Applying the Power Rule (AP Calculus AB 2.5)

Run the limit definition on $x^2$ and the answer is $2x$ ; run it on $x^3$ and the answer is $3x^2$ . (Topic 2.2's balloon, $V = \frac{4}{3}\pi r^3$ , came out to $\frac{dV}{dr} = 4\pi r^2$ , which is $\frac{4}{3}\pi$ times $3r^2$ .) Each time, the exponent comes down in front and the new exponent is one less. That pattern is the power rule, and it replaces the limit computation for every function of the form $x^r$ .

Why it works for whole-number exponents

Take a positive integer $n$ and use the form of the definition in which $x$ approaches a fixed point $a$ (topic 2.1):

f'(a) = \lim_{x \to a}\frac{x^n - a^n}{x - a}.

The numerator always has $x - a$ as a factor. Multiplying out shows that

x^n - a^n = (x - a)\left(x^{n - 1} + x^{n - 2}a + x^{n - 3}a^2 + \cdots + x\,a^{n - 2} + a^{n - 1}\right),

because in the product every middle term cancels with a neighbor, leaving only $x^n$ and $-a^n$ . (For $n = 2$ this is $x^2 - a^2 = (x - a)(x + a)$ , and for $n = 3$ it is $x^3 - a^3 = (x - a)(x^2 + xa + a^2)$ .) Cancel the factor $x - a$ , which is allowed since $x \neq a$ inside the limit:

f'(a) = \lim_{x \to a}\left(x^{n - 1} + x^{n - 2}a + \cdots + a^{n - 1}\right).

The bracket has $n$ terms, and as $x \to a$ each one becomes $a^{n - 1}$ . So the limit is $n\,a^{n - 1}$ . Since $a$ was any number, $\frac{d}{dx}[x^n] = n\,x^{n - 1}$ .

The small cases make sense on their own. A constant function $f(x) = c$ has a horizontal graph, so its slope is $0$ everywhere; the difference quotient is $\frac{c - c}{h} = 0$ for every $h$ . The function $f(x) = x$ is a line of slope 1, and the rule gives $1 \cdot x^0 = 1$ .

Every other exponent

The rule holds for negative, fractional and irrational exponents too. Two cases can be checked against the definition right now:

$\frac{1}{x} = x^{-1}$ . The rule predicts $-1 \cdot x^{-2} = -\frac{1}{x^2}$ . From the definition, for $x \neq 0$ , combining the fractions gives $\frac{1}{h}\left(\frac{1}{x + h} - \frac{1}{x}\right) = \frac{1}{h}\cdot\frac{x - (x + h)}{x(x + h)} = \frac{-1}{x(x + h)}$ , which tends to $-\frac{1}{x^2}$ as $h \to 0$ .
$\sqrt{x} = x^{1/2}$ . The rule predicts $\frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$ . From the definition, for $x > 0$ , multiplying by the conjugate gives $\frac{\sqrt{x + h} - \sqrt{x}}{h} = \frac{(x + h) - x}{h\left(\sqrt{x + h} + \sqrt{x}\right)} = \frac{1}{\sqrt{x + h} + \sqrt{x}}$ , which tends to $\frac{1}{2\sqrt{x}}$ as $h \to 0$ . At $x = 9$ this is $\frac{1}{6}$ , the value topic 2.1 found.

A proof for every real exponent uses tools from Unit 3, but the rule itself is stated for all $r$ and used that way from now on.

Rewrite first, then differentiate

The power rule only applies to a single power of $x$ , so roots, reciprocals and products of powers must be rewritten as $x^r$ first. These identities do all the work:

Form	Rewritten
$\sqrt[n]{x^m}$	$x^{m/n}$
$\dfrac{1}{x^n}$	$x^{-n}$
$x^p \cdot x^q$	$x^{p + q}$
$\dfrac{x^p}{x^q}$	$x^{p - q}$

After differentiating, rewriting the answer back with roots and positive exponents is optional on the exam; $-\frac{1}{2}x^{-3/2}$ and $-\frac{1}{2x\sqrt{x}}$ are both correct.

Where the rule's answer is undefined

The derivative formula $r\,x^{r - 1}$ can be undefined at a point where $x^r$ itself is fine, and when it is, the function has no derivative there. For $x^{1/3}$ , the rule gives $\frac{1}{3}x^{-2/3} = \frac{1}{3\sqrt[3]{x^2}}$ , which is undefined at $x = 0$ : that is the vertical tangent of topic 2.4. For $x^{2/3}$ , the rule gives $\frac{2}{3}x^{-1/3}$ , undefined at $0$ : the cusp. For $\sqrt{x}$ , the derivative $\frac{1}{2\sqrt{x}}$ exists only for $x > 0$ , so $\sqrt{x}$ is not differentiable at the endpoint $0$ of its domain.

