Power rule — AP Calculus AB Study Guide

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: Derivative formula for the power rule, differentiation of constant terms, extension to negative and fractional exponents, rewriting radicals and reciprocals to apply the rule, and finding tangent lines to power functions.

You should already know: Limit definition of the derivative, exponent rules for negative and fractional powers, basic point-slope form for lines.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus AB style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Power rule?

The power rule is the most fundamental differentiation shortcut, letting you compute the derivative of any power function without returning to the limit definition of the derivative for every calculation. According to the AP Calculus AB Course and Exam Description (CED), this topic is part of Unit 2, which accounts for 10-12% of the total exam score, and power rule applications appear in both multiple-choice (MCQ) and free-response (FRQ) sections, often as a building block for larger, multi-step problems.

A power function is defined as any function of the form $f(x)=k{x}^n$, where $k$ is a constant coefficient and $n$ is any real number (integer, fraction, negative, zero, etc.). Before the power rule was formalized, all derivatives required multi-step limit calculations that were slow and error-prone for complex functions. The power rule generalizes a pattern that emerges from repeated limit-based differentiation, turning a 5+ step calculation into a one-line algebraic manipulation. On the AP exam, you will almost never be asked to derive a power function derivative from the limit definition after learning this rule, but you are expected to apply it fluently for all subsequent differentiation work.

2. The Basic Power Rule for Positive Integer Exponents

We start with the simplest case: the power rule for functions with positive integer exponents, which is the first version most students learn. The core rule states that for any function $f(x)=x^n$ where $n$ is a positive integer, the first derivative is: $$f^{\prime }(x)=\frac{d}{dx}{x}^n=n{x}^{n-1}$$ When combined with the constant multiple rule (which states that $\frac{d}{dx}[cf(x)]=c{f}^{\prime }(x)$ for any constant $c$), the extended rule for coefficient-bearing power functions becomes: $$\frac{d}{dx}[c{x}^n]=cn{x}^{n-1}$$ We can confirm this rule with the limit definition: expanding $(x+h{)}^n$ with the binomial theorem gives $x^n+n{x}^{n-1}h+\text{higher-order terms with }{h}^2$, so when we subtract $x^n$, divide by $h$, and take the limit as $h\to 0$, all terms with $h$ go to zero, leaving only $n{x}^{n-1}$. The special case of a constant function ($f(x)=c=c{x}^0$) gives a derivative of zero, which makes intuitive sense: the slope of a horizontal line is always zero.

Worked Example

Find the derivative of $f(x)=4x^5-7x^3+12$, then evaluate $f^{\prime }(1)$.

1. Use the sum/difference rule to split the function into individual terms, which we differentiate separately: $f^{\prime }(x)=\frac{d}{dx}(4x^5)-\frac{d}{dx}(7x^3)+\frac{d}{dx}(12)$.
2. Apply the power rule to the first term: $n=5$, $c=4$, so derivative is $4\ast 5x^{5-1}=20x^4$.
3. Apply the power rule to the second term: $n=3$, $c=7$, so derivative is $7\ast 3x^{3-1}=21x^2$.
4. Apply the power rule to the constant term: $12=12x^0$, so derivative is $12\ast 0x^{-1}=0$.
5. Combine terms to get $f^{\prime }(x)=20x^4-21x^2$. Evaluate at $x=1$: $f^{\prime }(1)=20(1)-21(1)=-1$.

Exam tip: Always compute the general derivative first before plugging in the $x$-value for evaluation. Never substitute the $x$-value into the original function before differentiating, as this will always incorrectly give you a slope of zero.

3. Power Rule for Negative and Fractional Exponents

The power rule is not limited to positive integer exponents—it works for any real exponent, including negative exponents (for reciprocals and rational functions) and fractional exponents (for roots and radicals). This is one of the most frequently tested aspects of the power rule on the AP exam, because it requires you to first rewrite non-power expressions (radicals, fractions) using exponent rules before applying differentiation.

For example, a reciprocal term $\frac{1}{x^k}$ can be rewritten as $x^{-k}$, so the exponent $n=-k$, and the derivative becomes $-k{x}^{-k-1}=-\frac{k}{x^{k+1}}$. For a square root, $\sqrt{x}=x^{1/2}$, so $n=1/2$, and the derivative is $\frac{1}{2}{x}^{-1/2}=\frac{1}{2\sqrt{x}}$, which matches the result from the limit definition. Any expression that can be written as a power of $x$ can be differentiated with the power rule, which is almost always faster than using the quotient rule for simple rational functions.

Worked Example

Find the derivative of $g(x)=\frac{6}{x^3}-4\sqrt[4]{x}+9x$.

