The Power Rule

In words: Bring down the exponent, then subtract 1 from the exponent.

How It Works

Step 1: Multiply by the exponent Step 2: Decrease the exponent by 1

Basic Examples

Example 1: $\frac{d}{dx}[x^3]$

Bring down the 3, subtract 1 from exponent: $= 3x^{3-1} = 3x^2$

Example 2: $\frac{d}{dx}[x^5]$

$= 5x^{5-1} = 5x^4$

Example 3: $\frac{d}{dx}[x^{10}]$

$= 10x^{10-1} = 10x^9$

Special Cases

First power: $\frac{d}{dx}[x^1] = \frac{d}{dx}[x]$

$= 1 \cdot x^{1-1} = 1 \cdot x^0 = 1$

The derivative of x is always 1!

Constant: $\frac{d}{dx}[c] = \frac{d}{dx}[x^0]$

$= 0 \cdot x^{-1} = 0$

The derivative of any constant is 0!

Negative Exponents

The power rule works for negative exponents too!

Example: $\frac{d}{dx}[x^{-2}]$

$= -2x^{-2-1} = -2x^{-3} = -\frac{2}{x^3}$

Example: $\frac{d}{dx}\left[\frac{1}{x}\right] = \frac{d}{dx}[x^{-1}]$

$= -1 \cdot x^{-1-1} = -x^{-2} = -\frac{1}{x^2}$

Fractional Exponents

Works for fractions too!

Example: $\frac{d}{dx}[x^{1/2}] = \frac{d}{dx}[\sqrt{x}]$

$= \frac{1}{2}x^{1/2 - 1} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$

Example: $\frac{d}{dx}[x^{2/3}]$

$= \frac{2}{3}x^{2/3 - 1} = \frac{2}{3}x^{-1/3} = \frac{2}{3x^{1/3}}$

Converting Radicals and Fractions

Before using the power rule, convert to exponent form:

Why It Works

The power rule comes from the limit definition:

$f'(x) = \lim_{h \to 0} \frac{(x+h)^n - x^n}{h}$

Using the binomial theorem and simplifying, we get $nx^{n-1}$.

But thankfully, we don't need to do that every time!

Common Mistakes to Avoid

❌ Wrong: $\frac{d}{dx}[x^3] = x^2$ (forgot to bring down the 3)

✓ Right: $\frac{d}{dx}[x^3] = 3x^2$

❌ Wrong: $\frac{d}{dx}[x^5] = 5x^5$ (forgot to subtract 1)

✓ Right: $\frac{d}{dx}[x^5] = 5x^4$

❌ Wrong: $\frac{d}{dx}[3] = 3$ (derivative of constant ≠ the constant)

✓ Right: $\frac{d}{dx}[3] = 0$

Practice Strategy

1. Identify the exponent on x
2. Multiply by that exponent
3. Subtract 1 from the exponent
4. Simplify if needed

Multiple Terms (Preview)

For polynomials, differentiate term by term:

$\frac{d}{dx}[x^3 + x^2] = \frac{d}{dx}[x^3] + \frac{d}{dx}[x^2] = 3x^2 + 2x$

We'll formalize this in the next lesson!

Bring down the exponent (7) and subtract 1:

$= 7x^{7-1}$

$= 7x^6$

Answer: f'(x) = 7x⁶

$f'(x) = 7x^6$

Part (b): Use power rule on each term:

$g'(x) = 5(3x^2) - 2(2x) + 8(1) - 0$

$g'(x) = 15x^2 - 4x + 8$

Part (c): Rewrite using negative exponent: $h(x) = x^{-4}$

$h'(x) = -4x^{-5} = -\frac{4}{x^5}$

Apply power rule:

$f'(x) = \frac{1}{2}x^{-1/2} + 3 \cdot \left(-\frac{2}{3}\right)x^{-5/3}$

$f'(x) = \frac{1}{2}x^{-1/2} - 2x^{-5/3}$

Rewrite with radicals:

$f'(x) = \frac{1}{2\sqrt{x}} - \frac{2}{x^{5/3}} = \frac{1}{2\sqrt{x}} - \frac{2}{\sqrt[3]{x^5}}$