Let’s find the first derivatives of each of these functions one by one.

Problem a)

Function: $y = \ln(x^2 - x)$

Solution: To differentiate this, we use the chain rule. The derivative of $\ln(u)$ is $\frac{1}{u} \cdot \frac{du}{dx}$.

1. Let $u = x^2 - x$.
2. Then $\frac{du}{dx} = 2x - 1$.
3. So, $y' = \frac{1}{x^2 - x} \cdot (2x - 1) = \frac{2x - 1}{x^2 - x}$.

Thus, the first derivative is: $y' = \frac{2x - 1}{x^2 - x}$

Problem b)

Function: $p(a) = \log_3(a^2 - 1)$

Solution: For this problem, we use the change of base formula along with the chain rule. The derivative of $\log_b(u)$ is $\frac{1}{u \ln(b)} \cdot \frac{du}{da}$.

1. Let $u = a^2 - 1$.
2. Then $\frac{du}{da} = 2a$.
3. So, $p'(a) = \frac{1}{(a^2 - 1) \ln(3)} \cdot 2a = \frac{2a}{(a^2 - 1) \ln(3)}$.

Thus, the first derivative is: $p'(a) = \frac{2a}{(a^2 - 1) \ln(3)}$

Problem c)

Function: $h(z) = z^2 - \ln^4(1 - z^2)$

Solution: This requires the derivative of $z^2$ and the derivative of $\ln^4(1 - z^2)$, using the chain rule.

1. The derivative of $z^2$ is $2z$.
2. For the $\ln^4(1 - z^2)$ term, let $u = \ln(1 - z^2)$, so $\ln^4(1 - z^2) = u^4$.
   • Differentiate $u^4$: $4u^3 \cdot \frac{du}{dz}$.
   • $\frac{du}{dz} = \frac{-2z}{1 - z^2}$, because $u = \ln(1 - z^2)$.

Therefore, $h'(z) = 2z - 4 \left(\ln(1 - z^2)\right)^3 \cdot \frac{-2z}{1 - z^2}$ $= 2z + \frac{8z \left(\ln(1 - z^2)\right)^3}{1 - z^2}$

Problem d)

Function: $y = \frac{(6x^2 - 3x + 1)^{\frac{3}{2}}}{\sqrt{x^2 - 3x}}$

Solution: To differentiate this, we can use the quotient rule and chain rule.

1. Let $u = (6x^2 - 3x + 1)^{\frac{3}{2}}$ and $v = (x^2 - 3x)^{\frac{1}{2}}$.
2. Differentiate $u$ and $v$:
   • $u' = \frac{3}{2} (6x^2 - 3x + 1)^{\frac{1}{2}} \cdot (12x - 3) = \frac{3(12x - 3) (6x^2 - 3x + 1)^{\frac{1}{2}}}{2}$.
   • $v' = \frac{1}{2} (x^2 - 3x)^{-\frac{1}{2}} \cdot (2x - 3) = \frac{2x - 3}{2 \sqrt{x^2 - 3x}}$.