Power Rule for Derivatives: Formula, Proof and 10 Examples

The power rule: for any real constant \( n \),

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

Bring the exponent down in front, then lower the exponent by one. It is the single most used rule in differential calculus.

Almost every derivative you will ever compute uses the power rule somewhere, either directly (polynomials, roots, reciprocals) or as one step inside the chain rule, the product rule and the quotient rule. Learn it well and learn to rewrite functions so it applies, and a large share of first-semester calculus becomes routine.

The intuition: growing squares and cubes

Think of \( x^2 \) as the area of a square with side \( x \). If you lengthen the side by a small amount \( \Delta x \), the square gains two thin strips along two edges, each with area \( x\,\Delta x \), plus a tiny corner square of area \( (\Delta x)^2 \). The corner is negligible when \( \Delta x \) is small, so the area grows at about \( 2x \) per unit of side. That is exactly \( \frac{d}{dx}x^2 = 2x \).

The same picture works for a cube. A cube with side \( x \) has volume \( x^3 \). Grow the side slightly and three thin slabs appear on three faces, each with volume \( x^2\,\Delta x \). The rest is tiny edge and corner pieces. So the volume grows at about \( 3x^2 \), which is the power rule for \( n = 3 \). The exponent comes down because there are \( n \) faces growing at once, and the power drops by one because each face has one dimension fewer than the whole shape.

Why it works (for positive integers)

Expand \( (x + h)^n \) with the binomial theorem:

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

Put it in the limit definition of the derivative. Subtracting \( x^n \) cancels the first term, and every remaining term has at least one factor of \( h \), so you can divide by \( h \):

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

Every term after the first still contains an \( h \), so it vanishes. For fractional and negative exponents the rule follows from logarithmic differentiation: if \( y = x^n \), then \( \ln y = n\ln x \), so \( y'/y = n/x \) and \( y' = nx^{n-1} \).

Rules that travel with it

• Constant rule: \( \frac{d}{dx}c = 0 \)
• Constant multiple: \( \frac{d}{dx}\big(c\,f(x)\big) = c\,f'(x) \)
• Sum rule: differentiate term by term.

Together with the power rule, these three let you differentiate any polynomial in one line. A constant has a flat graph, so its slope is zero. A constant multiple stretches a graph vertically, which multiplies every slope by the same factor. And the slope of a sum is the sum of the slopes, because limits add.

10 worked examples

1. \( \frac{d}{dx}x^5 = 5x^4 \)
2. \( \frac{d}{dx}7x^3 = 21x^2 \)
3. \( \frac{d}{dx}(4x^3 - 5x^2 + 2x - 9) = 12x^2 - 10x + 2 \)
4. \( \frac{d}{dx}\frac{1}{x^3} = \frac{d}{dx}x^{-3} = -3x^{-4} = -\frac{3}{x^4} \)
5. \( \frac{d}{dx}\sqrt{x} = \frac{d}{dx}x^{1/2} = \frac{1}{2\sqrt x} \) (details in derivative of √x)
6. \( \frac{d}{dx}x^{2/3} = \frac23 x^{-1/3} = \frac{2}{3\sqrt[3]{x}} \)
7. \( \frac{d}{dx}\frac{1}{\sqrt x} = \frac{d}{dx}x^{-1/2} = -\frac{1}{2x^{3/2}} \)
8. \( \frac{d}{dx}\frac{x^2 + 1}{x} \): split first into \( x + x^{-1} \), giving \( 1 - \frac{1}{x^2} \)
9. \( \frac{d}{dx}x^{\pi} = \pi x^{\pi - 1} \). Any constant exponent works.
10. \( \frac{d}{dx}(2x+1)^2 \): expand to \( 4x^2 + 4x + 1 \), giving \( 8x + 4 \). (Or use the chain rule: \( 2(2x+1)\cdot2 \).)

