Students can use this worksheet to practice evaluating derivatives by rewriting expressions before applying basic differentiation rules.

Let's solve each problem on Worksheet 4.1B: Basic Differentiation Practice step by step, without using the product, quotient, or chain rules, as instructed.

We will use:
- Power rule: $\frac{d}{dx}[x^n] = nx^{n-1}$
- Constant multiple rule: $\frac{d}{dx}[c \cdot f(x)] = c \cdot f'(x)$
- Sum/difference rule: $\frac{d}{dx}[f(x) \pm g(x)] = f'(x) \pm g'(x)$
- Rewrite expressions (expand, simplify, etc.) when needed.

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1. $\frac{d}{dx} x^4 =$

Use power rule:

$$
\frac{d}{dx} x^4 = 4x^{3}
$$

✔ Answer: $4x^3$

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

Apply constant multiple rule:

$$
\frac{d}{dx} (-3x^6) = -3 \cdot \frac{d}{dx}(x^6) = -3 \cdot 6x^5 = -18x^5
$$

✔ Answer: $-18x^5$

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3. $\frac{d}{dx} \left[4(x+2)(x+3)\right] =$

First, expand the expression inside:

$$
(x+2)(x+3) = x^2 + 3x + 2x + 6 = x^2 + 5x + 6
$$

Now multiply by 4:

$$
4(x^2 + 5x + 6) = 4x^2 + 20x + 24
$$

Now differentiate term by term:

$$
\frac{d}{dx}(4x^2 + 20x + 24) = 8x + 20
$$

✔ Answer: $8x + 20$

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4. $\frac{d}{dx} \left[5x^5\right] =$

Constant multiple + power rule:

$$
5 \cdot \frac{d}{dx}(x^5) = 5 \cdot 5x^4 = 25x^4
$$

✔ Answer: $25x^4$

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5. $\frac{d}{dr} (-2r)^3 =$

First, simplify the expression:

$$
(-2r)^3 = (-2)^3 r^3 = -8r^3
$$

Now differentiate:

$$
\frac{d}{dr}(-8r^3) = -8 \cdot 3r^2 = -24r^2
$$

✔ Answer: $-24r^2$

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6. $\frac{d}{dx} \left[(x-2)^2(2x+1)\right] =$

We cannot use product rule — so expand first.

First, expand $(x-2)^2$:

$$
(x-2)^2 = x^2 - 4x + 4
$$

Now multiply by $(2x+1)$:

$$
(x^2 - 4x + 4)(2x + 1)
$$

Distribute:

$$
= x^2(2x + 1) - 4x(2x + 1) + 4(2x + 1)
$$
$$
= 2x^3 + x^2 - 8x^2 - 4x + 8x + 4
$$
$$
= 2x^3 - 7x^2 + 4x + 4
$$

Now differentiate:

$$
\frac{d}{dx}(2x^3 - 7x^2 + 4x + 4) = 6x^2 - 14x + 4
$$

✔ Answer: $6x^2 - 14x + 4$

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7. $\frac{d}{dx} \left[12x^{3/2}\right] =$

Power rule with fractional exponent:

$$
\frac{d}{dx}(12x^{3/2}) = 12 \cdot \frac{3}{2} x^{1/2} = 18x^{1/2}
$$

✔ Answer: $18\sqrt{x}$ or $18x^{1/2}$

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8. $\frac{d}{dx} \left[3(2x^2)^2\right] =$

Simplify inside first:

$$
(2x^2)^2 = 4x^4
$$

So:

$$
3(4x^4) = 12x^4
$$

Differentiate:

$$
\frac{d}{dx}(12x^4) = 48x^3
$$

✔ Answer: $48x^3$

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9. $\frac{d}{dx} \left(\frac{(4x-2)(6x+1)}{x^2}\right) =$

We cannot use quotient rule. So rewrite as a sum of terms.

First, expand numerator:

$$
(4x - 2)(6x + 1) = 4x(6x + 1) - 2(6x + 1) = 24x^2 + 4x - 12x - 2 = 24x^2 - 8x - 2
$$

Now divide by $x^2$:

$$
\frac{24x^2 - 8x - 2}{x^2} = 24 - \frac{8}{x} - \frac{2}{x^2} = 24 - 8x^{-1} - 2x^{-2}
$$

Now differentiate:

$$
\frac{d}{dx}(24 - 8x^{-1} - 2x^{-2}) = 0 + 8x^{-2} + 4x^{-3} = \frac{8}{x^2} + \frac{4}{x^3}
$$

✔ Answer: $\frac{8}{x^2} + \frac{4}{x^3}$

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10. $\frac{d}{dx} \left[4x^3(x+1)^2\right] =$

Cannot use product rule — expand first.

