Power Rule for Derivatives: Formula, Proof and 10 Examples

The power rule for derivatives applies to xⁿ. Here, n is any real constant exponent. We find the derivative of xⁿ with respect to x as d/dx(xⁿ) = n x^n-1. To differentiate x to a power, bring the exponent to the front. This becomes a multiplying coefficient. Then, decrease the original power by one.

This rule applies widely. For example, n can be a positive integer or a negative integer. It can also be a rational fraction or an irrational real number like π. This is the most used formula in differential calculus. It appears directly in polynomial problems. Also, it serves as a core step inside the product rule, quotient rule, and chain rule. You can make evaluating derivatives routine. To do this, learn to rewrite algebraic terms into clean exponent powers.

Applying the power rule derivative

The standard formula operates on power functions. These functions have a variable base x. Also, they have a fixed constant exponent n.

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

Applying the power rule involves three basic steps:

1. Identify the constant exponent: Locate the power n attached to the base variable x.
2. Multiply by the exponent: Bring n to the front of the term as a coefficient.
3. Decrease the power by one: Subtract 1 from the original exponent to form the new power, n – 1.

For example, differentiate x⁴ using the power rule derivative. First, bring 4 to the front. Then, subtract 1 from the exponent. This gives you 4x³. However, a constant multiplier c may sit in front of the term. For instance, consider c·xⁿ. You should multiply c by n. This results in c·n·x^n-1.

Geometric intuition: growing squares and cubes

Consider x² as a square’s area. Its side length is x. You can lengthen each side by a tiny amount, Δx. This adds two thin strips along adjacent edges. Each strip has an area of xΔx. Also, a tiny corner square appears with an area of (Δx)². As Δx approaches zero, the corner square becomes negligible. As a result, the area growth rate is 2x. This matches d/dx(x²) = 2x, a result of the power rule derivative.

Similarly, think of a cube with volume x³. You can increase each edge slightly. This adds three flat slabs on three faces. Each slab has a volume of x²Δx. There are also negligible edge and corner pieces. The instantaneous volume growth rate is 3x². So, d/dx(x³) = 3x², which is an example of the power rule derivative. The exponent comes down because there are n growing faces. The power drops by one because each face has one lower dimension than the solid figure.

Proofs for the power rule

The power rule holds true for every real constant exponent n. You can understand why it works by examining two proofs. First, examine one for positive integers using limits. A second proof for all real numbers additionally uses logarithms.

Proof for positive integer exponents using the limit definition

Start with the limit definition of the derivative for f(x) = xⁿ where n is a positive integer:

$$f'(x) = \lim_{h \to 0} \frac{(x + h)^n - x^n}{h}$$

Expand (x + h)ⁿ using the Binomial Theorem:

$$(x+h)^n = x^n + n x^{n-1}h + \binom{n}{2}x^{n-2}h^2 + \dots + h^n$$

Substitute this expansion back into the limit difference quotient:

$$\begin{aligned}\lim_{h\to0}\frac{(x+h)^n - x^n}{h} &= \lim_{h\to 0}\frac{x^n + n x^{n-1}h + \binom{n}{2}x^{n-2}h^2 + \dots + h^n - x^n}{h} &= \lim_{h\to 0}\left(nx^{n-1} + \binom{n}{2}x^{n-2}h + \dots + h^{n-1}\right) &= nx^{n-1}\end{aligned}$$

First, subtract xⁿ. This cancels the first term. All remaining terms contain at least one factor of h. Next, divide by h. This removes one factor of h from every term. As h approaches zero, every term with h vanishes. Therefore, n x^n-1 remains, which is the power rule derivative.

Proof for all real exponents using logarithmic differentiation

You cannot use binomial expansion directly for rational, negative, or irrational real exponents like π. Instead, use logarithmic differentiation. Let y = xⁿ (for x > 0). Then take the natural logarithm of both sides.

$$\ln(y) = \ln(x^n) = n \ln(x)$$

Differentiate both sides with respect to x using implicit differentiation:

$$\frac{1}{y} \frac{dy}{dx} = n \cdot \frac{1}{x}$$

Multiply both sides by y = xⁿ to solve for dy/dx:

$$\frac{dy}{dx} = y \cdot \frac{n}{x} = x^n \cdot \frac{n}{x} = n x^{n-1}$$

This confirms that the power rule applies universally across all real constant exponents.

Rules that travel with the power rule

You can combine the power rule with three basic linearity rules. This helps you differentiate complex polynomials. You can also differentiate algebraic expressions line by line.

• Constant Rule: The derivative of any constant is zero (d/dx(c) = 0). A constant function is a horizontal line with a slope of zero everywhere.
• Constant Multiple Rule: d/dx[c · f(x)] = c · f'(x). A constant multiplier scales the function vertically, multiplying its slope by the same factor.
• Sum and Difference Rule: d/dx[f(x) ± g(x)] = f'(x) ± g'(x). Because limits add, you can differentiate terms individually.

