To find the second derivative \( y'' \) given that \( \frac{dy}{dx} = y' = 1 - 2xy \), we'll differentiate the first derivative with respect to \( x \).

Given:
\[
y' = 1 - 2xy
\]

### Step 1: Differentiate both sides with respect to \( x \).
We apply the product rule to differentiate the term \( -2xy \):

\[
y'' = \frac{d}{dx}(1) - 2 \cdot \frac{d}{dx}(xy)
\]

Since the derivative of a constant (1) is 0:

\[
y'' = 0 - 2 \cdot \left( \frac{d}{dx}(x \cdot y) \right)
\]

### Step 2: Apply the product rule to \( \frac{d}{dx}(xy) \).
The product rule states:

\[
\frac{d}{dx}(xy) = \frac{dx}{dx} \cdot y + x \cdot \frac{dy}{dx} 
\]

Substituting into the equation:

\[
y'' = -2 \cdot (1 \cdot y + x \cdot y')
\]

Simplifying:

\[
y'' = -2y - 2x \cdot y'
\]

### Step 3: Substitute \( y' = 1 - 2xy \) back into the equation:

\[
y'' = -2y - 2x \cdot (1 - 2xy)
\]

Expanding the equation:

\[
y'' = -2y - 2x + 4x^2y
\]

### Final Answer:
\[
y'' = 4x^2y - 2y - 2x
\]

This is the second derivative \( y'' \) in terms of \( x \) and \( y \).

Would you like further details on any step or have any other questions?

Here are 5 related questions to consider:
1. How does the product rule apply to differentiate products of functions?
2. What is the geometric interpretation of the second derivative \( y'' \)?
3. How would \( y'' \) change if the original differential equation had a different form?
4. How do you determine if the function \( y(x) \) is concave up or down using \( y'' \)?
5. How can implicit differentiation be applied in different contexts?

**Tip:** Always check if any simplifications can be made after differentiating, especially when dealing with products or quotients.

Learn how to find the second derivative y'' = 4x^2y - 2y - 2x step-by-step from the given differential equation dy/dx = y' = 1 - 2xy. This problem involves differential calculus, product rule, and algebraic manipulation. Suitable for advanced high school and undergraduate students.

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