# Power Rule Practice Problems with Answers and Solutions

> 12 power rule practice problems with answers and step-by-step solutions: polynomials, negative and fractional exponents, roots and tangent lines.

Source: https://practicecalculus.com/practice-problems/power-rule
Updated: 2026-11-06

$$ \frac{d}{dx}\,x^n = n\,x^{n-1} $$

## At a glance

- **Problems:** 12, with answers and full solutions
- **Level:** Calculus 1 and AP Calculus AB, Unit 2
- **You need:** Exponent rules, especially $\frac{1}{x^n} = x^{-n}$ and $\sqrt[n]{x^m} = x^{m/n}$
- **Main idea:** Rewrite as a power of x, bring the exponent down, then subtract 1 from it
- **Practice with new numbers:** [Power rule lessons](/learn/calculus-1/derivative-rules/power-rule)

## Key takeaways

- Rewrite roots and fractions as powers of $x$ before you differentiate.
- Subtracting 1 from a negative exponent makes it more negative: $x^{-2}$ becomes $-2x^{-3}$.
- Expand products and split quotients over a single term first. Then the power rule handles each term on its own.

These **power rule practice problems** start with single powers and polynomials, then move to negative and fractional exponents, rewriting before you differentiate, and tangent lines. Try each one first. The answers come right after the list, then the full solutions.

The rule:

$$\frac{d}{dx}\,x^n = n\,x^{n-1}$$

You will also use the [sum and difference rule](/rules/sum-and-difference-rule) and the [constant multiple rule](/rules/constant-multiple-rule) on almost every problem. For a refresher, read the [power rule](/rules/power-rule) page.

## Problems

**Powers and polynomials**

1. $y = x^7$
2. $y = 4x^5$
3. $y = 3x^4 - 2x^3 + 5x - 9$

**Negative and fractional exponents**

4. $y = \dfrac{1}{x^3}$
5. $y = 6\sqrt{x}$
6. $y = \sqrt[3]{x^2}$
7. $y = \dfrac{5}{x^2} - \dfrac{2}{\sqrt{x}}$

**Rewrite first**

8. $y = (2x + 1)(x - 3)$
9. $y = \dfrac{x^4 - 3x^2 + 2}{x^2}$
10. $y = x^\pi + \pi^2$

**Tangent lines**

11. Find the tangent line to $y = x^3 - 2x$ at $x = 2$.
12. Find the points where $y = x^3 - 3x^2$ has a horizontal tangent line.

## Answers

| Problem | Answer |
| --- | --- |
| 1 | $7x^6$ |
| 2 | $20x^4$ |
| 3 | $12x^3 - 6x^2 + 5$ |
| 4 | $-\dfrac{3}{x^4}$ |
| 5 | $\dfrac{3}{\sqrt{x}}$ |
| 6 | $\dfrac{2}{3\sqrt[3]{x}}$ |
| 7 | $-\dfrac{10}{x^3} + \dfrac{1}{x^{3/2}}$ |
| 8 | $4x - 5$ |
| 9 | $2x - \dfrac{4}{x^3}$ |
| 10 | $\pi x^{\pi - 1}$ |
| 11 | $y = 10x - 16$ |
| 12 | $(0, 0)$ and $(2, -4)$ |

## Step-by-step solutions

**1.** Bring the $7$ down and subtract $1$ from the exponent:

$$y' = 7x^6$$

**2.** The constant $4$ just comes along:

$$y' = 4\cdot 5x^4 = 20x^4$$

**3.** Differentiate term by term. The derivative of $5x$ is $5$ and the derivative of the constant $-9$ is $0$:

$$y' = 12x^3 - 6x^2 + 5$$

**4.** Rewrite as $y = x^{-3}$. Then $-3 - 1 = -4$:

$$y' = -3x^{-4} = -\frac{3}{x^4}$$

**5.** Rewrite as $y = 6x^{1/2}$:

$$y' = 6\cdot\frac{1}{2}x^{-1/2} = 3x^{-1/2} = \frac{3}{\sqrt{x}}$$

**6.** Rewrite as $y = x^{2/3}$. Then $\frac{2}{3} - 1 = -\frac{1}{3}$:

$$y' = \frac{2}{3}x^{-1/3} = \frac{2}{3\sqrt[3]{x}}$$

**7.** Rewrite both terms as powers: $y = 5x^{-2} - 2x^{-1/2}$.

$$y' = -10x^{-3} - 2\left(-\frac{1}{2}\right)x^{-3/2} = -\frac{10}{x^3} + \frac{1}{x^{3/2}}$$

**8.** Expand first: $(2x + 1)(x - 3) = 2x^2 - 5x - 3$.

$$y' = 4x - 5$$

The derivative of a product is not the product of the derivatives. Expanding avoids that trap, or you can use the [product rule](/rules/product-rule).

**9.** The bottom is a single term, so split the fraction:

$$y = x^2 - 3 + 2x^{-2} \quad\Rightarrow\quad y' = 2x - 4x^{-3} = 2x - \frac{4}{x^3}$$

**10.** $\pi$ is a constant exponent, so the power rule applies to $x^\pi$. The term $\pi^2 \approx 9.87$ is just a number, so its derivative is $0$:

$$y' = \pi x^{\pi - 1}$$

**11.** The point is $y(2) = 8 - 4 = 4$. The slope is $y' = 3x^2 - 2$, so $y'(2) = 10$:

$$y - 4 = 10(x - 2) \quad\Rightarrow\quad y = 10x - 16$$

**12.** A horizontal tangent has slope $0$:

$$y' = 3x^2 - 6x = 3x(x - 2) = 0 \quad\Rightarrow\quad x = 0 \text{ or } x = 2$$

The points are $(0, 0)$ and $(2, 8 - 12) = (2, -4)$.

## Common mistakes on these problems

- **Going the wrong way with negative exponents.** In problem 4, $x^{-3}$ becomes $-3x^{-4}$, not $-3x^{-2}$. Subtracting 1 always moves left on the number line.
- **Differentiating a root without rewriting.** Write $\sqrt[3]{x^2}$ as $x^{2/3}$ before you start.
- **Differentiating a product factor by factor.** In problem 8, $(2)(1) = 2$ is wrong. Expand or use the product rule.
- **Treating $\pi^2$ like a variable.** In problem 10, $\pi^2$ is a constant, so its derivative is $0$, not $2\pi$.

## Keep practicing

These 12 problems use fixed numbers. For unlimited practice with new numbers every time and instant feedback, try the [power rule lesson](/learn/calculus-1/derivative-rules/power-rule), then the [power rule practice](/learn/calculus-1/derivative-rules/power-rule-practice). For more examples and the reasoning behind the rule, read the [power rule guide](/blog/power-rule). Next, try the [product and quotient rule practice problems](/practice-problems/product-and-quotient-rules).

## Frequently asked questions

### What is the power rule for derivatives?

For any real number $n$, $\frac{d}{dx}x^n = nx^{n-1}$. Bring the exponent down as a coefficient, then lower the exponent by one. It works for whole, negative and fractional exponents.

### How do you use the power rule with square roots?

Rewrite the root as a fractional power, for example $\sqrt{x} = x^{1/2}$ and $\sqrt[3]{x^2} = x^{2/3}$. Then apply the rule: $\frac{d}{dx}x^{1/2} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$.

### Does the power rule work for e to the x?

No. The power rule is for a variable base and a constant exponent, like $x^3$. In $e^x$ the variable is in the exponent, so you use $\frac{d}{dx}e^x = e^x$. For a constant like $\pi^2$, the derivative is $0$.