Prompt Title: Mat

To find the partial derivatives of the given function $W = \cos(x^2 + 2y) - e^{4x - z^4y} + Y^3$, we'll compute the partial derivatives with respect to each of the variables $x$, $y$, and $z$. We'll treat $Y$ as a constant since it doesn't involve $x$, $y$, or $z$.

1. Partial Derivative with Respect to $x$:

$\frac{\partial W}{\partial x} = -\sin(x^2 + 2y) \cdot (2x) - e^{4x - z^4y} \cdot 4 + 0$

Simplifying:

$\frac{\partial W}{\partial x} = -2x\sin(x^2 + 2y) - 4e^{4x - z^4y}$

2. Partial Derivative with Respect to $y$:

$\frac{\partial W}{\partial y} = -\sin(x^2 + 2y) \cdot 2 + e^{4x - z^4y} \cdot z^4 + 0$

Simplifying:

$\frac{\partial W}{\partial y} = -2\sin(x^2 + 2y) + z^4e^{4x - z^4y}$

3. Partial Derivative with Respect to $z$:

$\frac{\partial W}{\partial z} = 0 - e^{4x - z^4y} \cdot (-4yz^3) + 0$

Simplifying:

$\frac{\partial W}{\partial z} = 4yz^3e^{4x - z^4y}$

So, the partial derivatives of $W$ with respect to $x$, $y$, and $z$ are:

$\frac{\partial W}{\partial x} = -2x\sin(x^2 + 2y) - 4e^{4x - z^4y}$

$\frac{\partial W}{\partial y} = -2\sin(x^2 + 2y) + z^4e^{4x - z^4y}$

$\frac{\partial W}{\partial z} = 4yz^3e^{4x - z^4y}$