Power Of A Product Worksheet - Free Printable

Let's solve each problem step-by-step using the Power Rule, Product Rule, and Quotient Rule as needed. We'll also use basic derivative rules for trigonometric functions, exponential functions, and logarithmic differentiation where applicable.

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1. $ f(x) = 3x^2 + 5x - 2 $

Use Power Rule:
$$
f'(x) = 3(2x) + 5(1) - 0 = 6x + 5
$$

✔ Answer: $ \boxed{6x + 5} $

---

2. $ g(x) = -4x^4 + 5x^3 - 2x + 3 $

Apply Power Rule to each term:
$$
g'(x) = -4(4x^3) + 5(3x^2) - 2(1) + 0 = -16x^3 + 15x^2 - 2
$$

✔ Answer: $ \boxed{-16x^3 + 15x^2 - 2} $

---

3. $ f(x) = 2\sqrt{x} + 7\sqrt{x^3} - \frac{2}{x^2} $

Rewrite in exponents:
- $ \sqrt{x} = x^{1/2} $
- $ \sqrt{x^3} = x^{3/2} $
- $ \frac{2}{x^2} = 2x^{-2} $

So:
$$
f(x) = 2x^{1/2} + 7x^{3/2} - 2x^{-2}
$$

Differentiate:
$$
f'(x) = 2 \cdot \frac{1}{2}x^{-1/2} + 7 \cdot \frac{3}{2}x^{1/2} - 2(-2)x^{-3}
= x^{-1/2} + \frac{21}{2}x^{1/2} + 4x^{-3}
$$

Rewriting:
$$
f'(x) = \frac{1}{\sqrt{x}} + \frac{21}{2}\sqrt{x} + \frac{4}{x^3}
$$

✔ Answer: $ \boxed{\frac{1}{\sqrt{x}} + \frac{21}{2}\sqrt{x} + \frac{4}{x^3}} $

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4. $ g(x) = 8\sqrt{x^5} - 7x^4 + \frac{5}{\sqrt{x}} $

Rewrite:
- $ \sqrt{x^5} = x^{5/2} $
- $ \frac{5}{\sqrt{x}} = 5x^{-1/2} $

So:
$$
g(x) = 8x^{5/2} - 7x^4 + 5x^{-1/2}
$$

Differentiate:
$$
g'(x) = 8 \cdot \frac{5}{2}x^{3/2} - 7(4x^3) + 5 \cdot \left(-\frac{1}{2}\right)x^{-3/2}
= 20x^{3/2} - 28x^3 - \frac{5}{2}x^{-3/2}
$$

Rewriting:
$$
g'(x) = 20\sqrt{x^3} - 28x^3 - \frac{5}{2\sqrt{x^3}}
$$

✔ Answer: $ \boxed{20x^{3/2} - 28x^3 - \frac{5}{2}x^{-3/2}} $

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5. $ f(x) = \frac{x^2 + 7x - 18}{x + 9} $

Try factoring numerator:
$$
x^2 + 7x - 18 = (x + 9)(x - 2)
$$

So:
$$
f(x) = \frac{(x+9)(x-2)}{x+9} = x - 2 \quad \text{(for } x \ne -9\text{)}
$$

Then:
$$
f'(x) = 1
$$

✔ Answer: $ \boxed{1} $

---

6. $ f(x) = \frac{x^2 - 5x - 24}{x - 8} $

Factor numerator:
$$
x^2 - 5x - 24 = (x - 8)(x + 3)
$$

So:
$$
f(x) = \frac{(x-8)(x+3)}{x-8} = x + 3 \quad \text{(for } x \ne 8\text{)}
$$

Then:
$$
f'(x) = 1
$$

✔ Answer: $ \boxed{1} $

---

7. $ f(x) = \frac{x^{-3} + 7\sqrt{x^3} - 4x^2}{2\sqrt{x}} $

First, simplify numerator and denominator:

Numerator:
- $ x^{-3} $
- $ 7\sqrt{x^3} = 7x^{3/2} $
- $ -4x^2 $

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

So:
$$
f(x) = \frac{x^{-3} + 7x^{3/2} - 4x^2}{2x^{1/2}} = \frac{1}{2} \left( x^{-3} \cdot x^{-1/2} + 7x^{3/2} \cdot x^{-1/2} - 4x^2 \cdot x^{-1/2} \right)
$$

