Given the function $f(x, y) = x + x^2y^2 + y$, let's compute the partial derivatives with respect to $x$ and $y$.

(i) Partial derivative of $f(x, y)$ with respect to $x$:

To find $\frac{\partial f}{\partial x}$, we treat $y$ as a constant and differentiate the function with respect to $x$:

$f(x, y) = x + x^2y^2 + y$

Now, differentiate each term:

1. $\frac{\partial}{\partial x} (x) = 1$
2. $\frac{\partial}{\partial x} (x^2y^2) = 2xy^2$ (since $y^2$ is treated as a constant)
3. $\frac{\partial}{\partial x} (y) = 0$ (since $y$ is independent of $x$)

So, the partial derivative of $f$ with respect to $x$ is:

$\frac{\partial f}{\partial x} = 1 + 2xy^2$

(ii) Partial derivative of $f(x, y)$ with respect to $y$:

To find $\frac{\partial f}{\partial y}$, we treat $x$ as a constant and differentiate the function with respect to $y$:

$f(x, y) = x + x^2y^2 + y$

Now, differentiate each term:

1. $\frac{\partial}{\partial y} (x) = 0$ (since $x$ is independent of $y$)
2. $\frac{\partial}{\partial y} (x^2y^2) = 2x^2y$ (since $x^2$ is treated as a constant)
3. $\frac{\partial}{\partial y} (y) = 1$

So, the partial derivative of $f$ with respect to $y$ is:

$\frac{\partial f}{\partial y} = 2x^2y + 1$

Summary:

• $\frac{\partial f}{\partial x} = 1 + 2xy^2$
• $\frac{\partial f}{\partial y} = 2x^2y + 1$

Would you like more details or have any further questions? Here are some related questions:

1. What is the second-order partial derivative of $f(x, y)$ with respect to $x$?
2. What is the second-order partial derivative of $f(x, y)$ with respect to $y$?
3. Can you compute the mixed partial derivative $\frac{\partial^2 f}{\partial x \partial y}$?
4. How does the total derivative of $f(x, y)$ behave if $x$ and $y$ are functions of another variable $t$?
5. Can you find the critical points of $f(x, y)$ using the first derivatives?

Tip: To check for critical points of a function of two variables, you need to set the first-order partial derivatives equal to zero and solve the system of equations.