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To find the equation of the tangent line to the function \( y = \left(\frac{x + 1}{x - 3}\right)^2 \) at \( x = 4 \), we need to follow these steps:

1. **Find the derivative of \( y \)**: This will give us the slope of the tangent at any point \( x \).
2. **Evaluate the derivative at \( x = 4 \)**: This gives the slope of the tangent line at that point.
3. **Find \( y \) at \( x = 4 \)**: This will provide the point through which the tangent line passes.
4. **Use the point-slope form of the line**: \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) \) is the point of tangency.

Let's proceed with these steps.

### Step 1: Find the Derivative of \( y \)
Let \( y = \left(\frac{x + 1}{x - 3}\right)^2 \). To differentiate this, use the chain rule:
\[
\frac{dy}{dx} = 2 \left(\frac{x + 1}{x - 3}\right) \cdot \frac{d}{dx}\left(\frac{x + 1}{x - 3}\right).
\]
Now, find the derivative of \( \frac{x + 1}{x - 3} \) using the quotient rule:
\[
\frac{d}{dx}\left(\frac{x + 1}{x - 3}\right) = \frac{(x - 3) \cdot 1 - (x + 1) \cdot 1}{(x - 3)^2} = \frac{-4}{(x - 3)^2}.
\]

Substitute this back into the derivative:
\[
\frac{dy}{dx} = 2 \left(\frac{x + 1}{x - 3}\right) \cdot \frac{-4}{(x - 3)^2} = -\frac{8(x + 1)}{(x - 3)^3}.
\]

### Step 2: Evaluate the Derivative at \( x = 4 \)
Substitute \( x = 4 \) into the derivative:
\[
\frac{dy}{dx} \bigg|_{x=4} = -\frac{8(4 + 1)}{(4 - 3)^3} = -\frac{8 \cdot 5}{1} = -40.
\]

So, the slope of the tangent at \( x = 4 \) is \( -40 \).

### Step 3: Find \( y \) at \( x = 4 \)
Substitute \( x = 4 \) into the original function:
\[
y = \left(\frac{4 + 1}{4 - 3}\right)^2 = \left(\frac{5}{1}\right)^2 = 25.
\]

So, the point of tangency is \( (4, 25) \).

### Step 4: Find the Equation of the Tangent Line
Using the point-slope form \( y - y_1 = m(x - x_1) \):
\[
y - 25 = -40(x - 4).
\]
Simplify this:
\[
y - 25 = -40x + 160 \implies y = -40x + 185.
\]

### Final Answer
The equation of the tangent line to \( y = \left(\frac{x + 1}{x - 3}\right)^2 \) at \( x = 4 \) is:
\[
\boxed{y = -40x + 185}.
\]

Would you like further details on any of these steps or have any other questions?

Here are 5 related questions:
1. How do you use the chain rule in calculus?
2. What is the quotient rule, and how does it apply to derivatives?
3. How can you find the tangent line to a curve at a point without using derivatives?
4. What is the geometric interpretation of a derivative?
5. Why is the point-slope form useful for writing equations of lines?

**Tip:** When differentiating composite functions, always remember to apply the chain rule. It ensures that every component of the function is accounted for in the differentiation process.

Find the tangent to y = ((x + 1)/(x - 3))^2 at x = 4.

Learn how to find the tangent line to y = ((x + 1)/(x - 3))^2 at x = 4 using the derivative, applying the quotient and chain rule. This problem includes detailed steps for solving with calculus, suitable for Grades 11-12 students.

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