To find the partial derivative of $f(x, y) = |xy|$ with respect to $x$, we need to treat $y$ as a constant and differentiate with respect to $x$. However, we need to carefully address the absolute value function, since its behavior changes depending on the sign of $xy$.

We can break the absolute value function into two cases:

1. If $xy \geq 0$, then $|xy| = xy$.
2. If $xy < 0$, then $|xy| = -xy$.

Thus, the function $f(x, y)$ can be written as:

\begin{cases} xy & \text{if } xy \geq 0 \\ -xy & \text{if } xy < 0 \end{cases}$$ Now, let's differentiate each case with respect to $$x$$. ### Case 1: $$xy \geq 0$$ If $$xy \geq 0$$, then: $$f(x, y) = xy$$ The partial derivative with respect to $$x$$ is: $$\frac{\partial}{\partial x} (xy) = y$$ ### Case 2: $$xy < 0$$ If $$xy < 0$$, then: $$f(x, y) = -xy$$ The partial derivative with respect to $$x$$ is: $$\frac{\partial}{\partial x} (-xy) = -y$$ ### Conclusion: So, the partial derivative of $$f(x, y) = |xy|$$ with respect to $$x$$ is: $$\frac{\partial f}{\partial x} = \begin{cases} y & \text{if } xy \geq 0 \\ -y & \text{if } xy < 0 \end{cases}$$ Would you like more details or have any questions? ### Related questions: 1. How would the partial derivative change if we had $$|x^2y|$$ instead of $$|xy|$$? 2. What happens if we take the partial derivative of $$f(x, y) = \sqrt{x^2 + y^2}$$ with respect to $$x$$? 3. Can you explain how the sign of $$xy$$ affects the differentiation process? 4. How do we handle absolute values when differentiating a multivariable function in general? 5. What would happen if we took the second partial derivative of $$|xy|$$? **Tip:** When dealing with absolute values in differentiation, consider breaking the function into piecewise cases based on the sign of the argument inside the absolute value.