To find the gradient vector field of the function $f(x, y) = x e^{5xy}$, we need to compute the partial derivatives of $f$ with respect to $x$ and $y$.

The gradient of $f$, denoted $\nabla f(x, y)$, is defined as: $\nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)$

Let's compute each partial derivative.

Step 1: Compute $\frac{\partial f}{\partial x}$

Given: $f(x, y) = x e^{5xy}$ To find $\frac{\partial f}{\partial x}$, apply the product rule, since $f(x, y) = x \cdot e^{5xy}$.

1. Differentiate $x$ with respect to $x$: $1$.
2. Differentiate $e^{5xy}$ with respect to $x$: $\frac{\partial}{\partial x} e^{5xy} = e^{5xy} \cdot 5y$

So: $\frac{\partial f}{\partial x} = 1 \cdot e^{5xy} + x \cdot (e^{5xy} \cdot 5y) = e^{5xy} + 5xy e^{5xy}$ or, simplifying, $\frac{\partial f}{\partial x} = e^{5xy}(1 + 5xy)$

Step 2: Compute $\frac{\partial f}{\partial y}$

Now, let’s find $\frac{\partial f}{\partial y}$.

To do this, we treat $x$ as a constant: $f(x, y) = x e^{5xy}$

The derivative of $e^{5xy}$ with respect to $y$ is: $\frac{\partial}{\partial y} e^{5xy} = e^{5xy} \cdot 5x$

So: $\frac{\partial f}{\partial y} = x \cdot (e^{5xy} \cdot 5x) = 5x^2 e^{5xy}$

Final Answer

The gradient vector field $\nabla f(x, y)$ is: $\nabla f(x, y) = \left( e^{5xy}(1 + 5xy), 5x^2 e^{5xy} \right)$

Would you like further details on the steps?

Here are some related questions to expand on this topic:

1. How do you interpret the direction and magnitude of a gradient vector field?
2. What does it mean when the gradient vector field is zero at a point?
3. How would the gradient vector change if $f(x, y)$ were a function of three variables?
4. Can you find the divergence of this gradient vector field?
5. What are practical applications of gradient vector fields in physics or engineering?

Tip: The gradient vector field points in the direction of the steepest increase of the function and its magnitude represents the rate of increase at any given point.