Simplify exponents:
- $ x^{-3 - 1/2} = x^{-7/2} $
- $ x^{3/2 - 1/2} = x^1 $
- $ x^{2 - 1/2} = x^{3/2} $

So:
$$
f(x) = \frac{1}{2} \left( x^{-7/2} + 7x - 4x^{3/2} \right)
$$

Now differentiate:
$$
f'(x) = \frac{1}{2} \left( -\frac{7}{2}x^{-9/2} + 7 - 4 \cdot \frac{3}{2}x^{1/2} \right)
= \frac{1}{2} \left( -\frac{7}{2}x^{-9/2} + 7 - 6x^{1/2} \right)
$$

$$
= -\frac{7}{4}x^{-9/2} + \frac{7}{2} - 3x^{1/2}
$$

✔ Answer: $ \boxed{-\frac{7}{4}x^{-9/2} + \frac{7}{2} - 3\sqrt{x}} $

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8. $ h(x) = \frac{2 + 5x^2 - 8\sqrt{x^7}}{-3\sqrt{x}} $

Rewrite all terms:
- $ \sqrt{x^7} = x^{7/2} $
- $ \sqrt{x} = x^{1/2} $

So:
$$
h(x) = \frac{2 + 5x^2 - 8x^{7/2}}{-3x^{1/2}} = -\frac{1}{3} \left( 2x^{-1/2} + 5x^{2 - 1/2} - 8x^{7/2 - 1/2} \right)
= -\frac{1}{3} \left( 2x^{-1/2} + 5x^{3/2} - 8x^3 \right)
$$

Differentiate:
$$
h'(x) = -\frac{1}{3} \left( 2 \cdot (-\frac{1}{2})x^{-3/2} + 5 \cdot \frac{3}{2}x^{1/2} - 8(3x^2) \right)
= -\frac{1}{3} \left( -x^{-3/2} + \frac{15}{2}x^{1/2} - 24x^2 \right)
$$

Distribute:
$$
= \frac{1}{3}x^{-3/2} - \frac{5}{2}x^{1/2} + 8x^2
$$

✔ Answer: $ \boxed{\frac{1}{3x^{3/2}} - \frac{5}{2}\sqrt{x} + 8x^2} $

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9. $ s(x) = 2x^{-3} \sec(x) $

Use Product Rule: $ (uv)' = u'v + uv' $

Let:
- $ u = 2x^{-3} $ → $ u' = 2(-3)x^{-4} = -6x^{-4} $
- $ v = \sec(x) $ → $ v' = \sec(x)\tan(x) $

So:
$$
s'(x) = u'v + uv' = (-6x^{-4})\sec(x) + (2x^{-3})(\sec(x)\tan(x))
$$

Factor:
$$
= -6x^{-4}\sec(x) + 2x^{-3}\sec(x)\tan(x)
= \sec(x) \left( -\frac{6}{x^4} + \frac{2\tan(x)}{x^3} \right)
$$

✔ Answer: $ \boxed{\sec(x)\left( -\frac{6}{x^4} + \frac{2\tan(x)}{x^3} \right)} $

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10. $ f(x) = 3x^4 e^x $

Product Rule:
- $ u = 3x^4 $ → $ u' = 12x^3 $
- $ v = e^x $ → $ v' = e^x $

$$
f'(x) = 12x^3 e^x + 3x^4 e^x = 3x^3 e^x (4 + x)
$$

✔ Answer: $ \boxed{3x^3 e^x (x + 4)} $

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11. $ f(x) = -7x^3 e^x $

Product Rule:
- $ u = -7x^3 $ → $ u' = -21x^2 $
- $ v = e^x $ → $ v' = e^x $

$$
f'(x) = -21x^2 e^x - 7x^3 e^x = -7x^2 e^x (3 + x)
$$

✔ Answer: $ \boxed{-7x^2 e^x (x + 3)} $

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12. $ f(x) = 5x^2 \cos(x) $

Product Rule:
- $ u = 5x^2 $ → $ u' = 10x $
- $ v = \cos(x) $ → $ v' = -\sin(x) $

$$
f'(x) = 10x \cos(x) + 5x^2 (-\sin(x)) = 10x\cos(x) - 5x^2\sin(x)
$$

✔ Answer: $ \boxed{10x\cos(x) - 5x^2\sin(x)} $

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13. $ h(x) = 2e^x \sqrt{x} $

