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Let's go through both problems in detail.

### Problem 36:
**Given function: \( y = (2x + 1)^2 \)**

**A. Use the Power Rule to find the slope formula.**

Using the Power Rule, where \(\frac{d}{dx}[u^n] = n \cdot u^{n-1} \cdot \frac{du}{dx}\):

1. Here, \(u = 2x + 1\) and \(n = 2\).
2. Derivative of \(u\) is \(\frac{du}{dx} = 2\).
3. Applying the Power Rule:

\[
\frac{dy}{dx} = 2 \cdot (2x + 1) \cdot 2 = 4(2x + 1)
\]

So, the slope formula is:

\[
\frac{dy}{dx} = 4(2x + 1)
\]

**B. Simplify \((2x + 1)^2\) by squaring the binomial.**

Expanding the square:

\[
(2x + 1)^2 = 4x^2 + 4x + 1
\]

**C. Use the simplified expression from Part B to find the slope formula.**

1. \( y = 4x^2 + 4x + 1 \)
2. Derivative:

\[
\frac{dy}{dx} = 8x + 4
\]

**D. Compare answers from A and C and determine the correct one.**

The answers from Parts A and C are different:

- Part A: \( \frac{dy}{dx} = 4(2x + 1) = 8x + 4 \).
- Part C: \( \frac{dy}{dx} = 8x + 4 \).

They are actually the same. The reason they appear different is because Part A is derived using the chain rule directly on the original function, and Part C simplifies the function first and then differentiates. Both are valid, so they agree in their result.

### Problem 37:
**Given function: \( y = \frac{\sqrt[3]{x}}{x^5} \) and \( x < 0 \)**.

**Prove \( y > 0 \).**

1. **Simplify the function:**

\[
y = \frac{x^{1/3}}{x^5} = x^{1/3 - 5} = x^{-14/3}
\]

2. Since \( x < 0 \) (always negative), \( x^{-14/3} \) would be positive because the power \(-14/3\) is an even value (\(-14\) is even).

3. Therefore, \( y = x^{-14/3} > 0 \).

So, the function \( y \) is always positive for \( x < 0 \).

---

Would you like to go into further detail on any of these solutions? Here are some related questions for further exploration:

1. How would the slope formula change if the function in Problem 36 was \( (3x + 1)^2 \)?
2. Can you provide an example of a function where the answers from parts A and C would not match?
3. What would happen to the slope if \( x \) were negative in Problem 36?
4. Can you find the second derivative for the function in Problem 37?
5. How would you interpret the meaning of the derivative obtained in Part A of Problem 36?

**Tip:** Understanding the process of differentiation from both the chain rule and the expanded form helps in verifying complex derivatives effectively.

Consider the function y = (2x + 1)^2. Use the Power Rule to see what the slope formula might be. Simplify (2x + 1)^2 by squaring the binomial. Compare answers from A and C. Consider the function y = cube root of x / x^5. Prove that y > 0 when x < 0.

This problem set explores the use of the Power Rule and binomial expansion to find the derivative of y = (2x+1)^2. It also proves that y > 0 for x < 0 in the function y = cube root of x / x^5. Suitable for advanced calculus students.

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