A few notes on the reasoning. In example 4 the exponent goes from \( -3 \) to \( -4 \), not to \( -2 \): subtracting one from a negative number makes it more negative. In example 6 the new exponent is \( \frac23 - 1 = -\frac13 \), and a negative exponent means the result belongs in the denominator. Example 8 shows the most useful habit of all: split a fraction with a single-term denominator into separate powers before differentiating.

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 \( \frac52x^{3/2} \), and the second gives \( -2\cdot\frac12x^{-1/2} \):

$$\frac{d}{dx}\left(x^{5/2} - 2x^{1/2}\right) = \frac52x^{3/2} - \frac{1}{\sqrt x}$$

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 is done. Turn every root and fraction into a power of \( x \):

Expression	Rewrite as
\( \sqrt[3]{x^2} \)	\( x^{2/3} \)
\( \frac{5}{x^2} \)	\( 5x^{-2} \)
\( \frac{1}{2\sqrt{x}} \)	\( \tfrac12 x^{-1/2} \)
\( x^2\sqrt{x} \)	\( x^{5/2} \)

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 exponent contains \( x \): \( 2^x \), \( e^{x} \), \( x^x \). Use exponential rules or logarithmic differentiation.
• The base is not just \( x \): \( (3x^2+1)^{10} \) needs the chain rule.
• Products and quotients of functions: use the product rule or quotient rule (or simplify first).

Common mistakes

• Moving the exponent the wrong way. The derivative of \( x^{-3} \) is \( -3x^{-4} \), not \( -3x^{-2} \). Always subtract one, even when the exponent is negative.
• Using the power rule on an exponential. \( \frac{d}{dx}2^x \) is not \( x\cdot2^{x-1} \). The power rule needs a variable base and a constant exponent; \( 2^x \) is the other way around, and its derivative is \( 2^x\ln 2 \).
• Treating a constant like a power. \( \pi^2 \) is just a number, about 9.87, so its derivative is 0, not \( 2\pi \).
• Differentiating a product factor by factor. For \( x^2\cdot x^3 \), multiplying \( 2x \) by \( 3x^2 \) gives \( 6x^3 \), which is wrong. Combine first to \( x^5 \); the derivative is \( 5x^4 \).
• Differentiating a root in the denominator directly. Writing \( \frac{1}{\sqrt x} \) and “differentiating the bottom” goes nowhere. Rewrite it as \( x^{-1/2} \) first.

Where it’s used

The power rule is the engine behind finding critical points, where you set a polynomial derivative equal to zero, and behind tangent lines, linear approximation and Newton’s method. Run it backwards and you get the power rule for integrals, the first antiderivative formula everyone learns. It is also why Taylor series are so convenient: a series is a sum of powers, so you can differentiate it 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 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) \( 24x^3 - 3 \); (2) \( -\frac{10}{x^6} \); (3) \( \frac72x^{5/2} \); (4) \( \frac{1}{3x^{2/3}} \); (5) \( 2x - \frac{2}{x^2} \); (6) \( -4 \); (7) \( 30x^4 \).

Hints: in (3) combine to \( x^{7/2} \) first; in (5) split into \( x^2 + 2x^{-1} \); in (6) the derivative is \( 4x^3 \), and a negative number cubed stays negative.

FAQ

Does the power rule work for n = 0?

Yes. \( x^0 = 1 \) and the rule gives \( 0\cdot x^{-1} = 0 \), consistent with the constant rule (for \( x \neq 0 \)).

Does the power rule work for negative and fractional exponents?

Yes. It holds for every real constant exponent. Rewrite roots as fractional powers and reciprocals as negative powers, then apply the same rule.

What is the derivative of x?

It is 1. Here \( x = x^1 \), so the rule gives \( 1\cdot x^0 = 1 \). That matches the graph: the line \( y = x \) has slope 1 everywhere.

Is there a power rule for integrals?

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

Why is it called the power rule?

Because it applies to power functions, those of the form \( x^n \) with a constant exponent. It is sometimes called the power law or the exponent rule for derivatives.

Further reading

• Differentiation Formulas (Paul’s Online Math Notes) — more power rule examples with roots and negative exponents, plus 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.