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.