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To find the slope of the tangent line to the curve given by the equation

\[
2xy^2 + 2xy + x^2 y = 12
\]

at the point \((x, y) = (2, 1)\), we need to take the derivative of the equation with respect to \(x\) implicitly.

### Step 1: Differentiate Both Sides with Respect to \(x\)
We start by differentiating each term:

1. **Differentiate \(2xy^2\):**
   \[
   \frac{d}{dx}(2xy^2) = 2\left(y^2 + x \cdot 2y \frac{dy}{dx}\right) = 2y^2 + 4xy \frac{dy}{dx}
   \]

2. **Differentiate \(2xy\):**
   \[
   \frac{d}{dx}(2xy) = 2\left(y + x \frac{dy}{dx}\right) = 2y + 2x \frac{dy}{dx}
   \]

3. **Differentiate \(x^2 y\):**
   \[
   \frac{d}{dx}(x^2 y) = 2x y + x^2 \frac{dy}{dx}
   \]

So, differentiating the entire equation yields:
\[
2y^2 + 4xy \frac{dy}{dx} + 2y + 2x \frac{dy}{dx} + 2x y + x^2 \frac{dy}{dx} = 0
\]

### Step 2: Substitute \(x = 2\) and \(y = 1\) into the Equation
Now, substitute \(x = 2\) and \(y = 1\) to simplify:

1. \(2y^2 = 2 \cdot (1)^2 = 2\)
2. \(4xy \frac{dy}{dx} = 4 \cdot 2 \cdot 1 \cdot \frac{dy}{dx} = 8 \frac{dy}{dx}\)
3. \(2y = 2 \cdot 1 = 2\)
4. \(2x \frac{dy}{dx} = 2 \cdot 2 \cdot \frac{dy}{dx} = 4 \frac{dy}{dx}\)
5. \(2xy = 2 \cdot 2 \cdot 1 = 4\)
6. \(x^2 \frac{dy}{dx} = (2)^2 \frac{dy}{dx} = 4 \frac{dy}{dx}\)

Substituting these values into the differentiated equation, we get:

\[
2 + 8 \frac{dy}{dx} + 2 + 4 \frac{dy}{dx} + 4 + 4 \frac{dy}{dx} = 0
\]

Simplifying:

\[
8 + 16 \frac{dy}{dx} = 0
\]

### Step 3: Solve for \(\frac{dy}{dx}\)

\[
16 \frac{dy}{dx} = -8
\]

\[
\frac{dy}{dx} = -\frac{1}{2}
\]

### Answer
The slope of the tangent line at \((2, 1)\) is \(-\frac{1}{2}\).

### Correct Option
The correct answer is:
\[
\boxed{-\frac{1}{2}}
\]

Would you like a more detailed explanation on any step?

### Related Questions
1. How do you determine the slope of a tangent line using implicit differentiation?
2. What is the purpose of implicit differentiation in multivariable calculus?
3. How would the solution change if we needed the slope at a different point on the curve?
4. What are the practical applications of finding tangent slopes in physics and engineering?
5. Can we use this technique to find tangent slopes for higher-order polynomial equations?

### Tip
When differentiating implicitly, always remember to apply the chain rule for terms involving both \(x\) and \(y\) carefully to avoid errors.

For the equation: 2xy^2 + 2xy + x^2y = 12, what is the slope of the tangent line at (x, y) = (2, 1)?

Find the slope of the tangent line to the curve defined by 2xy^2 + 2xy + x^2y = 12 at the point (2, 1) using implicit differentiation. Learn how to apply the chain rule and calculate the derivative step-by-step.

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