Let's compute the partial derivatives of the given function $f(x, y) = 2x^2 - xy + xy^2$ with respect to $x$ and $y$.

Partial Derivative with respect to $x$:

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

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

Differentiating term by term:

1. $\frac{\partial}{\partial x}(2x^2) = 4x$
2. $\frac{\partial}{\partial x}(-xy) = -y$ (since $y$ is constant with respect to $x$)
3. $\frac{\partial}{\partial x}(xy^2) = y^2$ (since $y^2$ is treated as constant with respect to $x$)

Thus, the partial derivative with respect to $x$ is:

$\frac{\partial f}{\partial x} = 4x - y + y^2$

Partial Derivative with respect to $y$:

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

1. $\frac{\partial}{\partial y}(2x^2) = 0$ (since $x^2$ does not depend on $y$)
2. $\frac{\partial}{\partial y}(-xy) = -x$
3. $\frac{\partial}{\partial y}(xy^2) = 2xy$ (using the power rule)

Thus, the partial derivative with respect to $y$ is:

$\frac{\partial f}{\partial y} = -x + 2xy$

Final Answer:

• $\frac{\partial f}{\partial x} = 4x - y + y^2$
• $\frac{\partial f}{\partial y} = -x + 2xy$

Would you like to explore further details or have any questions about these derivatives?

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Here are 5 related questions to expand your understanding:

1. How do partial derivatives apply in multivariable optimization problems?
2. What is the concept of mixed partial derivatives, and how are they computed?
3. How do partial derivatives relate to the gradient of a function?
4. What is the meaning of the total derivative in the context of multivariable functions?
5. How can the chain rule be used with multivariable functions to compute derivatives?

Tip: Partial derivatives measure how a function changes as one variable changes while the others are held constant. This is important when working with functions of several variables!