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The problem presented is to evaluate the partial derivatives of the function \( f(x, y) = x^2 + x^2y^3 - 2y^2 \), with respect to \( x \) and \( y \).

Let's compute the partial derivatives:

### 1. Partial derivative with respect to \( x \) (\( f_x \)):

To find \( f_x \), we differentiate \( f(x, y) = x^2 + x^2y^3 - 2y^2 \) with respect to \( x \), treating \( y \) as a constant:

\[
f_x = \frac{\partial}{\partial x} \left( x^2 + x^2y^3 - 2y^2 \right)
\]

Step by step:

- The derivative of \( x^2 \) with respect to \( x \) is \( 2x \).
- The derivative of \( x^2y^3 \) (where \( y^3 \) is treated as a constant) is \( 2xy^3 \).
- The derivative of \( -2y^2 \) with respect to \( x \) is \( 0 \), since it doesn't depend on \( x \).

So, we have:

\[
f_x = 2x + 2xy^3
\]

### 2. Partial derivative with respect to \( y \) (\( f_y \)):

To find \( f_y \), we differentiate \( f(x, y) = x^2 + x^2y^3 - 2y^2 \) with respect to \( y \), treating \( x \) as a constant:

\[
f_y = \frac{\partial}{\partial y} \left( x^2 + x^2y^3 - 2y^2 \right)
\]

Step by step:

- The derivative of \( x^2 \) with respect to \( y \) is \( 0 \), since it doesn't depend on \( y \).
- The derivative of \( x^2y^3 \) (where \( x^2 \) is treated as a constant) is \( 3x^2y^2 \).
- The derivative of \( -2y^2 \) with respect to \( y \) is \( -4y \).

So, we have:

\[
f_y = 3x^2y^2 - 4y
\]

### Final results:

- \( f_x = 2x + 2xy^3 \)
- \( f_y = 3x^2y^2 - 4y \)

Would you like more details or explanations?

### Relative questions:
1. What is the significance of partial derivatives in multivariable functions?
2. How can we interpret \( f_x \) and \( f_y \) geometrically?
3. What are the higher-order partial derivatives for this function?
4. How do partial derivatives apply to optimization problems in calculus?
5. What is the difference between a total derivative and a partial derivative?

### Tip:
When finding partial derivatives, always treat all other variables as constants except for the one you're differentiating with respect to.

Given f(x, y) = x^2 + x^2y^3 - 2y^2. Evaluate fx, fy.

Learn how to evaluate the partial derivatives of the function f(x, y) = x^2 + x^2y^3 - 2y^2 with respect to x and y. This problem covers key concepts in multivariable calculus, including the use of partial derivatives and the chain rule.

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