What if you could differentiate $x^{100}$ without computing a single limit? The Power Rule makes this possible. It is the workhorse of calculus that you will use more than any other formula.
The pattern is elegant: to find the derivative of $x^n$, you bring down the exponent and reduce it by one. Once you see it, you'll never forget it.
This rule, combined with the rules in the rest of this section, allows you to differentiate any polynomial instantly.
| Property | Value |
|---|---|
| Concept | Differentiation Formulas |
| Chapter | 2.3 |
| Difficulty | Beginner |
| Time | ~15 minutes |
If $n$ is any real number, then:
$$\boxed{\frac{d}{dx}(x^n) = nx^{n-1}}$$
In words: "Bring down the power, reduce it by one."
Constant Function: $\frac{d}{dx}(c) = 0$ for any constant $c$
Why? A constant doesn't change, so its rate of change is zero. Geometrically, $y = c$ is a horizontal line with slope 0.
Identity Function: $\frac{d}{dx}(x) = 1$
The line $y = x$ has slope 1 everywhere.
Verify for $f(x) = x^2$:
$$f'(x) = \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h} = \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x^2}{h}$$
$$= \lim_{h \to 0} \frac{2xh + h^2}{h} = \lim_{h \to 0}(2x + h) = 2x$$
This matches the Power Rule: $\frac{d}{dx}(x^2) = 2x^{2-1} = 2x$ ✓
| Function | Derivative | Pattern |
|---|---|---|
| $x^2$ | $2x$ | Bring down 2, power becomes 1 |
| $x^3$ | $3x^2$ | Bring down 3, power becomes 2 |
| $x^4$ | $4x^3$ | Bring down 4, power becomes 3 |
| $x^{100}$ | $100x^{99}$ | Bring down 100, power becomes 99 |
The Power Rule works for all real exponents, not just positive integers:
| Function | Rewrite | Derivative |
|---|---|---|
| $\frac{1}{x}$ | $x^{-1}$ | $-x^{-2} = -\frac{1}{x^2}$ |
| $\frac{1}{x^2}$ | $x^{-2}$ | $-2x^{-3} = -\frac{2}{x^3}$ |
| $\sqrt{x}$ | $x^{1/2}$ | $\frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$ |
| $\sqrt[3]{x}$ | $x^{1/3}$ | $\frac{1}{3}x^{-2/3} = \frac{1}{3\sqrt[3]{x^2}}$ |
Key insight: Always convert roots and fractions to exponential form before differentiating!
Find $f'(x)$ if $f(x) = x^7$.
Differentiate $g(x) = \sqrt[4]{x}$.
Find $\frac{d}{dt}\left(\frac{5}{t^3}\right)$.
Differentiate $h(x) = \frac{1}{\sqrt[3]{x^2}}$ and simplify your answer.
Prove the Power Rule for $f(x) = x^3$ using the limit definition of the derivative.
Hint: You'll need the algebraic identity $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$.
If $A(r) = \pi r^2$ represents the area of a circle with radius $r$, what does $A'(r) = 2\pi r$ represent physically?
Without computing, which function grows faster as $x \to \infty$: $f(x) = x^4$ or $g(x) = x^5$? How does this relate to their derivatives?
Think of it as "multiply and decrease":
The exponent tells you how many times $x$ is multiplied together. When you differentiate, one of those $x$'s "comes down" as a coefficient, leaving one fewer $x$ in the product.
For $x^5 = x \cdot x \cdot x \cdot x \cdot x$:
| Previous | Up | Next |
|---|---|---|
| Derivative Definition | Skills Index | Sum/Difference Rules |
Last updated: 2026-01-22