To solve for \(d''(3)\) where \(d(x) = \frac{x}{x - 2}\), we need to find the second derivative of the function and then evaluate it at \(x = 3\).

### Step 1: Simplify the Function

Given the function:
\[
d(x) = \frac{x}{x - 2}
\]

### Step 2: First Derivative

To find the first derivative \(d'(x)\), we can use the quotient rule:
\[
\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}
\]
where \(u = x\) and \(v = x - 2\).

Calculate \(u'\) and \(v'\):
- \(u' = 1\)
- \(v' = 1\)

Apply the quotient rule:
\[
d'(x) = \frac{(1)(x - 2) - (x)(1)}{(x - 2)^2} = \frac{x - 2 - x}{(x - 2)^2} = \frac{-2}{(x - 2)^2}
\]

### Step 3: Second Derivative

Now, find the second derivative \(d''(x)\) by differentiating \(d'(x)\):
\[
d'(x) = \frac{-2}{(x - 2)^2}
\]

Rewrite \(d'(x)\) as:
\[
d'(x) = -2(x - 2)^{-2}
\]

Differentiate using the chain rule:
\[
d''(x) = -2 \times (-2)(x - 2)^{-3} \cdot (1) = 4(x - 2)^{-3}
\]
\[
d''(x) = \frac{4}{(x - 2)^3}
\]

### Step 4: Evaluate \(d''(3)\)

Substitute \(x = 3\) into \(d''(x)\):
\[
d''(3) = \frac{4}{(3 - 2)^3} = \frac{4}{1^3} = 4
\]

### Final Answer

\[
d''(3) = 4
\]

Would you like more details, or do you have any questions?

Here are 5 questions that extend from this problem:

1. What is the general formula for the nth derivative of \(d(x) = \frac{x}{x-2}\)?
2. How does the quotient rule compare to the product rule in terms of application and complexity?
3. What happens to \(d(x) = \frac{x}{x-2}\) as \(x\) approaches 2?
4. Can you find the critical points and analyze the concavity of \(d(x) = \frac{x}{x-2}\)?
5. How would you approach finding the Taylor series expansion of \(d(x) = \frac{x}{x-2}\) around a given point?

**Tip:** When dealing with rational functions, always consider simplifying the function first and checking for any potential division by zero or undefined points.

Learn how to find the second derivative of the rational function d(x) = x/(x - 2) and evaluate it at x = 3. This solution covers the application of the quotient rule and chain rule in calculus. Suitable for students in Grades 11-12.

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