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The problem asks to show that for the function \( y = x^2 - x \), the derivative \( \frac{dy}{dx} \) satisfies the equation:

\[
\frac{dy}{dx} = 1 + \frac{2y}{x}.
\]

Let's go through the steps.

### Step 1: Differentiate \( y = x^2 - x \)
The function is given as:

\[
y = x^2 - x.
\]

Taking the derivative of \( y \) with respect to \( x \):

\[
\frac{dy}{dx} = \frac{d}{dx}(x^2 - x) = 2x - 1.
\]

### Step 2: Express \( y \) in terms of \( x \)
We know that:

\[
y = x^2 - x.
\]

We need to manipulate the equation to match the required form. First, solve for \( 2y \):

\[
2y = 2(x^2 - x) = 2x^2 - 2x.
\]

Now, divide by \( x \) to get \( \frac{2y}{x} \):

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

### Step 3: Show the desired relationship
We need to show:

\[
\frac{dy}{dx} = 1 + \frac{2y}{x}.
\]

From Step 1, we have \( \frac{dy}{dx} = 2x - 1 \), and from Step 2, we have \( \frac{2y}{x} = 2x - 2 \).

Now, substitute \( \frac{2y}{x} \) into the equation:

\[
1 + \frac{2y}{x} = 1 + (2x - 2) = 2x - 1.
\]

This matches the expression we found for \( \frac{dy}{dx} \), so the given equation is true.

Thus, we have shown that:

\[
\frac{dy}{dx} = 1 + \frac{2y}{x}.
\]

---

Would you like more details on any of the steps? Here are some related questions:

1. How do we apply the power rule in differentiation?
2. What is the purpose of differentiating implicit functions?
3. How would the derivative change if the function had additional terms?
4. Can we use another method (like implicit differentiation) to verify the result?
5. How does this problem relate to solving differential equations?

**Tip:** Always check the form of your final result and ensure it matches the required format in such proofs or verifications.

If y = x^2 - x, show that dy/dx = 1 + 2y/x.

In this problem, you will learn how to differentiate y = x^2 - x and show that dy/dx equals 1 + 2y/x. This step-by-step solution applies differentiation rules, including the power rule, and simplifies the equation for a clearer understanding.

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