To find the gradient of the function $f(x, y, z) = e^{xy} - x \cos(yz^2)$, we need to compute the partial derivatives of $f$ with respect to each variable $x$, $y$, and $z$.

Step 1: Partial derivative of $f$ with respect to $x$

$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left( e^{xy} - x \cos(yz^2) \right)$ For the first term $e^{xy}$, we use the chain rule: $\frac{\partial}{\partial x} \left( e^{xy} \right) = y e^{xy}$ For the second term $-x \cos(yz^2)$, apply the product rule: $\frac{\partial}{\partial x} \left( -x \cos(yz^2) \right) = -\cos(yz^2)$ Thus, $\frac{\partial f}{\partial x} = y e^{xy} - \cos(yz^2)$

Step 2: Partial derivative of $f$ with respect to $y$

$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left( e^{xy} - x \cos(yz^2) \right)$ For the first term $e^{xy}$, apply the chain rule: $\frac{\partial}{\partial y} \left( e^{xy} \right) = x e^{xy}$ For the second term $-x \cos(yz^2)$, use the chain rule: $\frac{\partial}{\partial y} \left( -x \cos(yz^2) \right) = x \sin(yz^2) \cdot z^2$ Thus, $\frac{\partial f}{\partial y} = x e^{xy} + x z^2 \sin(yz^2)$

Step 3: Partial derivative of $f$ with respect to $z$

$\frac{\partial f}{\partial z} = \frac{\partial}{\partial z} \left( e^{xy} - x \cos(yz^2) \right)$ For the first term $e^{xy}$, there is no $z$ dependence, so: $\frac{\partial}{\partial z} \left( e^{xy} \right) = 0$ For the second term $-x \cos(yz^2)$, apply the chain rule: $\frac{\partial}{\partial z} \left( -x \cos(yz^2) \right) = x \sin(yz^2) \cdot 2yz$ Thus, $\frac{\partial f}{\partial z} = 2xyz \sin(yz^2)$

Gradient of $f$

Now, combining all the partial derivatives, we get the gradient of $f$: $\nabla f = \left( y e^{xy} - \cos(yz^2), \; x e^{xy} + xz^2 \sin(yz^2), \; 2xyz \sin(yz^2) \right)$

Let me know if you'd like further clarification or more details!

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Here are five related questions you might find interesting:

1. How do you compute the divergence of a vector field?
2. What is the physical meaning of the gradient in a scalar field?
3. How does the gradient relate to directional derivatives?
4. How can you extend this calculation to higher dimensions?
5. What are the applications of gradients in optimization problems?

Tip: The gradient always points in the direction of the steepest increase of a function.