Calculus derivative matching worksheet for practice and assessment.

Let's solve each derivative step-by-step and match it with the correct option from the list (a to i).

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1. $\frac{d}{dx}(x^3 + 4x^2 + 7x + 2)$

Differentiate term by term:
$$
\frac{d}{dx}(x^3) = 3x^2,\quad \frac{d}{dx}(4x^2) = 8x,\quad \frac{d}{dx}(7x) = 7,\quad \frac{d}{dx}(2) = 0
$$
So,
$$
3x^2 + 8x + 7
$$
→ This matches f. $3x^2 + 8x + 7$

✔ 1 → f

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2. $\frac{d}{dx}\left(\frac{7}{x^{2/3}}\right)$

Rewrite: $7x^{-2/3}$

Differentiate:
$$
7 \cdot \left(-\frac{2}{3}\right)x^{-5/3} = -\frac{14}{3}x^{-5/3} = \frac{-14}{3x^{5/3}}
$$
→ This matches a. $\frac{-14}{3x^{5/3}}$

✔ 2 → a

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

Rewrite $\sqrt{x} = x^{1/2}$, so:
$$
\frac{2x^2 - 3x + 1}{x^{1/2}} = 2x^{2 - 1/2} - 3x^{1 - 1/2} + x^{-1/2} = 2x^{3/2} - 3x^{1/2} + x^{-1/2}
$$

Now differentiate:
- $\frac{d}{dx}(2x^{3/2}) = 2 \cdot \frac{3}{2}x^{1/2} = 3x^{1/2}$
- $\frac{d}{dx}(-3x^{1/2}) = -3 \cdot \frac{1}{2}x^{-1/2} = -\frac{3}{2}x^{-1/2}$
- $\frac{d}{dx}(x^{-1/2}) = -\frac{1}{2}x^{-3/2}$

So total:
$$
3x^{1/2} - \frac{3}{2}x^{-1/2} - \frac{1}{2}x^{-3/2}
= 3\sqrt{x} - \frac{3}{2\sqrt{x}} - \frac{1}{2x\sqrt{x}}
$$

→ This matches i. $3\sqrt{x} - \frac{3}{2\sqrt{x}} - \frac{1}{2x\sqrt{x}}$

✔ 3 → i

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

Use chain rule:
Let $u = 2x - 3$, then $\frac{d}{dx}(u^2) = 2u \cdot u' = 2(2x - 3)(2) = 4(2x - 3) = 8x - 12$

→ This matches e. $8x - 12$

✔ 4 → e

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5. $\frac{d}{dx}(\sqrt{3x + 2})$

Let $u = 3x + 2$, so $\frac{d}{dx}(u^{1/2}) = \frac{1}{2}u^{-1/2} \cdot u' = \frac{1}{2\sqrt{3x+2}} \cdot 3 = \frac{3}{2\sqrt{3x+2}}$

→ This matches g. $\frac{3}{2\sqrt{3x+2}}$

✔ 5 → g

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6. $\frac{d}{dx}((ax + b)(cx + d))$

Use product rule:
$$
\frac{d}{dx}[(ax + b)(cx + d)] = (a)(cx + d) + (ax + b)(c) = a(cx + d) + c(ax + b)
$$
$$
= acx + ad + acx + bc = 2acx + ad + bc
$$
But in options, we have b. $c(ax + b) + a(cx + d)$, which is exactly:
$$
c(ax + b) + a(cx + d) = acx + bc + acx + ad = 2acx + ad + bc
$$
Same as above.

So even though it's not simplified, this expression is equivalent.

✔ 6 → b

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7. $\frac{d}{dx}\left(\frac{2x + 5}{3x - 2}\right)$

Use quotient rule:
$$
\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}
$$
Let $u = 2x + 5$, $u' = 2$
$v = 3x - 2$, $v' = 3$

So:
$$
\frac{2(3x - 2) - (2x + 5)(3)}{(3x - 2)^2} = \frac{6x - 4 - (6x + 15)}{(3x - 2)^2} = \frac{6x - 4 - 6x - 15}{(3x - 2)^2} = \frac{-19}{(3x - 2)^2}
$$

→ This matches c. $\frac{-19}{(3x - 2)^2}$

✔ 7 → c

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8. $\frac{d}{dx}\left(\frac{3x + 2}{(x + 5)(2x + 1) + 3}\right)$

First simplify denominator:
$$
(x + 5)(2x + 1) = 2x^2 + x + 10x + 5 = 2x^2 + 11x + 5
$$
Add 3: $2x^2 + 11x + 8$

So function becomes:
$$
\frac{3x + 2}{2x^2 + 11x + 8}
$$

Now use quotient rule:
Let $u = 3x + 2$, $u' = 3$
$v = 2x^2 + 11x + 8$, $v' = 4x + 11$

Then:
$$
\frac{u'v - uv'}{v^2} = \frac{3(2x^2 + 11x + 8) - (3x + 2)(4x + 11)}{(2x^2 + 11x + 8)^2}
$$

Compute numerator:
- $3(2x^2 + 11x + 8) = 6x^2 + 33x + 24$
- $(3x + 2)(4x + 11) = 12x^2 + 33x + 8x + 22 = 12x^2 + 41x + 22$

So:
$$
6x^2 + 33x + 24 - (12x^2 + 41x + 22) = 6x^2 + 33x + 24 - 12x^2 - 41x - 22 = -6x^2 - 8x + 2
$$

Thus:
$$
\frac{-6x^2 - 8x + 2}{(2x^2 + 11x + 8)^2}
$$

→ This matches d. $\frac{-6x^2 - 8x + 2}{(2x^2 + 11x + 8)^2}$

✔ 8 → d

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9. $\frac{d}{dx}\left(\frac{2x - 3}{3x + 4}\right)$

Quotient rule again:

$u = 2x - 3$, $u' = 2$
$v = 3x + 4$, $v' = 3$

$$
\frac{2(3x + 4) - (2x - 3)(3)}{(3x + 4)^2} = \frac{6x + 8 - (6x - 9)}{(3x + 4)^2} = \frac{6x + 8 - 6x + 9}{(3x + 4)^2} = \frac{17}{(3x + 4)^2}
$$

→ This matches h. $\frac{17}{(3x + 4)^2}$

✔ 9 → h

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✔ Final Matching:

| Number | Derivative | Matches |
|--------|------------|--------|
| 1 | $\frac{d}{dx}(x^3 + 4x^2 + 7x + 2)$ | f |
| 2 | $\frac{d}{dx}\left(\frac{7}{x^{2/3}}\right)$ | a |
| 3 | $\frac{d}{dx}\left(\frac{2x^2 - 3x + 1}{\sqrt{x}}\right)$ | i |
| 4 | $\frac{d}{dx}(2x - 3)^2$ | e |
| 5 | $\frac{d}{dx}(\sqrt{3x + 2})$ | g |
| 6 | $\frac{d}{dx}((ax + b)(cx + d))$ | b |
| 7 | $\frac{d}{dx}\left(\frac{2x + 5}{3x - 2}\right)$ | c |
| 8 | $\frac{d}{dx}\left(\frac{3x + 2}{(x+5)(2x+1)+3}\right)$ | d |
| 9 | $\frac{d}{dx}\left(\frac{2x - 3}{3x + 4}\right)$ | h |

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🟩 Answer Summary:

1 → f
2 → a
3 → i
4 → e
5 → g
6 → b
7 → c
8 → d
9 → h

Let me know if you'd like this formatted as a table or for printing!