To evaluate \(\frac{dy}{dx}\) for the equation \(x + y + xy = 1\), we will differentiate the equation implicitly with respect to \(x\).

Given:
\[
x + y + xy = 1
\]

### Step 1: Differentiate each term with respect to \(x\).
- The derivative of \(x\) with respect to \(x\) is \(1\).
- The derivative of \(y\) with respect to \(x\) is \(\frac{dy}{dx}\).
- The derivative of \(xy\) with respect to \(x\) requires the product rule: \(\frac{d}{dx}(xy) = y \cdot \frac{d}{dx}(x) + x \cdot \frac{d}{dx}(y)\).

So, differentiating both sides with respect to \(x\):
\[
\frac{d}{dx}(x) + \frac{d}{dx}(y) + \frac{d}{dx}(xy) = \frac{d}{dx}(1)
\]

### Step 2: Apply the derivatives.
\[
1 + \frac{dy}{dx} + (y + x\frac{dy}{dx}) = 0
\]

### Step 3: Simplify the equation.
Combine like terms:
\[
1 + \frac{dy}{dx} + y + x\frac{dy}{dx} = 0
\]
\[
\frac{dy}{dx}(1 + x) + (1 + y) = 0
\]

### Step 4: Solve for \(\frac{dy}{dx}\).
\[
\frac{dy}{dx}(1 + x) = -(1 + y)
\]
\[
\frac{dy}{dx} = \frac{-(1 + y)}{1 + x}
\]

Thus, the derivative \(\frac{dy}{dx}\) is:
\[
\frac{dy}{dx} = \frac{-(1 + y)}{1 + x}
\]

Would you like to explore this further or have any questions?

### Related Questions:
1. How do you apply the product rule in differentiation?
2. Can you solve a similar implicit differentiation problem for a different equation?
3. What are some common applications of implicit differentiation?
4. How would the solution change if the equation were \(x + y + xy = c\), where \(c\) is a constant?
5. Can you explain the geometric interpretation of \(\frac{dy}{dx}\) in this context?

### Tip:
Implicit differentiation is especially useful when the equation cannot be easily solved for \(y\) as a function of \(x\).

Learn how to evaluate the derivative dy/dx for the implicit equation x + y + xy = 1. This problem involves implicit differentiation and the product rule for differentiation, suitable for advanced high school students.

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