CalculusConcept reference

Power rule

The power rule differentiates x to a power as the original exponent times x to one lower power, wherever the stated real-domain rule applies.

On this page 8 sections

01Understand the idea#

The power rule gives a shortcut for $f(x)=x^n$ : use the original exponent as a multiplier, then subtract one from the exponent.

f\prime(x)=nx^{n-1}

For positive integers, this works at every real input.

For $f(x)=x^3$ , expansion gives $f(x+h)-f(x)=3x^2h+3xh^2+h^3$ . After division by nonzero $h$ :

The difference quotient is

3x^2+3xh+h^2

As $h\to0$ , the last two terms vanish, leaving $3x^2$ .

Check your reasoning

$f(x)=x^{4}$ . Find $f\prime(x)$ .

Show answer and explanation

$4x^{3}$

Multiplier $4$ ; new exponent $3$ .

Negative integer powers follow the same rule where $x\ne0$ . For $f(x)=x^{-2}$ , the multiplier is $-2$ and the new exponent is $-2-1=-3$ .

f\prime(x)=-2x^{-3}

Zero is excluded because the original reciprocal power is undefined there.

Check your reasoning

A response is $r(x)=x^{3}$ . Find $r\prime(-2)$ .

Show answer and explanation

Use $3x^{2}$ at $x=-2$ .

For $x>0$ , write $x^{1/2}=\sqrt{x}$ and $x^{3/2}=(\sqrt{x})^3$ . The rule also works for these fractional exponents.

\frac{d}{dx}x^{1/2}=\frac12x^{-1/2}

The new exponent is $-1/2$ : here $x^{-1/2}=1/\sqrt{x}$ . Use this formula for $x>0$ .

Check your reasoning

$f(x)=x^{-2}$ , $x\ne0$ . Claim: $f\prime(0)=0$ . Repair.

Show answer and explanation

No derivative: $f(0)$ is undefined.

Zero is outside the domain.

For $x^5$ , keep both operations: multiplier $5$ , exponent $4$ . Dropping either changes the derivative.

Key takeaway