Let's solve each of these problems step by step using the chain rule where needed.

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1. $y = \frac{(3x^2 - 2)^2}{x^3}$

We will apply the quotient rule and chain rule here.

The quotient rule states:

$\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{[v(x)]^2}$

Let:

• $u(x) = (3x^2 - 2)^2$
• $v(x) = x^3$

Step 1: Differentiate $u(x) = (3x^2 - 2)^2$

Apply the chain rule:

$u'(x) = 2(3x^2 - 2) \cdot \frac{d}{dx}(3x^2 - 2)$

Differentiate $3x^2 - 2$:

$\frac{d}{dx}(3x^2 - 2) = 6x$

So:

$u'(x) = 2(3x^2 - 2) \cdot 6x = 12x(3x^2 - 2)$

Step 2: Differentiate $v(x) = x^3$

$v'(x) = 3x^2$

Step 3: Apply the quotient rule

Now, applying the quotient rule:

$y'(x) = \frac{x^3 \cdot 12x(3x^2 - 2) - (3x^2 - 2)^2 \cdot 3x^2}{(x^3)^2}$

Simplifying:

$y'(x) = \frac{12x^4(3x^2 - 2) - 3x^2(3x^2 - 2)^2}{x^6}$

Now, you can expand and simplify this further if needed.

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2. $y = \frac{x^2}{(x^2 - 1)(x - 1)}$

Here, we again apply the quotient rule.

Let:

• $u(x) = x^2$
• $v(x) = (x^2 - 1)(x - 1)$

Step 1: Differentiate $u(x) = x^2$

$u'(x) = 2x$

Step 2: Differentiate $v(x) = (x^2 - 1)(x - 1)$

We need to apply the product rule here.

Let:

• $p(x) = x^2 - 1$
• $q(x) = x - 1$

The product rule states:

$\frac{d}{dx} [p(x) \cdot q(x)] = p'(x) \cdot q(x) + p(x) \cdot q'(x)$

First, find the derivatives:

$p'(x) = 2x \quad \text{and} \quad q'(x) = 1$

Now, apply the product rule:

$v'(x) = 2x(x - 1) + (x^2 - 1)(1)$

Simplifying:

$v'(x) = 2x(x - 1) + (x^2 - 1)$

$v'(x) = 2x^2 - 2x + x^2 - 1 = 3x^2 - 2x - 1$

Step 3: Apply the quotient rule

Now, applying the quotient rule:

$y'(x) = \frac{(x^2)(3x^2 - 2x - 1) - (x^2 - 1)(x - 1)(2x)}{[(x^2 - 1)(x - 1)]^2}$

This is the derivative. You can expand and simplify if needed.

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Summary of steps:

1. Use the quotient rule for both problems.
2. Apply the chain rule when differentiating composite functions (like $(3x^2 - 2)^2$).
3. Apply the product rule where necessary.

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Would you like me to expand these derivatives further or provide more details?