10 worked examples

Example 1: Basic integer exponent
Find d/dx(x⁵).
Apply the power rule directly with n = 5:
$$\frac{d}{dx} x^5 = 5 x^{5-1} = 5x^4$$

Example 2: Constant coefficient
Find d/dx(7x³).
Multiply the constant coefficient 7 by 3:
$$\frac{d}{dx}(7x^3) = 7 \cdot 3x^{3-1} = 21x^2$$

Example 3: Polynomial function
Find d/dx(4x³ – 5x² + 2x – 9).
Differentiate term by term:
$$\frac{d}{dx}(4x^3 - 5x^2 + 2x - 9) = 12x^2 - 10x + 2$$

Example 4: Reciprocal function with a negative exponent
Find d/dx(1/x³). First, rewrite 1/x³ as x^-3.
Then, you can differentiate it using the power rule derivative.
$$\frac{d}{dx} \left(\frac{1}{x^3}\right) = \frac{d}{dx} x^{-3} = -3x^{-3-1} = -3x^{-4} = -\frac{3}{x^4}$$
Subtracting 1 from -3 yields -4. So, the exponent becomes more negative.

Example 5: Square root function
Find d/dx(√x).
Rewrite √x as x^1/2 (details in derivative of √x):
$$\frac{d}{dx} \sqrt{x} = \frac{d}{dx} x^{1/2} = \frac{1}{2}x^{1/2 - 1} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$$

Example 6: Fractional rational exponent
Find d/dx(x^2/3).
Bring 2/3 down as a coefficient and subtract 1 from 2/3, giving -1/3:
$$\frac{d}{dx} x^{2/3} = \frac{2}{3} x^{-1/3} = \frac{2}{3\sqrt[3]{x}}$$

Radical in denominator

Example 7: Radical in a denominator
Find d/dx(1/√x).
Rewrite 1/√x as x^-1/2:
$$\frac{d}{dx} \left(\frac{1}{\sqrt{x}}\right) = \frac{d}{dx} x^{-1/2} = -\frac{1}{2}x^{-3/2} = -\frac{1}{2x^{3/2}}$$

Example 8: Rational expression with a single denominator
Find d/dx((x² + 1)/x).
Split the fraction before you take the derivative:
$$\frac{x^2 + 1}{x} = x + x^{-1} \quad \Rightarrow \quad \frac{d}{dx}(x + x^{-1}) = 1 - x^{-2} = 1 - \frac{1}{x^2}$$
Splitting single-term denominators avoids unnecessary quotient rule calculations. As a result, this simplifies your work for a power rule derivative.

Example 9: Irrational constant exponent
Find d/dx(x^π).
Because π is a constant real number, apply the standard power rule formula:
$$\frac{d}{dx} x^{\pi} = \pi x^{\pi - 1}$$

Example 10: Expanded binomial quadratic
You can find d/dx((2x + 1)²).
First, expand it to 4x² + 4x + 1.
Then, differentiate it term by term using the power rule derivative. As a result, you get:
$$\frac{d}{dx}(4x^2 + 4x + 1) = 8x + 4$$
Alternatively, apply the chain rule. This gives you 2(2x + 1) · 2 = 8x + 4.

Exam-level examples

11. A quotient that is secretly a sum of powers. Differentiate \(\frac{x^3 - 2x}{\sqrt{x}}\). Divide each term of the numerator by \(x^{1/2}\) first:

$$\frac{x^3 - 2x}{\sqrt x} = x^{5/2} - 2x^{1/2}$$

Now apply the power rule to each term. The first gives \(\frac{5}{2}x^{3/2}\), and the second gives \(-2 \cdot \frac{1}{2}x^{-1/2}\):

$$\frac{d}{dx}\left(x^{5/2} - 2x^{1/2}\right) = \frac{5}{2}x^{3/2} - \frac{1}{\sqrt x}$$

No quotient rule needed, and the answer is already simplified.

12. A tangent line. Find the tangent line to \(y = x^3 - 2x\) at \(x = 2\). The point is \((2, 4)\), since \(8 - 4 = 4\). The derivative is \(3x^2 - 2\), so the slope at \(x = 2\) is \(12 - 2 = 10\). Point-slope form gives

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

The tangent line calculator will draw this for you if you want to see it.

13. Velocity from position. A particle moves along a line with position \(s(t) = t^3 - 6t^2 + 9t\). Its velocity is the derivative:

$$v(t) = 3t^2 - 12t + 9 = 3(t - 1)(t - 3)$$

The particle is momentarily at rest when \(v = 0\), which happens at \(t = 1\) and \(t = 3\). Differentiating once more gives the acceleration \(a(t) = 6t - 12\). Every step was the power rule.

