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Find an equation of the tangent line to the given curve at the specified point.

To find the equation of the tangent line to the curve \( y = \frac{1 + x}{1 + e^x} \) at the point \( (0, \frac{1}{2}) \), we need to follow these steps:

### Step 1: Find the derivative \( y' \) (the slope of the tangent line).
The equation for the curve is a quotient, so we can apply the **quotient rule** for derivatives:
\[
y = \frac{1 + x}{1 + e^x}
\]
The quotient rule states:
\[
\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}
\]
where \( u = 1 + x \) and \( v = 1 + e^x \).

Now, let's differentiate each part:

- \( u = 1 + x \), so \( u' = 1 \).
- \( v = 1 + e^x \), so \( v' = e^x \).

Using the quotient rule, we get:
\[
y' = \frac{(1)(1 + e^x) - (1 + x)(e^x)}{(1 + e^x)^2}
\]
Simplifying the numerator:
\[
y' = \frac{1 + e^x - e^x - x e^x}{(1 + e^x)^2} = \frac{1 - x e^x}{(1 + e^x)^2}
\]

### Step 2: Evaluate the derivative at \( x = 0 \).
We substitute \( x = 0 \) into the derivative:
\[
y'(0) = \frac{1 - 0 \cdot e^0}{(1 + e^0)^2} = \frac{1}{(1 + 1)^2} = \frac{1}{4}
\]
Thus, the slope of the tangent line at \( (0, \frac{1}{2}) \) is \( \frac{1}{4} \).

### Step 3: Use the point-slope form to find the equation of the tangent line.
The point-slope form of a line is given by:
\[
y - y_1 = m(x - x_1)
\]
where \( m \) is the slope, and \( (x_1, y_1) \) is the point. We know \( m = \frac{1}{4} \), and the point is \( (0, \frac{1}{2}) \). Substituting these values:
\[
y - \frac{1}{2} = \frac{1}{4}(x - 0)
\]
Simplifying:
\[
y - \frac{1}{2} = \frac{1}{4}x
\]
\[
y = \frac{1}{4}x + \frac{1}{2}
\]

### Final Answer:
The equation of the tangent line to the curve at \( (0, \frac{1}{2}) \) is:
\[
y = \frac{1}{4}x + \frac{1}{2}
\]

Would you like more details or have any further questions? Here are a few related questions:

1. How would the derivative change if the function had a different denominator?
2. What if the point was at \( x = 1 \)? How would the equation of the tangent line differ?
3. Can you explain why the point-slope form is used here?
4. How does the quotient rule apply when there are more complex terms in both numerator and denominator?
5. Could you solve for the tangent line using implicit differentiation in a different case?

**Tip:** When applying the quotient rule, always simplify the numerator before plugging in values to make the calculation easier.

Find an equation of the tangent line to the given curve at the specified point: y = (1 + x) / (1 + e^x), at the point (0, 1/2).

Learn how to find the equation of the tangent line to the curve y = (1 + x) / (1 + e^x) at the point (0, 1/2). This solution covers the quotient rule for differentiation and the point-slope form of a line. Suitable for students in Grades 11-12.

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