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To find the gradient \( \nabla u \) of the function \( u = x y^2 z^3 \), we will compute the partial derivatives of \( u \) with respect to \( x \), \( y \), and \( z \).

Given:
\[
u = x y^2 z^3
\]

The gradient \( \nabla u \) is:
\[
\nabla u = \frac{\partial u}{\partial x} \hat{i} + \frac{\partial u}{\partial y} \hat{j} + \frac{\partial u}{\partial z} \hat{k}
\]

### Step 1: Compute \( \frac{\partial u}{\partial x} \)
\[
\frac{\partial u}{\partial x} = \frac{\partial}{\partial x} (x y^2 z^3) = y^2 z^3
\]
Thus, the \( \hat{i} \)-component of \( \nabla u \) is:
\[
y^2 z^3 \hat{i}
\]

### Step 2: Compute \( \frac{\partial u}{\partial y} \)
\[
\frac{\partial u}{\partial y} = \frac{\partial}{\partial y} (x y^2 z^3) = 2x y z^3
\]
So, the \( \hat{j} \)-component of \( \nabla u \) is:
\[
2x y z^3 \hat{j}
\]

### Step 3: Compute \( \frac{\partial u}{\partial z} \)
\[
\frac{\partial u}{\partial z} = \frac{\partial}{\partial z} (x y^2 z^3) = 3x y^2 z^2
\]
Thus, the \( \hat{k} \)-component of \( \nabla u \) is:
\[
3x y^2 z^2 \hat{k}
\]

### Final Result
Combining these results, we get:
\[
\nabla u = y^2 z^3 \hat{i} + 2x y z^3 \hat{j} + 3x y^2 z^2 \hat{k}
\]

Now, let's compare this with the answer choices:

- **Option (a)**: \( x y^2 (y z \hat{i} + z x \hat{j} + x y \hat{k}) \) — Does not match.
- **Option (b)**: \( 2 y z^2 (y z \hat{i} + z x \hat{j} + x y \hat{k}) \) — Does not match.
- **Option (c)**: \( y z^2 (y z \hat{i} + 2 z x \hat{j} + 3 x y \hat{k}) \) — Matches.
- **Option (d)**: \( 2 x y z (y z \hat{i} + z x \hat{j} + x y \hat{k}) \) — Does not match.

### Answer
The correct answer is **(c)**. 

Would you like a breakdown of each option? 

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#### Related Questions
1. How does the gradient operation change for functions of more than three variables?
2. What does the gradient of a scalar function represent geometrically?
3. How would the process differ if we were calculating the divergence of a vector field instead?
4. Can you explain why the coefficients of each unit vector are partial derivatives?
5. How would this change if the function \( u \) involved trigonometric functions?

#### Tip
When calculating gradients, each component corresponds to the rate of change of the function in that specific direction, making it crucial to compute each partial derivative carefully.

Learn how to calculate the gradient ∇u of the function u = xy^2z^3. This solution involves finding partial derivatives with respect to x, y, and z and is suitable for calculus students.

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