General Power Functions
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 3.3: “Differentiation Rules” |
| Book URL | https://openstax.org/details/books/calculus-volume-1 |
Freely available and openly licensed.
Try This First (Predict, Then Check)
Before reading on, predict: what do you think $\dfrac{d}{dx}[x^{-2}]$ should be?
The power rule for positive integers says $\dfrac{d}{dx}[x^n] = nx^{n-1}$. Try applying that pattern with $n = -2$:
Predicted answer: $\underline{\hspace{3cm}}$.
Now verify using the definition: compute $\lim_{h \to 0} \dfrac{(x+h)^{-2} - x^{-2}}{h}$, simplify by combining fractions over a common denominator, and cancel $h$.
The pattern $\dfrac{d}{dx}[x^r] = rx^{r-1}$ works for all real exponents $r$, whether negative, fractional, or irrational. Knowing this lets you spot when rewriting a function as $x^r$ makes differentiation easier.
Prerequisite Check
Before this lesson, make sure you can do all of the following:
Quick Reference
General Power Rule. \[ \frac{d}{dx}[x^r] = r x^{r-1}, \quad r \in \mathbb{R} \quad (x > 0 \text{ if } r \text{ is not an integer}) \]
The same formula $rx^{r-1}$ works for any real exponent: positive integer, negative, fractional, or irrational.
Rewriting strategy. Before differentiating, convert:
- $\dfrac{1}{x^n} \to x^{-n}$
- $\sqrt[n]{x} \to x^{1/n}$
- $\dfrac{1}{\sqrt[n]{x^m}} \to x^{-m/n}$
Key Concepts
1. Negative Exponents
Example 1. Differentiate $f(x) = \dfrac{1}{x^3}$.
Rewrite: $f(x) = x^{-3}$.
Apply the power rule with $r = -3$: \[ f'(x) = -3 x^{-3-1} = -3x^{-4} = \frac{-3}{x^4}. \]
The pattern works: exponent multiplies, exponent decreases by 1.
| $f(x)$ | Rewrite | $r$ | $f'(x)$ |
|---|---|---|---|
| $1/x$ | $x^{-1}$ | $-1$ | $-x^{-2} = -1/x^2$ |
| $1/x^2$ | $x^{-2}$ | $-2$ | $-2x^{-3} = -2/x^3$ |
| $1/x^5$ | $x^{-5}$ | $-5$ | $-5x^{-6}$ |
2. Fractional Exponents (Root Functions)
Example 2. Differentiate $g(x) = \sqrt{x}$.
Rewrite: $g(x) = x^{1/2}$. \[ g'(x) = \frac{1}{2}x^{1/2 - 1} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}. \]
Example 3. Differentiate $h(x) = \sqrt[3]{x^2}$.
Rewrite: $h(x) = x^{2/3}$. \[ h'(x) = \frac{2}{3}x^{2/3 - 1} = \frac{2}{3}x^{-1/3} = \frac{2}{3\sqrt[3]{x}}. \]
3. Multiple Representations
Graphical picture. The derivative of $x^r$ at $x = 1$ is always $r$ (since $r \cdot 1^{r-1} = r$). For $r > 1$, the slope of $x^r$ at $(1, 1)$ is greater than 1 (curve rises steeply). For $r < 1$ (but $r > 0$), the slope is less than 1 (curve rises gently). For $r < 0$, the slope is negative (function is decreasing).
| $f(x)$ | At $x = 1$: $f(1)$ | $f'(1)$ |
|---|---|---|
| $x^3$ | 1 | 3 |
| $x^{1/2}$ | 1 | $1/2$ |
| $x^{-1}$ | 1 | $-1$ |
| $x^{-2}$ | 1 | $-2$ |
Verbal picture. Negative exponents make functions that decrease as $x$ increases (for $x > 0$); fractional exponents make functions that increase but flatten out; these are all related by the same formula.
4. Combining with Other Rules
Example 4. Differentiate $f(x) = 3x^{-2} + 5x^{1/3} - 7$.
$f'(x) = 3(-2x^{-3}) + 5\left(\dfrac{1}{3}x^{-2/3}\right) - 0 = -6x^{-3} + \dfrac{5}{3}x^{-2/3}$.
