 8y^2 \cos(xy)$ $f_y(x, y) = 8 \sin(xy) + 8xy \cos(xy)$ $\nabla f(x, y) = \left( 8y^2 \cos(xy), \, 8 \sin(xy) + 8xy \cos(xy) \right)$

Would you like further details or have any questions?

Here are 5 related questions:

1. How would we find higher-order partial derivatives of $f(x, y)$?
2. What is the interpretation of the gradient vector field in a physical context?
3. How would we find the directional derivative of $f(x, y)$ at a point?
4. What does setting $f_x(x, y) = 0$ and $f_y(x, y) = 0$ imply about the function $f$?
5. How can we visualize the gradient vector field of $f(x, y)$ for specific values of $x$ and $y$?

Tip: Remember, the gradient vector points in the direction of the steepest increase of a function.