https://file.aiquickdraw.com/mathbot/file/1730437657578zqxfn9yc.jpg

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



Find the first derivatives of the following functions:

(a) y = ln(x^2 - x)
(b) p(a) = log_3(a^2 - 1)
(c) h(z) = z^2 - ln^4(1 - z^2)
(d) y = (6x^2 - 3x + 1)^(3/2) / sqrt(x^2 - 3x)

Learn how to find the first derivatives of complex logarithmic and rational functions, including examples such as y = ln(x^2 - x) and y = (6x^2 - 3x + 1)^(3/2) / sqrt(x^2 - 3x). This guide covers differentiation techniques like the chain rule, logarithmic differentiation, and the quotient rule. Suitable for high school and college-level calculus students.

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