Blog · Calculus guides · 22 October 2026 · 4 min read

Power Rule: Derivatives, Examples & Practice Problems

Q: Does the power rule work for negative and fractional exponents?

Yes, for any real power n. d/dxx^(-1) = -x^(-2) and d/dxx^(2/3) = 2/3x^(-1/3). Just keep track of the subtraction.

Q: Why does the power rule work?

For a whole number n, expand (x+h)^n: it is x^n + nx^(n-1)h plus terms with h^2 or higher. Subtract x^n, divide by h, and let h→ 0. Everything with an h left in it disappears, and nx^(n-1) is all that survives.

Q: What is the derivative of a constant?

Zero. A constant is c· x^0, and the power rule brings down that 0. Or, more simply: something that never changes has a rate of change of nothing.

Q: What is the derivative of x?

One. x = x^1, so the rule gives 1· x^0 = 1. A straight line y = x has slope 1 everywhere, which is reassuring.

The power rule says the derivative of $x^{n}$ is $n x^{n - 1}$ : bring the power down in front, then lower the power by one. It works for any real power, including fractions and negatives, so it covers roots and $\frac{1}{x}$ -type terms once you rewrite them as powers. For example, $\frac{d}{d x} x^{5} = 5 x^{4}$ and $\frac{d}{d x} x = \frac{1}{2} x^{- 1/2}$ .

The formula

\frac{d}{d x} [x^{n}] = n x^{n - 1} and \frac{d}{d x} [a x^{n}] = an x^{n - 1}

Down in front, down by one. The shortest rule in calculus, and the one doing quiet background work in almost every other rule on this blog.

When to use it

Use the power rule on any term that is a number times $x$ to a fixed power: $7 x^{3}$ , $x^{- 2}$ , $3 x$ , $\frac{4}{x ^{5}}$ . It works term by term across a sum, so a whole polynomial goes in one pass. The power must be a constant and the base must be plain $x$ . If the base is a bracket, as in $(3 x + 1)^{4}$ , you need the chain rule as well. If the power has the variable in it, as in $2^{x}$ or $e^{x}$ , the power rule does not apply at all; those have their own rules.

How to do it, step by step

Rewrite every term as a power of $x$ . Roots become fractional powers ( $x = x^{1/2}$ , $3 x^{2} = x^{2/3}$ ) and anything over $x$ becomes a negative power ( $\frac{2}{x ^{3}} = 2 x^{- 3}$ ).
Bring the power down and multiply. In $4 x^{5}$ , the 5 multiplies the 4 to give 20.
Subtract one from the power. $x^{5}$ becomes $x^{4}$ . For fractions, do the subtraction carefully: $\frac{1}{2} - 1 = - \frac{1}{2}$ .
Deal with the easy terms. $a x$ becomes $a$ , and a constant on its own becomes 0.
Rewrite the answer neatly. Turn negative powers back into fractions and fractional powers back into roots if you like. Either form is correct.

Worked examples

Example 1: A polynomial

Easy

Differentiate:

y = 4 x^{5} - 3 x^{2} + 7 x - 2

Term by term. $4 x^{5}$ gives $4 \cdot 5 x^{4} = 20 x^{4}$ . $- 3 x^{2}$ gives $- 6 x$ . $7 x$ gives $7$ . And $- 2$ , a constant, gives 0: a constant does not change, so its rate of change is nothing.

\frac{d y}{d x} = 20 x^{4} - 6 x + 7 + 0

Answer

\frac{d y}{d x} = 20 x^{4} - 6 x + 7

Example 2: Square roots and cube roots

Medium

Differentiate:

y = 6 x - 3 x

Rewrite the roots as powers first. This is the step that makes or breaks the problem.

y = 6 x^{1/2} - x^{1/3}

Now the rule. For the first term, $6 \cdot \frac{1}{2} = 3$ and $\frac{1}{2} - 1 = - \frac{1}{2}$ . For the second, the power comes down as $\frac{1}{3}$ and $\frac{1}{3} - 1 = - \frac{2}{3}$ .

\frac{d y}{d x} = 3 x^{- 1/2} - \frac{1}{3} x^{- 2/3}

Negative powers move to the bottom of a fraction, so tidy up.

Answer

\frac{d y}{d x} = \frac{3}{x} - \frac{1}{3 x ^{2/3}}

Example 3: Fractions with x on the bottom

Harder

Differentiate:

y = \frac{2}{x ^{3}} + \frac{x ^{2} + 1}{x}

No quotient rule needed. The first term is $2 x^{- 3}$ . For the second, split the fraction and divide each term on top by $x^{1/2}$ , subtracting powers:

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

So $y = 2 x^{- 3} + x^{3/2} + x^{- 1/2}$ , three plain powers. Differentiate each:

\frac{d y}{d x} = - 6 x^{- 4} + \frac{3}{2} x^{1/2} - \frac{1}{2} x^{- 3/2}

The trap here is the first term: the power goes from $- 3$ down to $- 4$ , not up to $- 2$ . Subtracting one from a negative number makes it more negative.