What the power rule does not cover

The rule needs the variable in the base and a constant in the exponent. It does not apply to:

Exponential functions, with the variable in the exponent: $2^x$ and $e^x$ are not powers of $x$ . Topic 2.7 handles $e^x$ .
Constants that look like powers: $\pi^3$ and $5^2$ are numbers, so their derivatives are $0$ .
Powers of other expressions: $(2x + 1)^3$ is a power of $2x + 1$ , and its derivative needs the chain rule (topic 3.1), which supplies a factor the power rule alone would miss.

Coefficients and sums, as in $4x^3 - 7\sqrt{x}$ , are handled term by term with the rules of topic 2.6.

Worked examples

Worked exampleExample 1: rewriting and differentiating

Find each derivative.

(a) $\dfrac{d}{dx}\left[\dfrac{1}{x^4}\right]$ (b) $\dfrac{d}{dx}\left[\sqrt[3]{x^5}\right]$ (c) $\dfrac{d}{dx}\left[\dfrac{1}{\sqrt{x}}\right]$ (d) $\dfrac{d}{dx}\left[x\sqrt{x}\right]$ (e) $\dfrac{d}{dx}\left[\pi^3\right]$

Show solution

(a) $\frac{1}{x^4} = x^{-4}$ , and $-4 - 1 = -5$ :

\frac{d}{dx}\left[x^{-4}\right] = -4x^{-5} = -\frac{4}{x^5}.

(b) $\sqrt[3]{x^5} = x^{5/3}$ , and $\frac{5}{3} - 1 = \frac{2}{3}$ :

\frac{d}{dx}\left[x^{5/3}\right] = \frac{5}{3}x^{2/3} = \frac{5}{3}\sqrt[3]{x^2}.

\frac{d}{dx}\left[x^{-1/2}\right] = -\frac{1}{2}x^{-3/2} = -\frac{1}{2x\sqrt{x}}.

(d) $x\sqrt{x} = x^1 \cdot x^{1/2} = x^{3/2}$ . One power, so no product rule is needed:

\frac{d}{dx}\left[x^{3/2}\right] = \frac{3}{2}x^{1/2} = \frac{3}{2}\sqrt{x}.

(e) $\pi^3$ is a constant (about 31.006), so its derivative is $0$ .

The step that decides each answer is the subtraction $r - 1$ . With a negative or fractional exponent, write it out rather than doing it in your head.

Worked exampleExample 2: a family of tangent lines

Let $f(x) = \dfrac{1}{x^2}$ .

(a) Write an equation for the line tangent to the graph of $f$ at $x = 1$ , and find where it crosses the $x$ -axis.

(b) Show that for every $a > 0$ , the line tangent to the graph of $f$ at $x = a$ crosses the $x$ -axis at $x = \frac{3a}{2}$ .

Show solution

Rewrite $f(x) = x^{-2}$ , so $f'(x) = -2x^{-3} = -\frac{2}{x^3}$ .

(a) $f(1) = 1$ and $f'(1) = -2$ , so the tangent line is $y = 1 - 2(x - 1)$ . Setting $y = 0$ gives $x - 1 = \frac{1}{2}$ , so it crosses at $x = \frac{3}{2}$ .

(b) At $x = a$ , the point is $\left(a, \frac{1}{a^2}\right)$ and the slope is $-\frac{2}{a^3}$ :

y = \frac{1}{a^2} - \frac{2}{a^3}(x - a).

Setting $y = 0$ and multiplying both sides by $a^3$ gives $0 = a - 2(x - a)$ , so $x - a = \frac{a}{2}$ and $x = \frac{3a}{2}$ .

Part (a) is the case $a = 1$ . The general version is only possible because the power rule gives $f'$ at every point at once; with the limit definition you would redo the limit for each $a$ .

Worked exampleExample 3: limits and exponents (exam level, no calculator)

(a) What is $\displaystyle\lim_{h \to 0}\frac{\sqrt[3]{8 + h} - 2}{h}$ ?

(A) $0$ (B) $\dfrac{1}{12}$ (C) $\dfrac{1}{3}$ (D) The limit does not exist.

(b) Let $g(x) = x^{4/3}$ and $k(x) = x^{1/3}$ . Is $g$ differentiable at $x = 0$ ? Is $k$ ? Justify your answers.