$ = 2e^x x^{1/2} $

Product Rule:
- $ u = 2e^x $ → $ u' = 2e^x $
- $ v = x^{1/2} $ → $ v' = \frac{1}{2}x^{-1/2} $

$$
h'(x) = 2e^x \cdot x^{1/2} + 2e^x \cdot \frac{1}{2}x^{-1/2}
= 2e^x \sqrt{x} + e^x \cdot \frac{1}{\sqrt{x}}
= e^x \left( 2\sqrt{x} + \frac{1}{\sqrt{x}} \right)
$$

✔ Answer: $ \boxed{e^x \left( 2\sqrt{x} + \frac{1}{\sqrt{x}} \right)} $

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14. $ f(x) = 4x^4 - 5x^3 + 2x^2 e^x $

Split into parts:
- $ f(x) = 4x^4 - 5x^3 + 2x^2 e^x $

Derivative of first two terms:
- $ 16x^3 - 15x^2 $

For $ 2x^2 e^x $: Product Rule
- $ u = 2x^2 $ → $ u' = 4x $
- $ v = e^x $ → $ v' = e^x $
- Derivative: $ 4x e^x + 2x^2 e^x $

Total:
$$
f'(x) = 16x^3 - 15x^2 + 4x e^x + 2x^2 e^x
$$

✔ Answer: $ \boxed{16x^3 - 15x^2 + 4x e^x + 2x^2 e^x} $

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15. $ f(x) = \frac{\tan(x)}{2x^2 + 1} $

Quotient Rule: $ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} $

- $ u = \tan(x) $ → $ u' = \sec^2(x) $
- $ v = 2x^2 + 1 $ → $ v' = 4x $

So:
$$
f'(x) = \frac{\sec^2(x)(2x^2 + 1) - \tan(x)(4x)}{(2x^2 + 1)^2}
$$

✔ Answer: $ \boxed{\frac{\sec^2(x)(2x^2 + 1) - 4x\tan(x)}{(2x^2 + 1)^2}} $

---

16. $ g(x) = \frac{\sin(x)}{e^x + 5} $

Quotient Rule:
- $ u = \sin(x) $ → $ u' = \cos(x) $
- $ v = e^x + 5 $ → $ v' = e^x $

$$
g'(x) = \frac{\cos(x)(e^x + 5) - \sin(x)e^x}{(e^x + 5)^2}
$$

✔ Answer: $ \boxed{\frac{\cos(x)(e^x + 5) - e^x \sin(x)}{(e^x + 5)^2}} $

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17. $ f(x) = \frac{x^2 - 3x + 2}{x + 3} $

We can try simplifying or use Quotient Rule.

Let’s use Quotient Rule:

- $ u = x^2 - 3x + 2 $ → $ u' = 2x - 3 $
- $ v = x + 3 $ → $ v' = 1 $

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

Expand numerator:
- $ (2x - 3)(x + 3) = 2x^2 + 6x - 3x - 9 = 2x^2 + 3x - 9 $
- Subtract $ x^2 - 3x + 2 $: $ 2x^2 + 3x - 9 - x^2 + 3x - 2 = x^2 + 6x - 11 $

So:
$$
f'(x) = \frac{x^2 + 6x - 11}{(x + 3)^2}
$$

✔ Answer: $ \boxed{\frac{x^2 + 6x - 11}{(x + 3)^2}} $

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18. $ f(x) = \frac{2e^x}{x - 2e^x} $

Quotient Rule:
- $ u = 2e^x $ → $ u' = 2e^x $
- $ v = x - 2e^x $ → $ v' = 1 - 2e^x $

$$
f'(x) = \frac{2e^x(x - 2e^x) - 2e^x(1 - 2e^x)}{(x - 2e^x)^2}
$$

Expand numerator:
- First term: $ 2e^x x - 4e^{2x} $
- Second term: $ -2e^x + 4e^{2x} $

Add:
$$
2xe^x - 4e^{2x} - 2e^x + 4e^{2x} = 2xe^x - 2e^x = 2e^x(x - 1)
$$

So:
$$
f'(x) = \frac{2e^x(x - 1)}{(x - 2e^x)^2}
$$

✔ Answer: $ \boxed{\frac{2e^x(x - 1)}{(x - 2e^x)^2}} $

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19. $ h(x) = \frac{2x^4 \cot(x)}{3x^2} $

Simplify first:
$$
h(x) = \frac{2}{3} x^{4-2} \cot(x) = \frac{2}{3} x^2 \cot(x)
$$

