Worksheet for computing derivatives of various functions at a point.

Let's solve each problem step by step. The goal is to find the derivative $ f'(p) $ for each given function $ f(x) $. This means we need to:

1. Differentiate the function $ f(x) $ with respect to $ x $, obtaining $ f'(x) $.
2. Then evaluate $ f'(x) $ at $ x = p $, giving us $ f'(p) $.

We’ll go through all 8 problems.

---

1. $ f(x) = -6x^2 - 9x + 7 $

Use the power rule:
- Derivative of $ x^n $ is $ nx^{n-1} $

$$
f'(x) = -6 \cdot 2x - 9 = -12x - 9
$$

Now evaluate at $ x = p $:

$$
f'(p) = -12p - 9
$$

✔ Answer: $ \boxed{-12p - 9} $

---

2. $ f(x) = -5x^2 - 9x - 2 $

Again, apply the power rule:

$$
f'(x) = -5 \cdot 2x - 9 = -10x - 9
$$

At $ x = p $:

$$
f'(p) = -10p - 9
$$

✔ Answer: $ \boxed{-10p - 9} $

---

3. $ f(x) = x^2 - 7x - 9 $

Differentiate:

$$
f'(x) = 2x - 7
$$

So,

$$
f'(p) = 2p - 7
$$

✔ Answer: $ \boxed{2p - 7} $

---

4. $ f(x) = \frac{3x + 3}{5x - 3} $

This is a quotient, so use the quotient rule:

If $ f(x) = \frac{u}{v} $, then

$$
f'(x) = \frac{u'v - uv'}{v^2}
$$

Let:
- $ u = 3x + 3 $ → $ u' = 3 $
- $ v = 5x - 3 $ → $ v' = 5 $

Now compute:

$$
f'(x) = \frac{(3)(5x - 3) - (3x + 3)(5)}{(5x - 3)^2}
$$

Simplify numerator:

$$
= \frac{15x - 9 - (15x + 15)}{(5x - 3)^2}
= \frac{15x - 9 - 15x - 15}{(5x - 3)^2}
= \frac{-24}{(5x - 3)^2}
$$

So,

$$
f'(p) = \frac{-24}{(5p - 3)^2}
$$

✔ Answer: $ \boxed{\frac{-24}{(5p - 3)^2}} $

---

5. $ f(x) = \frac{-3x + 4}{3x - 10} $

Quotient rule again.

Let:
- $ u = -3x + 4 $ → $ u' = -3 $
- $ v = 3x - 10 $ → $ v' = 3 $

$$
f'(x) = \frac{(-3)(3x - 10) - (-3x + 4)(3)}{(3x - 10)^2}
$$

Numerator:

$$
= -9x + 30 - (-9x + 12)
= -9x + 30 + 9x - 12
= 18
$$

So,

$$
f'(x) = \frac{18}{(3x - 10)^2}
\quad \Rightarrow \quad
f'(p) = \frac{18}{(3p - 10)^2}
$$

✔ Answer: $ \boxed{\frac{18}{(3p - 10)^2}} $

---

6. $ f(x) = \frac{x - 3}{8x - 7} $

Quotient rule:

- $ u = x - 3 $ → $ u' = 1 $
- $ v = 8x - 7 $ → $ v' = 8 $

$$
f'(x) = \frac{(1)(8x - 7) - (x - 3)(8)}{(8x - 7)^2}
$$

Numerator:

$$
= 8x - 7 - (8x - 24)
= 8x - 7 - 8x + 24
= 17
$$

So,

$$
f'(x) = \frac{17}{(8x - 7)^2}
\quad \Rightarrow \quad
f'(p) = \frac{17}{(8p - 7)^2}
$$

✔ Answer: $ \boxed{\frac{17}{(8p - 7)^2}} $

---

7. $ f(x) = 5\sqrt{x} $

Recall: $ \sqrt{x} = x^{1/2} $, so:

$$
f(x) = 5x^{1/2}
\quad \Rightarrow \quad
f'(x) = 5 \cdot \frac{1}{2}x^{-1/2} = \frac{5}{2\sqrt{x}}
$$

Then,

$$
f'(p) = \frac{5}{2\sqrt{p}}
$$

✔ Answer: $ \boxed{\frac{5}{2\sqrt{p}}} $

---

8. $ f(x) = -13\sqrt{x} $

Same idea: $ \sqrt{x} = x^{1/2} $

$$
f(x) = -13x^{1/2}
\quad \Rightarrow \quad
f'(x) = -13 \cdot \frac{1}{2}x^{-1/2} = -\frac{13}{2\sqrt{x}}
$$

So,

$$
f'(p) = -\frac{13}{2\sqrt{p}}
$$

✔ Answer: $ \boxed{-\frac{13}{2\sqrt{p}}} $

---

✔ Final Answers Summary:

| Problem | $ f'(p) $ |
|--------|-------------|
| 1 | $ -12p - 9 $ |
| 2 | $ -10p - 9 $ |
| 3 | $ 2p - 7 $ |
| 4 | $ \dfrac{-24}{(5p - 3)^2} $ |
| 5 | $ \dfrac{18}{(3p - 10)^2} $ |
| 6 | $ \dfrac{17}{(8p - 7)^2} $ |
| 7 | $ \dfrac{5}{2\sqrt{p}} $ |
| 8 | $ -\dfrac{13}{2\sqrt{p}} $ |

Let me know if you'd like these explained in more detail or formatted differently!