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The question asks to find the value of \( y' + \frac{dy}{dx} \) given that \( xy = 1 \).

Let's solve it step by step.

### Step 1: Differentiate \( xy = 1 \)

We have the equation:

\[
xy = 1
\]

To find \( y' \) (i.e., \( \frac{dy}{dx} \)), we need to apply implicit differentiation to both sides with respect to \( x \):

\[
\frac{d}{dx}(xy) = \frac{d}{dx}(1)
\]

Using the product rule on the left-hand side:

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

Simplifying:

\[
x \frac{dy}{dx} + y = 0
\]

### Step 2: Solve for \( \frac{dy}{dx} \)

Rearranging the equation:

\[
x \frac{dy}{dx} = -y
\]

\[
\frac{dy}{dx} = \frac{-y}{x}
\]

### Step 3: Substitute \( \frac{dy}{dx} \) into \( y' + \frac{dy}{dx} \)

The expression \( y' + \frac{dy}{dx} \) simplifies to:

\[
y' + \frac{dy}{dx} = \frac{dy}{dx} + \frac{dy}{dx}
\]

From step 2, we know that \( \frac{dy}{dx} = \frac{-y}{x} \), so:

\[
y' + \frac{dy}{dx} = 2 \times \frac{-y}{x} = \frac{-2y}{x}
\]

### Step 4: Use \( xy = 1 \) to simplify further

From the original equation \( xy = 1 \), we know that:

\[
y = \frac{1}{x}
\]

Substitute \( y = \frac{1}{x} \) into \( \frac{-2y}{x} \):

\[
\frac{-2y}{x} = \frac{-2 \times \frac{1}{x}}{x} = \frac{-2}{x^2}
\]

Thus, \( y' + \frac{dy}{dx} = \frac{-2}{x^2} \).

Since no specific value for \( x \) is given, and none of the options match this general expression, the correct answer is **d) none of these**.

---

Would you like more details on any part of the solution? Here are 5 related questions you might explore:

1. How do we apply implicit differentiation to more complex equations?
2. What is the product rule, and when do we use it?
3. How can we interpret the result \( \frac{-2}{x^2} \) geometrically?
4. What other techniques exist to solve similar implicit differentiation problems?
5. Why doesn't the expression simplify to one of the given options?

**Tip**: When dealing with implicit differentiation, always remember to apply the product rule when variables are multiplied.

Learn how to solve the problem of finding y' + dy/dx given the equation xy = 1. This solution involves using implicit differentiation and applying the product rule, suitable for students in Grades 11-12.

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