Applying the
Power Rule
Master the fundamental differentiation technique: if \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \). Learn to apply it to polynomials, negative exponents, and rational powers.
AP Calculus BC
What Is a Derivative?
The derivative of a function measures its instantaneous rate of change at any point. Geometrically, it gives the slope of the tangent line to the curve at that point.
Instead of computing limits every time, differentiation rules give us shortcuts. The Power Rule is the most fundamental of these shortcuts.
Visualizing Power Functions
Why the Power Rule Matters
Speed
Differentiate any polynomial in seconds without computing a single limit.
Foundation
Combined with the Chain, Product, and Quotient Rules, it handles nearly every algebraic function.
Universality
Works for integer, fractional, and negative exponents: \( x^{-3} \), \( x^{1/2} \), \( x^{\pi} \).
On the AP Calculus BC exam, the Power Rule appears in nearly every free-response and multiple-choice problem involving derivatives. Mastering it is non-negotiable.
The Power Rule: Formal Statement
Pairing with the Constant Multiple Rule
Constants simply ride along. They multiply the result but don't change the differentiation process. Combined with the Sum/Difference Rule, you can differentiate entire polynomials term by term.
1
Constant Multiple: \(\frac{d}{dx}[c \cdot f(x)] = c \cdot f'(x)\)
2
Sum Rule: \(\frac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x)\)
3
Constant Rule: \(\frac{d}{dx}[c] = 0\) (the derivative of a constant is always zero)
Power Rule vs. Other Differentiation Rules
| Rule | When to Use | Formula |
|---|---|---|
| Power Rule | \( x^n \) | \( nx^{n-1} \) |
| Product Rule | \( f(x) \cdot g(x) \) | \( f'g + fg' \) |
| Quotient Rule | \( \frac{f(x)}{g(x)} \) | \( \frac{f'g - fg'}{g^2} \) |
| Chain Rule | \( f(g(x)) \) | \( f'(g(x)) \cdot g'(x) \) |
How to Apply the Power Rule
1
Rewrite any roots, fractions, or reciprocals as power expressions: \( \sqrt{x} = x^{1/2} \), \( \frac{1}{x^3} = x^{-3} \).
2
Identify the coefficient \( c \) and exponent \( n \) for each term.
3
Multiply the coefficient by the exponent: new coefficient = \( c \cdot n \).
4
Subtract 1 from the exponent: new exponent = \( n - 1 \).
5
Simplify and combine terms. Rewrite in the form the problem requests (radical form, positive exponents, etc.).
Example 1: Basic Polynomial
1
\(\frac{d}{dx}[3x^4] = 3 \cdot 4 \cdot x^{4-1} =\) \(12x^3\)
2
\(\frac{d}{dx}[-5x^2] = -5 \cdot 2 \cdot x^{2-1} =\) \(-10x\)
3
\(\frac{d}{dx}[7x] = 7 \cdot 1 \cdot x^{1-1} = 7 \cdot x^0 =\) \(7\)
4
\(\frac{d}{dx}[-2] =\) \(0\) (constant)
Example 2: Negative and Fractional Exponents
1
Rewrite: \(\frac{4}{x^3} = 4x^{-3}\) and \(6\sqrt{x} = 6x^{1/2}\)
2
\(\frac{d}{dx}[4x^{-3}] = 4 \cdot (-3) \cdot x^{-3-1} =\) \(-12x^{-4}\)
3
\(\frac{d}{dx}[6x^{1/2}] = 6 \cdot \frac{1}{2} \cdot x^{1/2-1} =\) \(3x^{-1/2}\)
Example 3: Rewrite Before Differentiating
1
Simplify first: \(\frac{x^3 + 2x}{x} = \frac{x^3}{x} + \frac{2x}{x} = x^2 + 2\)
2
Now differentiate: \(\frac{d}{dx}[x^2 + 2] = 2x + 0 =\) \(2x\)
Graphical Interpretation
Works for All Real Exponents
Positive Integers
\(\frac{d}{dx}[x^5] = 5x^4\)
Negative Integers
\(\frac{d}{dx}[x^{-2}] = -2x^{-3}\)
Fractions
\(\frac{d}{dx}[x^{3/4}] = \frac{3}{4}x^{-1/4}\)
Irrational
\(\frac{d}{dx}[x^{\pi}] = \pi x^{\pi - 1}\)
Special Cases to Remember
1
\( n = 1 \): \(\frac{d}{dx}[x] = 1 \cdot x^{0} = 1\). The derivative of \( x \) is always 1.