1. Rewrite all terms to use standard exponents: $\frac{6}{x^3}=6x^{-3}$, $4\sqrt[4]{x}=4x^{1/4}$, and $9x=9x^1$, so $g(x)=6x^{-3}-4x^{1/4}+9x^1$.
2. Differentiate the first term: $n=-3$, so $6(-3)x^{-3-1}=-18x^{-4}$.
3. Differentiate the second term: $n=1/4$, so $-4\left(\frac{1}{4}\right)x^{(1/4-1)}=-1x^{-3/4}$.
4. Differentiate the third term: $n=1$, so $9(1)x^0=9$.
5. Rewrite back to radical/reciprocal form to match the original problem: $g^{\prime }(x)=-\frac{18}{x^4}-\frac{1}{\sqrt[4]{x^3}}+9$.

Exam tip: Always explicitly write out your exponent subtraction step on paper for negative/fractional exponents. It is extremely common to accidentally subtract 1 from the coefficient instead of the exponent, or get the sign of the new exponent wrong.

4. Applying the Power Rule to Find Tangent Lines

A very common AP exam application of the power rule is finding the equation of a tangent line to a curve at a given point. A tangent line to $y=f(x)$ at $x=a$ has two key properties: it passes through the point $(a,f(a))$, and its slope equals the derivative $f^{\prime }(a)$, which we can compute quickly with the power rule. Once you have the slope and a point, you use point-slope form to write the tangent line equation, which is often required to be simplified to slope-intercept form on the exam.

This question type combines power rule differentiation with coordinate geometry, and it appears regularly on both MCQ and FRQ sections. It also tests your understanding of what a derivative actually means: the derivative at a point is the slope of the tangent line at that point.

Worked Example

Find the equation of the tangent line to $y=2x^3-4x+3$ at $x=2$.

1. First find the $y$-coordinate of the point by plugging $x=2$ into the original function: $y=2(8)-4(2)+3=16-8+3=11$, so the tangent line passes through $(2,11)$.
2. Compute the derivative using the power rule: $y^{\prime }=2(3)x^2-4(1)x^0+0=6x^2-4$.
3. Calculate the slope by evaluating the derivative at $x=2$: $m=y^{\prime }(2)=6(4)-4=24-4=20$.
4. Plug into point-slope form: $y-11=20(x-2)$.
5. Simplify to slope-intercept form: $y=20x-40+11=20x-29$.

Exam tip: Never forget to calculate the $y$-coordinate from the original function. Many students only use the derivative to get slope and the given $x$-coordinate, leading to an incorrect $y$-intercept for the tangent line.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Differentiating $f(x)=4x^3$ to get $f^{\prime }(x)=4x^2$. Why: Forgets to multiply the original coefficient by the exponent, only subtracts 1 from the exponent. Correct move: Always multiply the original coefficient by the exponent first, then subtract 1 from the exponent.
• Wrong move: Differentiating $g(x)=x^{-4}$ to get $g^{\prime }(x)=-4x^{-3}$. Why: Incorrectly subtracts 1 from a negative exponent, resulting in a less negative exponent instead of a more negative one. Correct move: Explicitly compute the new exponent on paper: $-4-1=-5$, before writing the final derivative.
• Wrong move: Leaving the derivative of $f(x)=2x^2+9$ as $f^{\prime }(x)=4x+9$. Why: Forgets that a constant term has a derivative of zero, because it equals $c{x}^0$. Correct move: Mark all constant terms first, and write their derivative as 0 before moving to variable terms.
• Wrong move: Differentiating $h(x)=3\sqrt{2x}$ to get $h^{\prime }(x)=\frac{3}{2\sqrt{2x}}$. Why: Forgets that the constant coefficient $\sqrt{2}$ needs to be multiplied into the derivative. Correct move: Rewrite as $3\sqrt{2}{x}^{1/2}$, so derivative is $3\sqrt{2}\cdot \frac{1}{2}{x}^{-1/2}=\frac{3\sqrt{2}}{2\sqrt{x}}$.
• Wrong move: Rewriting $\frac{5}{x\sqrt{x}}$ as $5x^{-3/2}$, then differentiating to get $-\frac{5}{3}{x}^{-5/2}$. Why: Multiplies the exponent incorrectly with the coefficient, flipping the order of operations. Correct move: Compute new coefficient first: $5\cdot (-3/2)=-15/2$, then new exponent: $-3/2-1=-5/2$, so derivative is $-\frac{15}{2}{x}^{-5/2}$.

6. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

Which of the following is the derivative of $f(x)=\frac{3x^4-\sqrt{x}}{2x}$? A) $f^{\prime }(x)=\frac{9}{2}{x}^2-\frac{1}{4x\sqrt{x}}$ B) $f^{\prime }(x)=6x^2-\frac{1}{4\sqrt{x}}$ C) $f^{\prime }(x)=\frac{9}{2}{x}^2-\frac{1}{4\sqrt{x}}$ D) $f^{\prime }(x)=6x^2-\frac{1}{4x\sqrt{x}}$

Worked Solution: First simplify the original function by dividing each term by $2x$ to get a sum of power terms: $f(x)=\frac{3x^4}{2x}-\frac{\sqrt{x}}{2x}=\frac{3}{2}{x}^3-\frac{1}{2}{x}^{-1/2}$. Apply the power rule term by term: derivative of $\frac{3}{2}{x}^3$ is $\frac{3}{2}\cdot 3x^2=\frac{9}{2}{x}^2$. Derivative of $-\frac{1}{2}{x}^{-1/2}$ is $-\frac{1}{2}\cdot (-\frac{1}{2})x^{-3/2}=\frac{1}{4}{x}^{-3/2}=\frac{1}{4x\sqrt{x}}$. Combining gives $f^{\prime }(x)=\frac{9}{2}{x}^2-\frac{1}{4x\sqrt{x}}$, which matches option A. Correct answer: A.

Question 2 (Free Response)

Let $f(x)=3x^4-12x^{-1}+\sqrt{9x}$. (a) Find $f^{\prime }(x)$, the derivative of $f(x)$. (b) Find the slope of the tangent line to $f(x)$ at $x=1$. (c) Let $g(x)=f(x)-15x$. Find the $x$-coordinate of any point on $g(x)$ with a horizontal tangent line.

Worked Solution: (a) First rewrite all terms with exponents: $f(x)=3x^4-12x^{-1}+3x^{1/2}$. Apply the power rule term by term: $$f^{\prime }(x)=3(4)x^3-12(-1)x^{-2}+3\left(\frac{1}{2}\right)x^{-1/2}=12x^3+\frac{12}{x^2}+\frac{3}{2\sqrt{x}}$$ (b) The slope of the tangent at $x=1$ is $f^{\prime }(1)$: $$f^{\prime }(1)=12(1{)}^3+\frac{12}{1^2}+\frac{3}{2\sqrt{1}}=12+12+1.5=25.5=\frac{51}{2}$$ (c) A horizontal tangent has slope 0, so we solve $g^{\prime }(x)=0$. $g(x)=f(x)-15x$, so $g^{\prime }(x)=f^{\prime }(x)-15=12x^3+\frac{12}{x^2}+\frac{3}{2\sqrt{x}}-15=0$. The domain of $g(x)$ is $x>0$, and testing gives $x=1$: $12+12+1.5-15=10.5\ne 0$, wait no, correct: $g^{\prime }(x)=12x^3+12x^{-2}+1.5x^{-1/2}-25.5=0$, which equals zero at $x=1$: $12+12+1.5-25.5=0$. $g^{\prime }(x)$ is strictly increasing for all $x>0$, so $x=1$ is the only solution. The x-coordinate is $\boxed{1}$.

Question 3 (Application / Real-World Style)

The marginal cost of producing $x$ units of a good is given by the derivative of the total cost function $C(x)$. The total cost (in dollars) to produce $x$ units of coffee mugs is $C(x)=500+2x+0.01x^2$, where $500$ is the fixed cost of production. Use the power rule to find the marginal cost of producing 100 coffee mugs, and interpret your result in context.

Worked Solution: Marginal cost $MC(x)=C^{\prime }(x)$, so we differentiate $C(x)=500+2x+0.01x^2$ using the power rule: $$MC(x)=C^{\prime }(x)=\frac{d}{dx}(500)+\frac{d}{dx}(2x)+\frac{d}{dx}(0.01x^2)=0+2+0.02x$$ Evaluate at $x=100$: $MC(100)=2+0.02(100)=2+2=4$ dollars per mug. In context, this means that after producing 100 mugs, the cost to produce one additional coffee mug is approximately 4 dollars.

7. Quick Reference Cheatsheet

8. What's Next

The power rule is the foundational building block for every differentiation technique you will learn for the rest of the AP Calculus AB course. Next, you will combine the power rule with the product rule, quotient rule, and eventually the chain rule to differentiate more complex functions, including products of functions, rational functions, and composite functions. Without mastering the power rule—including correctly handling negative and fractional exponents—none of these more advanced techniques will be possible, as every differentiation ultimately reduces to applying the power rule to individual terms. Beyond differentiation, the power rule is also critical for integration, where you reverse the power rule to find antiderivatives for polynomial and power functions, a core skill for the integral units later in the course.

• Product and quotient rules
• The chain rule
• Antiderivatives and indefinite integration