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Let's solve each part of this question.

### Part (a)
We are given:
\[
y = (2x + 1)(2x^2 - x - 1)
\]
We need to find \(\frac{dy}{dx}\). We can use the **product rule** for derivatives, which states:
\[
\frac{d}{dx}[u \cdot v] = u'v + uv'
\]
Here, let:
- \( u = 2x + 1 \)
- \( v = 2x^2 - x - 1 \)

#### Step 1: Differentiate \( u \) and \( v \)
- \( u' = \frac{d}{dx}(2x + 1) = 2 \)
- \( v' = \frac{d}{dx}(2x^2 - x - 1) = 4x - 1 \)

#### Step 2: Apply the product rule
\[
\frac{dy}{dx} = u'v + uv' = 2(2x^2 - x - 1) + (2x + 1)(4x - 1)
\]

#### Step 3: Simplify the expression
First, expand both terms:
\[
2(2x^2 - x - 1) = 4x^2 - 2x - 2
\]
\[
(2x + 1)(4x - 1) = 8x^2 - 2x + 4x - 1 = 8x^2 + 2x - 1
\]

Now, combine both expressions:
\[
\frac{dy}{dx} = (4x^2 - 2x - 2) + (8x^2 + 2x - 1)
\]
\[
\frac{dy}{dx} = 12x^2 - 3
\]

Thus, the derivative is:
\[
\frac{dy}{dx} = 12x^2 - 3
\]

### Part (b)
We are given:
\[
y = \frac{u}{u + 1}, \quad \text{where} \quad u = 3x^2 - 1
\]
We need to find \(\frac{dy}{dx}\). This requires applying the **chain rule** and the **quotient rule**.

#### Step 1: Differentiate \( y = \frac{u}{u + 1} \)
Use the **quotient rule**, which states:
\[
\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}
\]
Here, let:
- \( f(u) = u \)
- \( g(u) = u + 1 \)

The derivatives are:
- \( f'(u) = \frac{d}{dx}(u) = \frac{du}{dx} = 6x \) (since \( u = 3x^2 - 1 \))
- \( g'(u) = \frac{d}{dx}(u + 1) = 6x \)

#### Step 2: Apply the quotient rule
\[
\frac{dy}{dx} = \frac{(6x)(u + 1) - u(6x)}{(u + 1)^2}
\]

#### Step 3: Simplify the expression
Substitute \( u = 3x^2 - 1 \):
\[
\frac{dy}{dx} = \frac{6x((3x^2 - 1) + 1) - 6x(3x^2 - 1)}{(3x^2 - 1 + 1)^2}
\]
Simplify:
\[
\frac{dy}{dx} = \frac{6x(3x^2) - 6x(3x^2 - 1)}{(3x^2)^2} = \frac{18x^3 - (18x^3 - 6x)}{9x^4}
\]
\[
\frac{dy}{dx} = \frac{6x}{9x^4} = \frac{2}{3x^3}
\]

Thus, the derivative is:
\[
\frac{dy}{dx} = \frac{2}{3x^3}
\]

---

Would you like further details on any of these steps or have additional questions? Here are some related problems for practice:

1. Use the product rule to find \(\frac{d}{dx}[(3x + 2)(x^3 - x)]\).
2. Apply the chain rule to differentiate \( y = (5x^2 - 1)^3 \).
3. Find \(\frac{dy}{dx}\) for \( y = \frac{x^2}{x + 3} \) using the quotient rule.
4. Differentiate \( y = \frac{1}{x^2 + 1} \) with respect to \(x\).
5. Use the product rule to differentiate \( y = (x^2 - 3x)(4x + 1) \).

**Tip:** When differentiating products or quotients of functions, always break them down into manageable parts and apply the appropriate rules step by step.

Find dy/dx for the given functions:
(a) If y = (2x + 1)(2x^2 - x - 1)
(b) If y = u/(u+1) where u = 3x^2 - 1

Learn how to apply the product and quotient rules to differentiate the functions y = (2x + 1)(2x^2 - x - 1) and y = u/(u+1) where u = 3x^2 - 1. Step-by-step solution provided, covering key differentiation techniques like the product, quotient, and chain rules. Suitable for Grades 11-12 or early college students.

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