First, expand $(x+1)^2 = x^2 + 2x + 1$

Now multiply by $4x^3$:

$$
4x^3(x^2 + 2x + 1) = 4x^5 + 8x^4 + 4x^3
$$

Differentiate:

$$
\frac{d}{dx}(4x^5 + 8x^4 + 4x^3) = 20x^4 + 32x^3 + 12x^2
$$

✔ Answer: $20x^4 + 32x^3 + 12x^2$

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11. $\frac{d}{dx} \left[\frac{2}{(3x)^2}\right] =$

Simplify denominator:

$$
(3x)^2 = 9x^2
\Rightarrow \frac{2}{9x^2} = \frac{2}{9}x^{-2}
$$

Differentiate:

$$
\frac{2}{9} \cdot (-2)x^{-3} = -\frac{4}{9}x^{-3} = -\frac{4}{9x^3}
$$

✔ Answer: $-\frac{4}{9x^3}$

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12. $\frac{d}{dx} \left(x^2\sqrt{x} - \frac{4}{\sqrt{x}}\right) =$

Rewrite radicals as exponents:

$$
\sqrt{x} = x^{1/2},\quad \frac{1}{\sqrt{x}} = x^{-1/2}
$$

So expression becomes:

$$
x^2 \cdot x^{1/2} - 4x^{-1/2} = x^{5/2} - 4x^{-1/2}
$$

Differentiate:

$$
\frac{d}{dx}(x^{5/2}) = \frac{5}{2}x^{3/2},\quad \frac{d}{dx}(-4x^{-1/2}) = -4 \cdot \left(-\frac{1}{2}\right)x^{-3/2} = 2x^{-3/2}
$$

So derivative is:

$$
\frac{5}{2}x^{3/2} + 2x^{-3/2}
$$

Or in radical form:

$$
\frac{5}{2}x\sqrt{x} + \frac{2}{x^{3/2}}
$$

✔ Answer: $\frac{5}{2}x^{3/2} + 2x^{-3/2}$

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13. $\frac{d}{dx} \left(\frac{5}{x}\right) =$

Rewrite: $\frac{5}{x} = 5x^{-1}$

Differentiate:

$$
5 \cdot (-1)x^{-2} = -\frac{5}{x^2}
$$

✔ Answer: $-\frac{5}{x^2}$

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14. $\frac{d}{dx} \left(\frac{-4}{(2x)^{-1}}\right) =$

Simplify expression:

$$
(2x)^{-1} = \frac{1}{2x},\quad \text{so } \frac{-4}{(2x)^{-1}} = -4 \cdot (2x) = -8x
$$

Now differentiate:

$$
\frac{d}{dx}(-8x) = -8
$$

✔ Answer: $-8$

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15. $\frac{d}{dx} \left(-\frac{1}{2}(x^3 - 3x^2 + 2x)\right) =$

Distribute the constant:

$$
-\frac{1}{2}x^3 + \frac{3}{2}x^2 - x
$$

Differentiate term by term:

$$
-\frac{1}{2} \cdot 3x^2 + \frac{3}{2} \cdot 2x - 1 = -\frac{3}{2}x^2 + 3x - 1
$$

✔ Answer: $-\frac{3}{2}x^2 + 3x - 1$

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✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1 | $4x^3$ |
| 2 | $-18x^5$ |
| 3 | $8x + 20$ |
| 4 | $25x^4$ |
| 5 | $-24r^2$ |
| 6 | $6x^2 - 14x + 4$ |
| 7 | $18x^{1/2}$ |
| 8 | $48x^3$ |
| 9 | $\frac{8}{x^2} + \frac{4}{x^3}$ |
| 10 | $20x^4 + 32x^3 + 12x^2$ |
| 11 | $-\frac{4}{9x^3}$ |
| 12 | $\frac{5}{2}x^{3/2} + 2x^{-3/2}$ |
| 13 | $-\frac{5}{x^2}$ |
| 14 | $-8$ |
| 15 | $-\frac{3}{2}x^2 + 3x - 1$ |

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