Calculate 
 
∂f
∂x
, 
∂f
∂y
, 
∂f
∂x
	 
(1, −1)
, and 
∂f
∂y
	 
(1, −1)
 
 when defined. (If an answer is undefined, enter UNDEFINED.)
f(x, y) = x−3y2 + xy2 + 2xy
∂f
∂x
=	
  
Correct: Your answer is correct.
 Perfect!
∂f
∂y
=	
  
Correct: Your answer is correct.
 You're right!
∂f
∂x
	 
(1, −1)
=	
-4
 
Correct: Your answer is correct.
 Fantastic work!
∂f
∂y
	 
(1, −1)
=	
 
Incorrect: Your answer is incorrect.

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.

Calculate ∂f/∂x, ∂f/∂y, ∂f/∂x at (1, −1), and ∂f/∂y at (1, −1) when defined. (If an answer is undefined, enter UNDEFINED.) f(x, y) = x−3y^2 + xy^2 + 2xy.

Learn how to calculate the partial derivatives of the function f(x, y) = x−3y² + xy² + 2xy with respect to x and y. This problem involves key multivariable calculus concepts such as partial differentiation and evaluating at specific points. Suitable for undergraduate-level students.

Math Problem Statement

when defined. (If an answer is undefined, enter UNDEFINED.) f(x, y) = x−3y2 + xy2 + 2xy ∂f ∂x

Correct: Your answer is correct. Perfect! ∂f ∂y

(1, −1)

(1, −1)

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