Rewrite before you differentiate

Most power-rule errors happen before any calculus. Therefore, turn every root and fraction into a power of x:

For instance, \(\frac{5}{x^2} = 5x^{-2}\) differentiates to \(-10x^{-3}\), which you can write back as \(-\frac{10}{x^3}\). Rewrite, differentiate, then rewrite again in the form your teacher expects.

When the power rule does NOT apply

The power rule requires a variable base and a constant exponent. It does not apply in these cases:

• The exponent contains a variable: Expressions like 2^x, e^x, or x^x are exponential or variable-power functions. Use exponential derivative rules or logarithmic differentiation.
• The base is a composite function: Differentiating (3x² + 1)¹⁰ requires the chain rule to account for the inner function derivative.
• Products and quotients of functions: Differentiating (x² + 1)(x³ – 4) requires expanding the product first or applying the product rule.

Common mistakes

• Moving negative exponents the wrong way: The derivative of x^-3 is -3x^-4, not -3x^-2. Always subtract 1 from the exponent.
• Using the power rule on an exponential function: d/dx(2^x) is not x · 2^x-1. The power rule requires a constant exponent; 2^x has a variable exponent, so its derivative is 2^x ln(2).
• Treating a constant like a variable power: Terms like π² or e³ are constant numbers. So, their derivative is 0. It is not 2π or 3e².
• Differentiating a product term by term: For x² · x³, do not multiply 2x by 3x². Combine powers first to get x⁵, whose derivative is 5x⁴.
• Differentiating a denominator term directly: Writing 1/√x and differentiating the bottom gives an incorrect result. Rewrite it as x^-1/2 before applying the rule.

Where it’s used

You use the power rule derivative to find critical points. You also use it for tangent lines, linear approximation, and Newton’s method. Reversing it gives you the power rule for integrals. In addition, it lets you differentiate Taylor series term by term.

Try it yourself

For any expression, including ones that mix the power rule with other rules, the derivative calculator shows each step.

Practice problems

Try these practice problems before checking the answers.

1. \(\frac{d}{dx}\left(6x^4 - 3x + 8\right)\)
2. \(\frac{d}{dx}\frac{2}{x^5}\)
3. \(\frac{d}{dx} x^3\sqrt{x}\)
4. \(\frac{d}{dx}\sqrt[3]{x}\)
5. \(\frac{d}{dx}\frac{x^4 + 2x}{x^2}\)
6. The slope of \(y = x^4\) at \(x = -1\)
7. The second derivative of \(x^6\)

Answers:

1. 24x³ – 3. Differentiate term by term: 6(4x³) – 3(1) + 0 = 24x³ – 3.
2. \(-\frac{10}{x^6}\). Rewrite as 2x^-5, then differentiate: 2(-5x^-6) = -10x^-6.
3. \(\frac{7}{2}x^{5/2}\). Combine powers first to get x^7/2, then differentiate to get \(\frac{7}{2}x^{5/2}\).
4. \(\frac{1}{3x^{2/3}}\). Rewrite as x^1/3; the derivative is \(\frac{1}{3}x^{-2/3} = \frac{1}{3x^{2/3}}\).
5. \(2x - \frac{2}{x^2}\). Split the quotient into \(x^2 + 2x^{-1}\) before taking the derivative.
6. -4. The derivative is 4x³; substituting \(x = -1\) yields \(4(-1)^3 = -4\).
7. 30x⁴. The first derivative is 6x⁵, and differentiating a second time yields 30x⁴.

Frequently asked questions

Does the power rule work for n = 0?

Yes. When n = 0, x⁰ = 1 (for x ≠ 0), and the rule yields 0 · x^-1 = 0, which matches the constant rule.

Does the power rule work for negative and fractional exponents?

Yes. It holds for every real constant exponent. Rewrite radicals as fractional powers and reciprocals as negative powers, then apply the standard formula.

What is the derivative of x?

The derivative of x is 1. Since x = x¹, the power rule gives 1 · x⁰ = 1, matching the slope of the line y = x.

Is there a power rule for integrals?

Yes, integration reverses the rule: \(\int x^n dx = \frac{x^{n+1}}{n+1} + C\) for \(n \neq -1\). The exception \(n = -1\) gives \(\ln|x|\); see integral of 1/x.

Why is it called the power rule?

We call it the power rule because it applies to power functions. These functions have the form xⁿ. They feature a variable base and a constant exponent.

Further reading

• Differentiation Formulas (Paul’s Online Math Notes) — This resource offers more power rule examples. These examples include roots and negative exponents. It also covers the sum and constant rules.
• Differentiation Rules (OpenStax Calculus Volume 1, Section 3.3) — a textbook treatment with proofs and exercises.
• Power rule (Wikipedia) — proofs for integer, rational and real exponents and some history of the rule.