2
\( n = 0 \): \( x^0 = 1 \) (a constant), so \(\frac{d}{dx}[1] = 0\). Constants vanish.
3
\( n = -1 \): \(\frac{d}{dx}\left[\frac{1}{x}\right] = \frac{d}{dx}[x^{-1}] = -x^{-2} = -\frac{1}{x^2}\).
4
\( n = \frac{1}{2} \): \(\frac{d}{dx}[\sqrt{x}] = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}\).
Common Mistakes to Avoid
✕ Forgetting to Subtract 1
Writing \(\frac{d}{dx}[x^4] = 4x^4\) instead of the correct \(4x^3\). Always reduce the exponent.
✕ Mishandling Negatives
For \(x^{-3}\): the new exponent is \(-3 - 1 = -4\), not \(-2\). Subtract, don't add.
✕ Confusing \( x^n \) with \( a^x \)
The Power Rule applies to \(x^n\) (variable base), NOT \(2^x\) (variable exponent). For \(a^x\), use the exponential rule.
Summary of Power Rule Applications
| Original \( f(x) \) | Rewritten | Derivative \( f'(x) \) |
|---|---|---|
| \( 5x^3 \) | already in form | \( 15x^2 \) |
| \( \sqrt[3]{x} \) | \( x^{1/3} \) | \( \frac{1}{3}x^{-2/3} \) |
| \( \frac{7}{x^2} \) | \( 7x^{-2} \) | \( -14x^{-3} \) |
| \( 4x^{3/2} \) | already in form | \( 6x^{1/2} \) |
| \( -\frac{1}{\sqrt{x}} \) | \( -x^{-1/2} \) | \( \frac{1}{2}x^{-3/2} \) |
Question 1 of 5
What is \(\displaystyle \frac{d}{dx}\left[7x^5\right]\)?
Apply the Power Rule: bring down the exponent and multiply by the coefficient: \( 7 \cdot 5 = 35 \). Then reduce the exponent: \( 5 - 1 = 4 \). So \( \frac{d}{dx}[7x^5] = 35x^4 \).
Question 2 of 5
What is \(\displaystyle \frac{d}{dx}\left[\frac{3}{x^4}\right]\)?
Rewrite as \( 3x^{-4} \). Apply the Power Rule: \( 3 \cdot (-4) \cdot x^{-4-1} = -12x^{-5} \). The negative exponent stays negative, and we subtract 1 to get \( -5 \).
Question 3 of 5
Find \(\displaystyle f'(x)\) if \( f(x) = 2x^3 - 4x^2 + x - 9 \).
Differentiate term by term: \( \frac{d}{dx}[2x^3] = 6x^2 \), \( \frac{d}{dx}[-4x^2] = -8x \), \( \frac{d}{dx}[x] = 1 \), and \( \frac{d}{dx}[-9] = 0 \). Combine to get \( 6x^2 - 8x + 1 \).
Question 4 of 5
What is \(\displaystyle \frac{d}{dx}\left[8\sqrt[3]{x}\right]\)?
Rewrite \( 8\sqrt[3]{x} = 8x^{1/3} \). Apply the Power Rule: \( 8 \cdot \frac{1}{3} \cdot x^{1/3 - 1} = \frac{8}{3} x^{-2/3} \). The exponent \( \frac{1}{3} - 1 = -\frac{2}{3} \).
Question 5 of 5
Which function does the Power Rule NOT directly apply to?
The Power Rule applies to \( x^n \) where the base is the variable. In \( 3^x \), the exponent is the variable, making it an exponential function: \( \frac{d}{dx}[3^x] = 3^x \ln 3 \). The other three are all power functions.