Why do we subtract 1 from the exponent in the Power Rule?

The Power Rule comes from applying the limit definition of the derivative to x^n. When you expand (x + h)^n using the binomial theorem and take the limit as h \to 0, all terms with h vanish, and you are left with nx^{n-1}. The -1 is not arbitrary—it emerges directly from the algebra of the binomial expansion.

Does the Power Rule work for negative exponents?

Yes, without exception. For example, \frac{d}{dx}x^{-2} = -2x^{-3}, or equivalently -\frac{2}{x^3}. You can also think of x^{-2} as \frac{1}{x^2} and verify using the quotient rule, but the Power Rule is faster and always correct for any real exponent.

What happens if the exponent is zero?

If f(x) = x^0 = 1 (a constant), the Power Rule gives f'(x) = 0 \cdot x^{-1} = 0. This is correct—the derivative of the constant function 1 is indeed zero everywhere. The Power Rule seamlessly includes this boundary case.

Can I use the Power Rule on 2x + 3?

Yes. Rewrite the expression as 2x^1 + 3x^0 to see the exponents explicitly. Then apply the Power Rule: 2 \cdot 1 \cdot x^0 + 3 \cdot 0 \cdot x^{-1} = 2 + 0 = 2. The Power Rule handles linear functions perfectly; it recovers the familiar slope.

Power Rule

The Power Rule states that the derivative of $x^n$ is $nx^{n-1}$ , making it the essential tool for differentiating any polynomial expression.

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

Solve a problem with Power Rule

Type the problem. The solver will use Power Rule where Power Rule is the right tool, and tell you when it is not.

What each symbol means

What Power Rule takes

Symbol	Meaning
$n$	The exponent (power) in $x^n$ ; a real-number constant that determines the degree of the term; if confused with the coefficient multiplying $x^n$ , you will apply the rule backwards.
$x$	The independent variable of the function being differentiated; if treated as a constant or confused with the exponent $n$ , the rule fails to produce the correct derivative.

When to use it

Reach for the Power Rule whenever you need to differentiate a term of the form $x^n$ where $n$ is a constant.

Level

Usually taught in: Calculus I · Appears on: AP Calculus

Worked examples

1. Differentiate a simple power

Problem

Find the derivative of

f(x) = x^5

$\frac{d}{dx}x^5$
We want to find the derivative of $x^5$ using the Power Rule.
$5 \cdot x^{5-1}$
Apply the Power Rule: bring down the exponent $n = 5$ and reduce the power by 1.
$5x^4$
Simplify the exponent: $5 - 1 = 4$ .

Answer: $5x^4$

This is a straightforward application of the Power Rule. We multiplied the coefficient (which is 1) by the exponent 5 and reduced the power from 5 to 4. This is the most direct use case for the rule.

2. Differentiate with negative and fractional exponents

Problem

Find the derivative of

g(x) = x^{-3} + 4x^{1/2}

$\frac{d}{dx}x^{-3} + \frac{d}{dx}(4x^{1/2})$
The derivative of a sum is the sum of derivatives, so we apply the Power Rule to each term separately.
$(-3) \cdot x^{-3-1} + 4 \cdot \frac{1}{2} \cdot x^{1/2 - 1}$
For the first term, $n = -3$ ; for the second, bring down $\frac{1}{2}$ while keeping the coefficient 4.
$-3x^{-4} + 2x^{-1/2}$
Simplify: $-3 \cdot x^{-4} = -3x^{-4}$ and $4 \cdot \frac{1}{2} = 2$ , giving $2x^{-1/2}$ .

Answer: $-3x^{-4} + 2x^{-1/2}$

This example shows that the Power Rule applies to negative exponents and fractional exponents alike. Many students worry these cases are different, but the rule works identically. Notice that we separately apply the rule to each term of the sum and keep coefficients in front.

3. Find a rate of change in a basketball context

Problem

A player's total points accumulated through

t

games in a season is modeled by

P(t) = 18t^2 + 5t

. What is the rate of change of total points when the player is in their 3rd game?

$P'(t) = \frac{d}{dt}(18t^2 + 5t)$
To find the instantaneous rate of change (points per game), we compute the derivative.
$18 \cdot 2 \cdot t^{2-1} + 5 \cdot 1 \cdot t^{1-1}$
Apply the Power Rule to each term: for $18t^2$ , the exponent is $n = 2$ ; for $5t$ , the exponent is $n = 1$ .
$36t + 5$
Simplify: $18 \cdot 2 = 36$ , $t^1 = t$ , and $t^0 = 1$ , so $5 \cdot 1 \cdot t^0 = 5$ .
$P'(3) = 36(3) + 5 = 108 + 5 = 113$
Substitute $t = 3$ to find the rate at the 3rd game: $36 \times 3 = 108$ , and $108 + 5 = 113$ .

