We are given the function $f(x, y) = x - 3y^2 + xy^2 + 2xy$, and we need to compute the partial derivatives $\frac{\partial f}{\partial x}$, $\frac{\partial f}{\partial y}$, and then evaluate them at the point $(1, -1)$.

Step 1: Compute $\frac{\partial f}{\partial x}$

To find the partial derivative of $f(x, y)$ with respect to $x$, treat $y$ as a constant and differentiate with respect to $x$.

$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left( x - 3y^2 + xy^2 + 2xy \right)$

• The derivative of $x$ with respect to $x$ is $1$.
• The derivative of $-3y^2$ with respect to $x$ is $0$ (since it’s independent of $x$).
• The derivative of $xy^2$ with respect to $x$ is $y^2$ (since $y^2$ is treated as a constant).
• The derivative of $2xy$ with respect to $x$ is $2y$.

Thus, we have:

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

Step 2: Compute $\frac{\partial f}{\partial y}$

To find the partial derivative of $f(x, y)$ with respect to $y$, treat $x$ as a constant and differentiate with respect to $y$.

$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left( x - 3y^2 + xy^2 + 2xy \right)$

• The derivative of $x$ with respect to $y$ is $0$.
• The derivative of $-3y^2$ with respect to $y$ is $-6y$.
• The derivative of $xy^2$ with respect to $y$ is $2xy$.
• The derivative of $2xy$ with respect to $y$ is $2x$.

Thus, we have:

$\frac{\partial f}{\partial y} = -6y + 2xy + 2x$

Step 3: Evaluate $\frac{\partial f}{\partial x}$ at $(1, -1)$

Now we substitute $x = 1$ and $y = -1$ into the expression for $\frac{\partial f}{\partial x}$.

$\frac{\partial f}{\partial x}(1, -1) = 1 + (-1)^2 + 2(-1) = 1 + 1 - 2 = 0$

So, $\frac{\partial f}{\partial x}(1, -1) = 0$.

Step 4: Evaluate $\frac{\partial f}{\partial y}$ at $(1, -1)$

Now we substitute $x = 1$ and $y = -1$ into the expression for $\frac{\partial f}{\partial y}$.

$\frac{\partial f}{\partial y}(1, -1) = -6(-1) + 2(1)(-1) + 2(1) = 6 - 2 + 2 = 6$

So, $\frac{\partial f}{\partial y}(1, -1) = 6$.

Final Answers:

• $\frac{\partial f}{\partial x} = 1 + y^2 + 2y$
• $\frac{\partial f}{\partial y} = -6y + 2xy + 2x$
• $\frac{\partial f}{\partial x}(1, -1) = 0$
• $\frac{\partial f}{\partial y}(1, -1) = 6$

Would you like any further clarification or details?

Here are 5 related questions to expand on this:

1. What is the geometric interpretation of partial derivatives?
2. How do you apply the chain rule for partial derivatives in multivariable calculus?
3. How can we compute the second-order partial derivatives of $f(x, y)$?
4. How can partial derivatives be used in optimization problems, such as finding local maxima and minima?
5. Can you explain the significance of mixed partial derivatives?

Tip: Always double-check your substitution values when evaluating functions at specific points to avoid small errors.