Answer

\frac{d y}{d x} = - \frac{6}{x ^{4}} + \frac{3}{2} x - \frac{1}{2 x ^{3/2}}

Your turn

Three to try, checked by the same machine that grades the game. Type the answer however you like; any correct form counts.

Problem 1

Easy

Differentiate:

y = 3 x^{4} - 2 x^{3} + 5 x - 7

Problem 2

Medium

Differentiate:

y = 4 x + \frac{3}{x ^{2}}

Problem 3

Harder

Differentiate:

y = 3 x^{4} - \frac{6}{3 x}

Common mistakes

Not rewriting roots first. Trying to differentiate $x$ as it stands is where most slips start. In our answer data, players get plain power-rule problems right 85% of the time but square roots only 67%. The rule is the same; the rewriting is the hard part.
Going the wrong way with negative powers. $\frac{d}{d x} x^{- 3} = - 3 x^{- 4}$ . One less than $- 3$ is $- 4$ .
Differentiating $\frac{1}{x ^{3}}$ as $\frac{1}{3 x ^{2}}$ . You cannot apply the rule to the bottom of a fraction. Rewrite as $x^{- 3}$ first, every time.
Using it on a bracket. $\frac{d}{d x} (3 x + 1)^{4}$ is not $4 (3 x + 1)^{3}$ ; the chain rule adds a factor of 3, giving $12 (3 x + 1)^{3}$ .
Using it on $e^{x}$ or $2^{x}$ . The variable is in the exponent, not the base. $\frac{d}{d x} e^{x} = e^{x}$ , not $x e^{x - 1}$ .

Practice it now

Five minutes, no account

These are the first derivatives a Sprint gives you: single powers and polynomials at difficulty 1, a derivative worked out at a point at 1 and 2, then roots and negative powers at 2 and 3. It is the warm-up, so it is also where the time for the hard ones is saved.

Play a Sprint See it on the formula sheet

FAQ

Does the power rule work for negative and fractional exponents?: Yes, for any real power $n$ . $\frac{d}{d x} x^{- 1} = - x^{- 2}$ and $\frac{d}{d x} x^{2/3} = \frac{2}{3} x^{- 1/3}$ . Just keep track of the subtraction.
Why does the power rule work?: For a whole number $n$ , expand $(x + h)^{n}$ : it is $x^{n} + n x^{n - 1} h$ plus terms with $h^{2}$ or higher. Subtract $x^{n}$ , divide by $h$ , and let $h \to 0$ . Everything with an $h$ left in it disappears, and $n x^{n - 1}$ is all that survives.
What is the derivative of a constant?: Zero. A constant is $c \cdot x^{0}$ , and the power rule brings down that 0. Or, more simply: something that never changes has a rate of change of nothing.
What is the power rule for integrals?: The reverse: add one to the power, then divide by the new power. $\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C$ , for every $n$ except $- 1$ , which would divide by zero. That one case is $\int \frac{1}{x} d x = ln ∣ x ∣ + C$ .
What is the derivative of x?: One. $x = x^{1}$ , so the rule gives $1 \cdot x^{0} = 1$ . A straight line $y = x$ has slope 1 everywhere, which is reassuring.
Where can I find power rule practice problems?: Right on this page: "Your turn" has three power rule practice problems, easy, medium and harder, checked as you type by the same checker that grades Asymptote, with the answer and the first step if you miss. For more practice against the clock, a five-minute Sprint is free and needs no account.

Power Rule: Derivatives, Examples & Practice Problems

When to use it

Use the power rule on any term that is a number times

x

to a fixed power:

7 x^{3}

x^{- 2}

3 x

\frac{4}{x ^{5}}

. It works term by term across a sum, so a whole polynomial goes in one pass. The power must be a constant and the base must be plain

x

. If the base is a bracket, as in

(3 x + 1)^{4}

, you need the chain rule as well. If the power has the variable in it, as in

2^{x}

e^{x}

, the power rule does not apply at all; those have their own rules.

How to do it, step by step

Rewrite every term as a power of $x$ . Roots become fractional powers (

x = x^{1/2}

3 x^{2} = x^{2/3}

) and anything over

x

becomes a negative power (

\frac{2}{x ^{3}} = 2 x^{- 3}

Bring the power down and multiply. In

4 x^{5}

, the 5 multiplies the 4 to give 20.

Subtract one from the power.

x^{5}

becomes

x^{4}

. For fractions, do the subtraction carefully:

\frac{1}{2} - 1 = - \frac{1}{2}

Deal with the easy terms.

a x

becomes

a

, and a constant on its own becomes 0.

Rewrite the answer neatly. Turn negative powers back into fractions and fractional powers back into roots if you like. Either form is correct.

