To find the partial derivative of $Z = f(x, y, u(x, y), v(x, y)) = xy + u(x, y)\sin{x} + v(x, y) e^{x^2}$ with respect to $x$, we'll apply the chain rule and differentiate each term in $Z$ with respect to $x$.

Step-by-step differentiation:

1. For the term $xy$: $\frac{\partial}{\partial x} (xy) = y$

2. For the term $u(x, y)\sin{x}$: Using the product rule, we treat $u(x, y)$ as a function of $x$ and $y$, so: $\frac{\partial}{\partial x} \left(u(x, y)\sin{x}\right) = \frac{\partial u}{\partial x} \sin{x} + u(x, y) \cos{x}$

3. For the term $v(x, y)e^{x^2}$: Again, using the product rule: $\frac{\partial}{\partial x} \left(v(x, y)e^{x^2}\right) = \frac{\partial v}{\partial x} e^{x^2} + v(x, y) \cdot \frac{d}{dx} \left(e^{x^2}\right)$ The derivative of $e^{x^2}$ with respect to $x$ is: $\frac{d}{dx} \left(e^{x^2}\right) = 2x e^{x^2}$ So, the derivative of $v(x, y)e^{x^2}$ is: $\frac{\partial v}{\partial x} e^{x^2} + v(x, y) \cdot 2x e^{x^2}$

Combining all terms:

Now, summing up all the partial derivatives, we get: $\frac{\partial Z}{\partial x} = y + \frac{\partial u}{\partial x} \sin{x} + u(x, y)\cos{x} + \frac{\partial v}{\partial x} e^{x^2} + 2x v(x, y) e^{x^2}$

This is the partial derivative of $Z$ with respect to $x$.

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Do you need further clarification or more details on any step?

Here are some related questions to consider:

1. What is the partial derivative of $Z$ with respect to $y$?
2. How would you compute the mixed partial derivative $\frac{\partial^2 Z}{\partial x \partial y}$?
3. How does the chain rule apply to functions with nested dependencies like $u(x, y)$ and $v(x, y)$?
4. What happens if $u(x, y)$ and $v(x, y)$ are given specific functional forms?
5. How would the solution change if we introduced higher-order terms in $x$?

Tip: Always use the product and chain rules when differentiating products of functions of multiple variables.