Now product rule:
- $ u = \frac{2}{3}x^2 $ → $ u' = \frac{4}{3}x $
- $ v = \cot(x) $ → $ v' = -\csc^2(x) $

$$
h'(x) = \frac{4}{3}x \cot(x) + \frac{2}{3}x^2 (-\csc^2(x))
= \frac{4}{3}x \cot(x) - \frac{2}{3}x^2 \csc^2(x)
$$

✔ Answer: $ \boxed{\frac{4}{3}x \cot(x) - \frac{2}{3}x^2 \csc^2(x)} $

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20. $ f(x) = \frac{\csc(x)}{-4x e^x} $

Simplify:
$$
f(x) = -\frac{1}{4} \cdot \frac{\csc(x)}{x e^x}
$$

Let’s write $ f(x) = -\frac{1}{4} \cdot \frac{\csc(x)}{x e^x} $

Let $ u = \csc(x) $, $ v = x e^x $

Use quotient rule on $ \frac{u}{v} $, then multiply by $ -\frac{1}{4} $

But better: write as product:
$$
f(x) = -\frac{1}{4} \csc(x) \cdot (x e^x)^{-1}
$$

But perhaps easier to apply Quotient Rule directly:

Let:
- $ u = \csc(x) $ → $ u' = -\csc(x)\cot(x) $
- $ v = -4x e^x $ → $ v' = -4 [e^x + x e^x] = -4e^x(1 + x) $

Now:
$$
f'(x) = \frac{u'v - uv'}{v^2}
= \frac{[-\csc(x)\cot(x)](-4x e^x) - [\csc(x)][-4e^x(1 + x)]}{(-4x e^x)^2}
$$

Simplify numerator:
- First term: $ +4x e^x \csc(x)\cot(x) $
- Second term: $ +4e^x(1 + x)\csc(x) $

Factor $ 4e^x \csc(x) $:
$$
= 4e^x \csc(x) [x \cot(x) + (1 + x)]
$$

Denominator: $ 16x^2 e^{2x} $

So:
$$
f'(x) = \frac{4e^x \csc(x) [x \cot(x) + 1 + x]}{16x^2 e^{2x}} = \frac{\csc(x) [x \cot(x) + x + 1]}{4x^2 e^x}
$$

✔ Answer: $ \boxed{\frac{\csc(x) (x \cot(x) + x + 1)}{4x^2 e^x}} $

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## ✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1 | $ 6x + 5 $ |
| 2 | $ -16x^3 + 15x^2 - 2 $ |
| 3 | $ \frac{1}{\sqrt{x}} + \frac{21}{2}\sqrt{x} + \frac{4}{x^3} $ |
| 4 | $ 20x^{3/2} - 28x^3 - \frac{5}{2}x^{-3/2} $ |
| 5 | $ 1 $ |
| 6 | $ 1 $ |
| 7 | $ -\frac{7}{4}x^{-9/2} + \frac{7}{2} - 3\sqrt{x} $ |
| 8 | $ \frac{1}{3x^{3/2}} - \frac{5}{2}\sqrt{x} + 8x^2 $ |
| 9 | $ \sec(x)\left( -\frac{6}{x^4} + \frac{2\tan(x)}{x^3} \right) $ |
| 10 | $ 3x^3 e^x (x + 4) $ |
| 11 | $ -7x^2 e^x (x + 3) $ |
| 12 | $ 10x\cos(x) - 5x^2\sin(x) $ |
| 13 | $ e^x \left( 2\sqrt{x} + \frac{1}{\sqrt{x}} \right) $ |
| 14 | $ 16x^3 - 15x^2 + 4x e^x + 2x^2 e^x $ |
| 15 | $ \frac{\sec^2(x)(2x^2 + 1) - 4x\tan(x)}{(2x^2 + 1)^2} $ |
| 16 | $ \frac{\cos(x)(e^x + 5) - e^x \sin(x)}{(e^x + 5)^2} $ |
| 17 | $ \frac{x^2 + 6x - 11}{(x + 3)^2} $ |
| 18 | $ \frac{2e^x(x - 1)}{(x - 2e^x)^2} $ |
| 19 | $ \frac{4}{3}x \cot(x) - \frac{2}{3}x^2 \csc^2(x) $ |
| 20 | $ \frac{\csc(x) (x \cot(x) + x + 1)}{4x^2 e^x} $ |

Let me know if you'd like any further explanation!