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.