Or equivalently: $f'(x) = -\dfrac{6}{x^3} + \dfrac{5}{3\sqrt[3]{x^2}}$.
Example 5. Differentiate $g(x) = \dfrac{4}{x^2} - \dfrac{3}{\sqrt{x}} + 2x$.
Rewrite: $g(x) = 4x^{-2} - 3x^{-1/2} + 2x$.
$g'(x) = -8x^{-3} + \dfrac{3}{2}x^{-3/2} + 2 = -\dfrac{8}{x^3} + \dfrac{3}{2x^{3/2}} + 2$.
5. Ask Why: Is the Pattern Always Valid?
The power rule was proved in this course (using the binomial theorem) for positive integer exponents. For negative integer exponents, you can verify it from the definition. For fractional exponents $r = p/q$, the proof uses implicit differentiation (covered later): if $y = x^{p/q}$, then $y^q = x^p$, and differentiating implicitly gives $qy^{q-1}y' = px^{p-1}$, so $y' = \dfrac{p}{q} \cdot \dfrac{x^{p-1}}{y^{q-1}} = \dfrac{p}{q}x^{p/q - 1}$.
The formula $rx^{r-1}$ is a theorem that requires proof for each class of exponents, not just an observed pattern. The same formula holds across every class of exponent.
Named Misconception: “Rewrite, Then Differentiate”
The most common error with negative and fractional exponents is trying to differentiate without rewriting first, then not knowing how to proceed.
Wrong: “The derivative of $\dfrac{1}{x^2}$ is $\dfrac{1}{2x}$” (treating it like a quotient and misapplying rules).
Right: Rewrite as $x^{-2}$, then differentiate: $-2x^{-3}$.
The rewriting step puts the function in the form $x^r$ that the power rule expects, so do it before you differentiate.
Common Errors
| Error | Example | Correction |
|---|---|---|
| Forgetting to rewrite | Differentiating $1/x^3$ as if it were $x^{-3}$ without the rewrite | Always convert to $x^{-3}$ form first, then apply rule |
| Sign error on exponent | $\frac{d}{dx}[x^{-2}] = -2x^{-1}$ (wrong) | $-2x^{-2-1} = -2x^{-3}$; the exponent decreases by 1: $-2 - 1 = -3$ |
| Leaving answer with negative exponents when asked for simplified form | $-3x^{-4}$ left as is | Rewrite as $-3/x^4$ when a simplified fraction is expected |
Leveled Practice
Level 1 -- Direct Application
Problem 1. Differentiate: (a) $x^{-4}$, (b) $x^{1/4}$, (c) $x^{-1/3}$.
Show answer
(a) $-4x^{-5}$. (b) $\frac{1}{4}x^{-3/4}$. (c) $-\frac{1}{3}x^{-4/3}$.
Problem 2. Rewrite and differentiate: (a) $\dfrac{5}{x^2}$, (b) $3\sqrt[4]{x}$, (c) $\dfrac{2}{\sqrt{x^3}}$.
Show answer
(a) $5x^{-2} \to -10x^{-3} = -10/x^3$.
(b) $3x^{1/4} \to \frac{3}{4}x^{-3/4}$.
(c) $2x^{-3/2} \to -3x^{-5/2} = -3/x^{5/2}$.
Level 2 -- Combined Expressions
Problem 3. Differentiate $f(x) = \sqrt{x} + \dfrac{1}{\sqrt{x}}$.
Show answer
$f(x) = x^{1/2} + x^{-1/2}$.
$f'(x) = \frac{1}{2}x^{-1/2} - \frac{1}{2}x^{-3/2} = \dfrac{1}{2\sqrt{x}} - \dfrac{1}{2x^{3/2}}$.
Factor: $\dfrac{1}{2x^{3/2}}(x - 1) = \dfrac{x-1}{2x^{3/2}}$.
Problem 4. Find the slope of the tangent to $y = x^{2/3}$ at $x = 8$. Interpret the result.
Show answer
$y'(x) = \frac{2}{3}x^{-1/3}$. At $x = 8$: $y'(8) = \frac{2}{3} \cdot 8^{-1/3} = \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{3}$.
