The Power Rule

The most fundamental differentiation rule for polynomials

The Power Rule

The Power Rule is the easiest and most-used differentiation rule. Master this, and you've mastered half of calculus!

The Rule

$\frac{d}{dx}[x^n] = nx^{n-1}$

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:

| Expression | Exponent Form | Derivative | |------------|---------------|------------| | $\sqrt{x}$ | $x^{1/2}$ | $\frac{1}{2}x^{-1/2}$ | | $\sqrt[3]{x}$ | $x^{1/3}$ | $\frac{1}{3}x^{-2/3}$ | | $\frac{1}{x}$ | $x^{-1}$ | $-x^{-2}$ | | $\frac{1}{x^2}$ | $x^{-2}$ | $-2x^{-3}$ |

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

Identify the exponent on x
Multiply by that exponent
Subtract 1 from the exponent
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!

📚 Practice Problems

1Problem 1easy

❓ Question:

Find the derivative of each function:

a) $f(x) = x^7$ b) $g(x) = 5x^3 - 2x^2 + 8x - 3$ c) $h(x) = \frac{1}{x^4}$

💡 Show Solution

Solution:

Part (a): Power rule: $\frac{d}{dx}[x^n] = nx^{n-1}$

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

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: $h(x) = x^{-4}$

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

2Problem 2easy

❓ Question:

Find the derivative of f(x) = x⁷

💡 Show Solution

Using the power rule:

$f'(x) = \frac{d}{dx}[x^7]$

Bring down the exponent (7) and subtract 1:

$= 7x^{7-1}$

$= 7x^6$

Answer: f'(x) = 7x⁶

3Problem 3easy

❓ Question:

Find the derivative of each function:

a) $f(x) = x^7$ b) $g(x) = 5x^3 - 2x^2 + 8x - 3$ c) $h(x) = \frac{1}{x^4}$

💡 Show Solution

Solution:

Part (a): Power rule: $\frac{d}{dx}[x^n] = nx^{n-1}$

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

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: $h(x) = x^{-4}$

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

4Problem 4medium

❓ Question:

Find the derivative of $f(x) = \sqrt{x} + \frac{3}{\sqrt[3]{x^2}}$ .

💡 Show Solution

Solution:

Rewrite using fractional exponents:

$f(x) = x^{1/2} + 3x^{-2/3}$

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:

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

5Problem 5medium

❓ Question:

Find the derivative of $f(x) = \sqrt{x} + \frac{3}{\sqrt[3]{x^2}}$ .

💡 Show Solution

Solution:

Rewrite using fractional exponents:

$f(x) = x^{1/2} + 3x^{-2/3}$

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:

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

6Problem 6medium

❓ Question:

Find the derivative of g(x) = 1/x⁴

💡 Show Solution

First, rewrite using negative exponents:

$g(x) = \frac{1}{x^4} = x^{-4}$

Now apply the power rule:

$g'(x) = \frac{d}{dx}[x^{-4}]$

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

$= -4x^{-5}$

We can rewrite in fraction form:

$= -\frac{4}{x^5}$

Answer: g'(x) = -4x⁻⁵ or -4/x⁵

7Problem 7medium

❓ Question:

Find the derivative of h(x) = ∛x

💡 Show Solution

First, convert the cube root to exponential form:

$h(x) = \sqrt[3]{x} = x^{1/3}$

Apply the power rule:

$h'(x) = \frac{d}{dx}[x^{1/3}]$

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

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

We can rewrite this:

$= \frac{1}{3x^{2/3}} = \frac{1}{3\sqrt[3]{x^2}}$

Answer: h'(x) = (1/3)x⁻²/³ or 1/(3∛(x²))

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