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Generalized Power Rule

Reference: Stewart 2.5  •  Chapter: 2  •  Section: 5

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The Most Common Chain Rule Pattern

You already know the Power Rule: $\frac{d}{dx}(x^n) = nx^{n-1}$. But what about $(x^2 + 1)^{10}$? You cannot just write $10(x^2+1)^9$. That ignores what is inside the parentheses.

The Generalized Power Rule fixes this by combining the Power Rule with the Chain Rule. It's arguably the most frequently used differentiation formula after the basic rules.

$$\frac{d}{dx}[g(x)]^n = n[g(x)]^{n-1} \cdot g'(x)$$

Key insight: Bring down the exponent, reduce by one, then multiply by the derivative of what's inside.

Prerequisite Map

This skillGeneralized Power Rule

Quick Reference

Property Value
Concept Chain Rule
Chapter 2.5
Difficulty Intermediate
Time ~15 minutes

Key Concepts

The Formula

For any differentiable function $g(x)$ and any real number $n$:

$$\boxed{\frac{d}{dx}[g(x)]^n = n[g(x)]^{n-1} \cdot g'(x)}$$

Why This Works

This is just the Chain Rule where the outer function is $f(u) = u^n$:

Component Value Derivative
Outer: $f(u)$ $u^n$ $nu^{n-1}$
Inner: $g(x)$ whatever's inside $g'(x)$

Chain Rule: $\frac{d}{dx}[g(x)]^n = f'(g(x)) \cdot g'(x) = n[g(x)]^{n-1} \cdot g'(x)$

The Three-Step Process

For any expression of the form $(\text{stuff})^n$:

  1. Bring down the exponent as a coefficient
  2. Reduce the exponent by 1, keeping the inside unchanged
  3. Multiply by the derivative of the inside

Example: $\frac{d}{dx}(x^3 + 2x)^7$

Step 1: Bring down 7         →  7
Step 2: Reduce exponent      →  7(x³ + 2x)⁶
Step 3: Multiply by inside'  →  7(x³ + 2x)⁶ · (3x² + 2)

Special Cases with Fractional and Negative Exponents

The formula works for all real exponents:

Expression Rewrite Derivative
$\sqrt{g(x)}$ $[g(x)]^{1/2}$ $\frac{1}{2}[g(x)]^{-1/2} \cdot g'(x) = \frac{g'(x)}{2\sqrt{g(x)}}$
$\sqrt[3]{g(x)}$ $[g(x)]^{1/3}$ $\frac{1}{3}[g(x)]^{-2/3} \cdot g'(x)$
$\frac{1}{g(x)}$ $[g(x)]^{-1}$ $-[g(x)]^{-2} \cdot g'(x) = \frac{-g'(x)}{[g(x)]^2}$
$\frac{1}{[g(x)]^2}$ $[g(x)]^{-2}$ $-2[g(x)]^{-3} \cdot g'(x)$

The Common Error

Wrong: $\frac{d}{dx}(x^2 + 1)^5 = 5(x^2+1)^4$ ✗

This forgets the Chain Rule! The derivative of $x^2 + 1$ is $2x$, not 1.

Correct: $\frac{d}{dx}(x^2 + 1)^5 = 5(x^2+1)^4 \cdot 2x = 10x(x^2+1)^4$ ✓

Rule of thumb: If there's anything other than just "$x$" inside the parentheses, you need to multiply by its derivative.

Practice Problems

Level 1 Basic Application

Find $\frac{d}{dx}(4x - 7)^3$.

Thought Process

Apply the three steps:

  1. Bring down 3
  2. Reduce exponent: $(4x-7)^2$
  3. Multiply by derivative of inside: $\frac{d}{dx}(4x-7) = 4$
Show Answer

$$\frac{d}{dx}(4x-7)^3 = 3(4x-7)^2 \cdot 4 = 12(4x-7)^2$$

Level 2 Polynomial Inside

Find $f'(x)$ if $f(x) = (x^2 - 3x + 1)^8$.

Thought Process

The inner function is $g(x) = x^2 - 3x + 1$, so $g'(x) = 2x - 3$.

Apply the Generalized Power Rule with $n = 8$.