Show solution

(a) Since $\sqrt[3]{8} = 2$ , the limit has the form $\frac{f(8 + h) - f(8)}{h}$ for $f(x) = x^{1/3}$ , so it equals $f'(8)$ (topic 2.1). By the power rule, $f'(x) = \frac{1}{3}x^{-2/3}$ , and $8^{-2/3} = \frac{1}{8^{2/3}} = \frac{1}{4}$ , so

f'(8) = \frac{1}{3}\cdot\frac{1}{4} = \frac{1}{12}.

The answer is (B). Choice (C) stops at the coefficient $\frac{1}{3}$ and forgets to evaluate $x^{-2/3}$ at 8; choice (D) comes from seeing $\frac{0}{0}$ and stopping.

(b) For $g$ , the power rule gives $g'(x) = \frac{4}{3}x^{1/3}$ , which is defined at $0$ : $g'(0) = 0$ . To confirm directly, the difference quotient at $0$ is $\frac{h^{4/3} - 0}{h} = h^{1/3}$ , which tends to $0$ as $h \to 0$ . So $g$ is differentiable at $x = 0$ , with a horizontal tangent there.

For $k$ , the difference quotient at $0$ is $\frac{h^{1/3}}{h} = \frac{1}{h^{2/3}}$ , which grows without bound as $h \to 0$ . So $k$ is not differentiable at $x = 0$ ; its graph has a vertical tangent there. The power rule's formula $\frac{1}{3}x^{-2/3}$ agrees: it is undefined at $0$ .

The exponent decides the outcome. With $r > 1$ , the new exponent $r - 1$ is positive and $x^{r - 1}$ is $0$ at $x = 0$ ; with $0 < r < 1$ , the new exponent is negative and $x^{r - 1}$ blows up there.

Where students lose points

On the exam

Questions that test the power rule directly present a root or reciprocal that must be rewritten first, or a limit in the shape of the derivative definition whose value the power rule gives at once, as in Example 3(a). On free-response work, an unsimplified derivative such as $-\frac{1}{2}x^{-3/2}$ is fully acceptable.

Check yourself

Question 1
$\dfrac{d}{dx}\left[\dfrac{1}{x^2\sqrt[3]{x}}\right] =$
1. A
  $-\dfrac{7}{3}x^{-10/3}$
2. B
  $-\dfrac{7}{3}x^{-4/3}$
3. C
  $-\dfrac{5}{3}x^{-8/3}$
4. D
  $\dfrac{1}{2x \cdot \frac{1}{3}x^{-2/3}}$
Show answer
A
The function is $x^{-2} \cdot x^{-1/3} = x^{-7/3}$ , and $-\frac{7}{3} - 1 = -\frac{10}{3}$ . Choice B adds 1 to the exponent instead of subtracting it, choice C combines the exponents as $-2 + \frac{1}{3}$ , and choice D differentiates the pieces of the denominator and leaves the fraction in place.
Question 2
What is the slope of the graph of $y = x^4$ at $x = -\frac{1}{2}$ ?
Show answer
$-\frac{1}{2}$
$\frac{dy}{dx} = 4x^3$ , and $4\left(-\frac{1}{2}\right)^3 = 4\left(-\frac{1}{8}\right) = -\frac{1}{2}$ . The cube keeps the negative sign, so the graph is falling there, as it is everywhere to the left of $x = 0$ .
Question 3
Find $f'(x)$ for $f(x) = x^{\pi}$ and $g'(x)$ for $g(x) = \pi^{\pi}$ .
Show answer
$f'(x) = \pi x^{\pi - 1}$ and $g'(x) = 0$ .
In $x^{\pi}$ the base is the variable and the exponent is a constant, so the power rule applies with $r = \pi$ . $\pi^{\pi}$ contains no variable at all; it is a number (about 36.462), so its derivative is 0.
Question 4
If $f(x) = \sqrt[4]{x^3}$ , what is $f'(16)$ ?
1. A
  $\dfrac{1}{2}$
2. B
  $\dfrac{3}{2}$
3. C
  $\dfrac{3}{8}$
4. D
  $6$
Show answer
C
$f(x) = x^{3/4}$ , so $f'(x) = \frac{3}{4}x^{-1/4}$ , and $16^{-1/4} = \frac{1}{2}$ , giving $f'(16) = \frac{3}{4}\cdot\frac{1}{2} = \frac{3}{8}$ . Choice D keeps the old exponent ( $\frac{3}{4}\cdot 16^{3/4} = 6$ ), choice B drops the negative sign ( $\frac{3}{4}\cdot 16^{1/4} = \frac{3}{2}$ ), and choice A forgets the coefficient $\frac{3}{4}$ .

Go deeper

OpenStax, Calculus Volume 1, 3.3 Differentiation Rules