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We are tasked with finding the equation of the tangent line to the curve \( y = (1 + 4x)^9 \) at the point \( (0, 1) \).

### Steps to solve:

1. **Find the derivative of the curve**:  
   The slope of the tangent line is given by the derivative of the function at the point. So, we need to compute the derivative \( \frac{dy}{dx} \).

   \[
   y = (1 + 4x)^9
   \]
   Using the chain rule, the derivative of \( y \) with respect to \( x \) is:
   \[
   \frac{dy}{dx} = 9(1 + 4x)^8 \cdot \frac{d}{dx}(1 + 4x)
   \]
   The derivative of \( 1 + 4x \) is 4, so:
   \[
   \frac{dy}{dx} = 9(1 + 4x)^8 \cdot 4
   \]
   \[
   \frac{dy}{dx} = 36(1 + 4x)^8
   \]

2. **Evaluate the derivative at the point \( (0, 1) \)**:  
   Now, substitute \( x = 0 \) into the derivative:
   \[
   \frac{dy}{dx} \bigg|_{x = 0} = 36(1 + 4(0))^8 = 36 \times 1^8 = 36
   \]
   So, the slope of the tangent line at \( (0, 1) \) is 36.

3. **Use the point-slope form of the equation of a line**:  
   The point-slope form of a line is:
   \[
   y - y_1 = m(x - x_1)
   \]
   where \( m \) is the slope and \( (x_1, y_1) \) is the point on the line. Here, \( m = 36 \), \( x_1 = 0 \), and \( y_1 = 1 \). Thus, the equation becomes:
   \[
   y - 1 = 36(x - 0)
   \]
   Simplifying:
   \[
   y - 1 = 36x
   \]
   \[
   y = 36x + 1
   \]

### Final Answer:
The equation of the tangent line to the curve at the point \( (0, 1) \) is:
\[
y = 36x + 1
\]

Would you like further details or clarifications?

Here are five related questions to explore:
1. What happens to the tangent line equation if the point is shifted to a different point on the curve?
2. How does the chain rule apply to functions raised to high powers, like in this problem?
3. What role does the point-slope form play in finding the tangent line?
4. How can we use the derivative to analyze the behavior of the curve near \( x = 0 \)?
5. What is the geometrical interpretation of the derivative at \( x = 0 \)?

**Tip**: The chain rule is especially useful for functions that involve compositions, such as powers of expressions. Make sure to differentiate both the outer and inner functions properly.

Find an equation of the tangent line to the curve at the given point: y = (1 + 4x)^9, (0, 1).

Learn how to find the equation of the tangent line to the curve y = (1 + 4x)^9 at the point (0, 1). This problem involves calculus concepts such as derivatives, chain rule, and the point-slope form of a line. Suitable for Grades 11-12.

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