Worked examples

Example 1: A polynomial

Easy

Differentiate:

y = 4 x^{5} - 3 x^{2} + 7 x - 2

\frac{d y}{d x} = 20 x^{4} - 6 x + 7 + 0

Answer

\frac{d y}{d x} = 20 x^{4} - 6 x + 7

Example 2: Square roots and cube roots

Medium

Differentiate:

y = 6 x - 3 x

Rewrite the roots as powers first. This is the step that makes or breaks the problem.

y = 6 x^{1/2} - x^{1/3}

Now the rule. For the first term, $6 \cdot \frac{1}{2} = 3$ and $\frac{1}{2} - 1 = - \frac{1}{2}$ . For the second, the power comes down as $\frac{1}{3}$ and $\frac{1}{3} - 1 = - \frac{2}{3}$ .

\frac{d y}{d x} = 3 x^{- 1/2} - \frac{1}{3} x^{- 2/3}

Negative powers move to the bottom of a fraction, so tidy up.

Answer

\frac{d y}{d x} = \frac{3}{x} - \frac{1}{3 x ^{2/3}}

Example 3: Fractions with x on the bottom

Harder

Differentiate:

y = \frac{2}{x ^{3}} + \frac{x ^{2} + 1}{x}

No quotient rule needed. The first term is $2 x^{- 3}$ . For the second, split the fraction and divide each term on top by $x^{1/2}$ , subtracting powers:

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

So $y = 2 x^{- 3} + x^{3/2} + x^{- 1/2}$ , three plain powers. Differentiate each:

\frac{d y}{d x} = - 6 x^{- 4} + \frac{3}{2} x^{1/2} - \frac{1}{2} x^{- 3/2}

The trap here is the first term: the power goes from $- 3$ down to $- 4$ , not up to $- 2$ . Subtracting one from a negative number makes it more negative.

Answer

\frac{d y}{d x} = - \frac{6}{x ^{4}} + \frac{3}{2} x - \frac{1}{2 x ^{3/2}}

Your turn

Three to try, checked by the same machine that grades the game. Type the answer however you like; any correct form counts.

Problem 1

Easy

Differentiate:

y = 3 x^{4} - 2 x^{3} + 5 x - 7

Problem 2

Medium

Differentiate:

y = 4 x + \frac{3}{x ^{2}}

Problem 3

Harder

Differentiate:

y = 3 x^{4} - \frac{6}{3 x}

Common mistakes

Not rewriting roots first. Trying to differentiate

x

as it stands is where most slips start. In our answer data, players get plain power-rule problems right 85% of the time but square roots only 67%. The rule is the same; the rewriting is the hard part.

Going the wrong way with negative powers.

\frac{d}{d x} x^{- 3} = - 3 x^{- 4}

. One less than

- 3

- 4

Differentiating $\frac{1}{x ^{3}}$ as $\frac{1}{3 x ^{2}}$ . You cannot apply the rule to the bottom of a fraction. Rewrite as

x^{- 3}

first, every time.

Using it on a bracket.

\frac{d}{d x} (3 x + 1)^{4}

is not

4 (3 x + 1)^{3}

; the chain rule adds a factor of 3, giving

12 (3 x + 1)^{3}

Using it on $e^{x}$ or $2^{x}$ . The variable is in the exponent, not the base.

\frac{d}{d x} e^{x} = e^{x}

, not

x e^{x - 1}

Practice it now

Five minutes, no account

Play a Sprint See it on the formula sheet

FAQ

Does the power rule work for negative and fractional exponents?

Yes, for any real power

n

\frac{d}{d x} x^{- 1} = - x^{- 2}

and

\frac{d}{d x} x^{2/3} = \frac{2}{3} x^{- 1/3}

. Just keep track of the subtraction.

Why does the power rule work?

For a whole number

n

, expand

(x + h)^{n}

: it is

x^{n} + n x^{n - 1} h

plus terms with

h^{2}

or higher. Subtract

x^{n}

, divide by

h

, and let

h \to 0

. Everything with an

h

left in it disappears, and

n x^{n - 1}

is all that survives.

What is the derivative of a constant?

Zero. A constant is

c \cdot x^{0}

, and the power rule brings down that 0. Or, more simply: something that never changes has a rate of change of nothing.

What is the power rule for integrals?

The reverse: add one to the power, then divide by the new power.

\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C

, for every

n

except

- 1

, which would divide by zero. That one case is

\int \frac{1}{x} d x = ln ∣ x ∣ + C

What is the derivative of x?

One.

x = x^{1}

, so the rule gives

1 \cdot x^{0} = 1

. A straight line

y = x

has slope 1 everywhere, which is reassuring.

Where can I find power rule practice problems?

Right on this page: "Your turn" has three power rule practice problems, easy, medium and harder, checked as you type by the same checker that grades Asymptote, with the answer and the first step if you miss. For more practice against the clock, a five-minute Sprint is free and needs no account.

Power Rule: Derivatives, Examples & Practice Problems

The formula

When to use it

How to do it, step by step

Worked examples

Example 1: A polynomial

Example 2: Square roots and cube roots

Example 3: Fractions with x on the bottom

Your turn

Common mistakes

FAQ

Keep going

Power Rule: Derivatives, Examples & Practice Problems

The formula

When to use it

How to do it, step by step

Worked examples

Example 1: A polynomial

Example 2: Square roots and cube roots

Example 3: Fractions with x on the bottom

Your turn

Common mistakes

FAQ

Keep going