Show Answer

$$f'(x) = 8(x^2 - 3x + 1)^7 \cdot (2x - 3)$$

$$= 8(2x-3)(x^2 - 3x + 1)^7$$

Level 3 Square Root (Fractional Exponent)

Find $\frac{dy}{dx}$ if $y = \sqrt{5x^2 + 3}$.

Thought Process

Rewrite: $\sqrt{5x^2+3} = (5x^2+3)^{1/2}$

Apply Generalized Power Rule with $n = 1/2$.

The inner function is $5x^2 + 3$ with derivative $10x$.

Show Answer

$$y = (5x^2 + 3)^{1/2}$$

$$\frac{dy}{dx} = \frac{1}{2}(5x^2+3)^{-1/2} \cdot 10x$$

$$= \frac{10x}{2\sqrt{5x^2+3}} = \frac{5x}{\sqrt{5x^2+3}}$$

Level 4 Negative Exponent (Reciprocal)

Find $g'(t)$ if $g(t) = \frac{1}{(2t^3 - t)^4}$.

Thought Process

Rewrite using negative exponent: $\frac{1}{(2t^3-t)^4} = (2t^3-t)^{-4}$

Apply Generalized Power Rule with $n = -4$.

The inner function is $2t^3 - t$ with derivative $6t^2 - 1$.

Show Answer

$$g(t) = (2t^3 - t)^{-4}$$

$$g'(t) = -4(2t^3-t)^{-5} \cdot (6t^2 - 1)$$

$$= \frac{-4(6t^2-1)}{(2t^3-t)^5}$$

$$= \frac{-24t^2 + 4}{(2t^3-t)^5}$$

Level 5 Combined with Product Rule

Find the derivative of $h(x) = (x+1)^3(x-2)^4$.

Thought Process

This is a product of two functions, each requiring the Generalized Power Rule.

Use Product Rule: $(uv)' = u'v + uv'$

For each factor, apply the Generalized Power Rule.

Show Answer

Let $u = (x+1)^3$ and $v = (x-2)^4$.

Find $u'$: $$u' = 3(x+1)^2 \cdot 1 = 3(x+1)^2$$

Find $v'$: $$v' = 4(x-2)^3 \cdot 1 = 4(x-2)^3$$

Apply Product Rule: $$h'(x) = u'v + uv'$$ $$= 3(x+1)^2 \cdot (x-2)^4 + (x+1)^3 \cdot 4(x-2)^3$$

Factor common terms: Both terms share $(x+1)^2$ and $(x-2)^3$:

$$= (x+1)^2(x-2)^3[3(x-2) + 4(x+1)]$$

$$= (x+1)^2(x-2)^3[3x - 6 + 4x + 4]$$

$$= (x+1)^2(x-2)^3(7x - 2)$$

CCI-Style Conceptual Questions

Level 2 Error Detection

A student claims that $\frac{d}{dx}(x^2 + 5)^3 = 3(x^2 + 5)^2$.

Is this correct? If not, what is the correct answer and what step did the student miss?

Thought Process

Check if the Chain Rule was applied. The student brought down the exponent and reduced it, but did they multiply by the derivative of the inside?

The inside is $x^2 + 5$. Its derivative is $2x$, not $1$.

Show Answer

Incorrect. The student forgot to multiply by the derivative of the inner function.

The inner function is $x^2 + 5$, and its derivative is $2x$.

Correct answer: $$\frac{d}{dx}(x^2+5)^3 = 3(x^2+5)^2 \cdot 2x = 6x(x^2+5)^2$$

Missing step: The student skipped multiplying by $g'(x) = 2x$.

Mastery Checklist

Mental Model

The Gift-Wrapping Analogy: Think of $[g(x)]^n$ as a gift inside $n$ layers of wrapping paper.

  1. Unwrap one layer (bring down $n$, reduce exponent by 1)
  2. Account for the gift inside (multiply by the derivative of what's wrapped)

If you forget step 2, you're only describing the wrapping paper, not what's actually changing inside!


Connections

Looking back:

Looking ahead:

Why this pattern is everywhere: Powers appear constantly in applications: areas ($r^2$), volumes ($r^3$), inverse-square laws ($1/r^2$), and growth models. Every time the base is a function, you need this rule.


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Last updated: 2026-01-22