The curve rises gently at $x = 8$: for every 1 unit increase in $x$, the output increases by about $1/3$ unit.
Level 3 -- Low-Floor-High-Ceiling Extension
Problem 5 (Extension). The power rule says $\dfrac{d}{dx}[x^r] = rx^{r-1}$ for any real $r$.
(a) (Floor) Verify this for $r = 0$: what is $\dfrac{d}{dx}[x^0] = \dfrac{d}{dx}[1]$?
(b) (Mid) Use the formula to differentiate $f(x) = x^{\pi}$ and $g(x) = x^{\sqrt{2}}$.
(c) (Ceiling) The power rule gives $\dfrac{d}{dx}[x^{1/n}]$ for any positive integer $n$. Use this to find the slope of the tangent to $y = \sqrt[n]{x}$ at $x = 1$ for any $n$. What happens as $n \to \infty$?
Show answer
(a) $r = 0$: $\frac{d}{dx}[x^0] = 0 \cdot x^{-1} = 0$. Matches $\frac{d}{dx}[1] = 0$. Consistent.
(b) $\frac{d}{dx}[x^\pi] = \pi x^{\pi - 1}$; $\frac{d}{dx}[x^{\sqrt{2}}] = \sqrt{2}\, x^{\sqrt{2}-1}$.
(c) $\frac{d}{dx}[x^{1/n}] = \frac{1}{n}x^{1/n - 1}$. At $x = 1$: slope $= \frac{1}{n}$.
As $n \to \infty$, the slope $\to 0$. This makes sense: $x^{1/n} \to 1$ as $n \to \infty$ for $x > 0$ (all these roots approach 1 at $x = 1$), and the curves all flatten out, so the tangent slope at $x = 1$ goes to zero.
Common Misconceptions
$\frac{d}{dx}[\sqrt{x}] = \frac{1}{2\sqrt{x}} \cdot \frac{d}{dx}[\sqrt{x}]$ (applying the power rule twice). The power rule $\frac{d}{dx}[x^n] = nx^{n-1}$ is applied once. Rewrite $\sqrt{x} = x^{1/2}$, then $\frac{d}{dx}[x^{1/2}] = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$. The chain rule would apply if the argument of the root were a function of $x$ other than $x$ itself; for $\sqrt{x}$, the argument is just $x$, and the chain rule factor is 1.
negative exponents produce negative derivatives. The power rule applies regardless of the sign of the exponent. For $f(x) = x^{-3}$: $f'(x) = -3x^{-4} = -3/x^4$. The derivative is negative here because the coefficient is $-3$, not because the original exponent was negative. For $f(x) = x^{-2}$: $f'(x) = -2x^{-3}$, which is also negative, but that is a property of this particular function, not of negative exponents in general. For $f(x) = 5x^{-1}$: $f'(x) = -5x^{-2}$, again negative. The sign of the derivative depends on the problem, not on the sign of the exponent alone.
Mastery Checklist
You have mastered this skill when you can do all of the following without referring to notes:
Mental Model
Every power function $x^r$, for any value of $r$, follows the same rule: bring down the exponent, subtract 1. The exponent multiplies the coefficient, and the exponent decreases by 1.
For negative $r$: the function is decreasing (slope is negative for $x > 0$), and the slope becomes more steeply negative as $x$ approaches 0 (the function blows up). The formula captures this: $-|r|x^{-|r|-1}$, which is large and negative near $x = 0$.
For fractional $r$ between 0 and 1: the function is increasing but concave down (flattening out). The derivative decreases, matching the “slowing growth” picture.
Connections
Within Calculus I (MATH161)
- Chain rule: The generalized power rule becomes $\frac{d}{dx}[g(x)^r] = r\,g(x)^{r-1}\,g'(x)$ with the chain rule, a pattern you will use in many applications.
- Implicit differentiation: The proof for fractional $r$ uses implicit differentiation; this lesson previews where that technique applies.
- Integration: $\int x^r\,dx = \dfrac{x^{r+1}}{r+1} + C$ for $r \neq -1$ is the antiderivative version. The single exception $r = -1$ (which gives $\ln|x|$) is the only case where the formula breaks down.
Back to Calculus I Skills | Previous: Power Rule | Next: Product Rule