Answer: $P'(3) = 113 \text{ points per game}$

In a real context, derivatives tell us how a quantity is changing at a specific instant. Here, after 3 games, the player is accumulating points at a rate of 113 points per game—in other words, the model predicts that at game 3, adding one more game would add roughly 113 points to the season total. This shows the Power Rule's use outside pure mathematics.

Common mistakes

Where Power Rule usually goes wrong

The mistake	Why it is wrong	The fix
$\frac{d}{dx}x^3 = 3x^3$	The exponent was brought down but not subtracted from the power; this violates the second, essential part of the Power Rule.	Always subtract 1 from the exponent: the correct derivative is $\frac{d}{dx}x^3 = 3 \cdot x^{3-1} = 3x^2$ .
$\frac{d}{dx}(5x^4) = 20x^4$	The coefficient 5 and exponent 4 were multiplied correctly, but the exponent in the final answer was not reduced by 1.	The coefficient multiplies the exponent, but the power itself must drop by 1: $\frac{d}{dx}(5x^4) = 5 \cdot 4x^{4-1} = 20x^3$ .
$\frac{d}{dx}x^{1/2} = \frac{1}{2}x^{1/2}$	The fractional exponent $\frac{1}{2}$ was brought down but not subtracted from itself; many students think fractions are exempt.	The Power Rule applies to all real exponents: $\frac{d}{dx}x^{1/2} = \frac{1}{2} \cdot x^{1/2 - 1} = \frac{1}{2}x^{-1/2}$ .

Tips and when to use something else

The Power Rule works for any constant exponent $n$ , including negative numbers, fractions, and zero; if $n$ itself depends on $x$ (like $x^x$ ), you will need logarithmic differentiation instead.
Always perform both steps: multiply by the exponent AND subtract 1 from it; skipping the second step is the most common mistake.
When you see $e^x$ , $\sin(x)$ , $\log(x)$ , or any other non-power function, the Power Rule does not apply; these special functions have their own derivative formulas.
For negative exponents, it often helps to rewrite $x^{-n}$ as $\frac{1}{x^n}$ before differentiating; you will still get the same answer, but it may feel more intuitive.

Related formulas

Definition of the DerivativeThe Definition of the Derivative finds the instantaneous rate of change at a point using limits; the foundation for all derivative calculations.Chain RuleThe Chain Rule tells you how to differentiate a composite function—when one function is plugged into another—by multiplying the outer and inner derivatives.Product RuleProduct Rule lets you find the derivative of a product of two functions without multiplying them out first; use it when the product is complex.Higher Order DerivativesHigher order derivatives describe how the rate of change is changing; use them to find concavity and acceleration in functions.Implicit DifferentiationUse implicit differentiation to find dy/dx when an equation defines y implicitly instead of as an explicit function of x.

Frequently asked questions

Why do we subtract 1 from the exponent in the Power Rule?: The Power Rule comes from applying the limit definition of the derivative to $x^n$ . When you expand $(x + h)^n$ using the binomial theorem and take the limit as $h \to 0$ , all terms with $h$ vanish, and you are left with $nx^{n-1}$ . The $-1$ is not arbitrary—it emerges directly from the algebra of the binomial expansion.
Does the Power Rule work for negative exponents?: Yes, without exception. For example, $\frac{d}{dx}x^{-2} = -2x^{-3}$ , or equivalently $-\frac{2}{x^3}$ . You can also think of $x^{-2}$ as $\frac{1}{x^2}$ and verify using the quotient rule, but the Power Rule is faster and always correct for any real exponent.
What happens if the exponent is zero?: If $f(x) = x^0 = 1$ (a constant), the Power Rule gives $f'(x) = 0 \cdot x^{-1} = 0$ . This is correct—the derivative of the constant function 1 is indeed zero everywhere. The Power Rule seamlessly includes this boundary case.
Can I use the Power Rule on $2x + 3$ ?: Yes. Rewrite the expression as $2x^1 + 3x^0$ to see the exponents explicitly. Then apply the Power Rule: $2 \cdot 1 \cdot x^0 + 3 \cdot 0 \cdot x^{-1} = 2 + 0 = 2$ . The Power Rule handles linear functions perfectly; it recovers the familiar slope.

Need a different method?

The full solver is not scoped to one formula — type any problem and it will pick the method.

Open the math solver

Reviewed 2026-09-18