The Power Rule: Why the Exponent Comes Down
Here's the single most useful fact you'll learn in calculus:
d/dx [xⁿ] = n·xⁿ⁻¹
That's it. If you have x raised to any power, the derivative is that power times x raised to one less than that power. Take the exponent, pull it down as a coefficient, then subtract one from the exponent.
Why does this matter? Because almost every function you'll ever differentiate can be broken down into powers of x. Polynomials, rational functions, radicals—they're all power functions in disguise. And the power rule handles all of them in one move.
You want the derivative of x³? It's 3x². Done.
You want the derivative of x¹⁰⁰? It's 100x⁹⁹. Done.
You want the derivative of 1/x? Rewrite it as x⁻¹. The derivative is -1·x⁻² = -1/x². Done.
You want the derivative of √x? Rewrite it as x^(1/2). The derivative is (1/2)·x^(-1/2) = 1/(2√x). Done.
The power rule doesn't just work for positive integers. It works for negative exponents, fractional exponents, even irrational exponents. If you can write it as xⁿ, the rule applies.
Let's figure out why it works, see it in action, and understand why this simple pattern unlocks the entire machinery of calculus.
What the Power Rule Does
The derivative measures how fast a function is changing at any given point. If you have a function f(x) = xⁿ, the derivative f'(x) tells you the instantaneous rate of change of that function.
For f(x) = x²:
- At x = 1: f'(1) = 2(1) = 2. The function is increasing at a rate of 2 units per unit of x.
- At x = 2: f'(2) = 2(2) = 4. The function is increasing twice as fast.
- At x = 3: f'(3) = 2(3) = 6. Faster still.
The derivative f'(x) = 2x tells you that the rate of change is proportional to x. The farther out you go, the steeper the parabola gets.
For f(x) = x³:
- f'(x) = 3x²
- At x = 1: f'(1) = 3. Increasing at rate 3.
- At x = 2: f'(2) = 12. Much steeper.
- At x = 3: f'(3) = 27. Explosive growth.
The higher the power, the faster the function accelerates as x increases.
Why the Power Rule Works: The Limit Definition
The derivative is defined as a limit:
f'(x) = lim (h→0) [f(x+h) - f(x)] / h
This says: To find the rate of change at x, look at how much f changes when you nudge x by a tiny amount h, divide by that nudge, and take the limit as h approaches zero.
Let's prove the power rule for f(x) = x² using this definition.
Start: f'(x) = lim (h→0) [(x+h)² - x²] / h
Expand (x+h)²: = lim (h→0) [x² + 2xh + h² - x²] / h
Simplify: = lim (h→0) [2xh + h²] / h
Factor out h: = lim (h→0) [h(2x + h)] / h
Cancel h: = lim (h→0) [2x + h]
Take the limit as h → 0: = 2x
Done. The derivative of x² is 2x.
Now let's do it for x³.
f'(x) = lim (h→0) [(x+h)³ - x³] / h
Expand (x+h)³ using the binomial theorem: (x+h)³ = x³ + 3x²h + 3xh² + h³
Substitute: = lim (h→0) [x³ + 3x²h + 3xh² + h³ - x³] / h
Simplify: = lim (h→0) [3x²h + 3xh² + h³] / h
Factor out h: = lim (h→0) [h(3x² + 3xh + h²)] / h
Cancel h: = lim (h→0) [3x² + 3xh + h²]
Take the limit as h → 0: = 3x²
The derivative of x³ is 3x².
See the pattern? When you expand (x+h)ⁿ, the binomial theorem gives you:
(x+h)ⁿ = xⁿ + n·xⁿ⁻¹·h + (terms with h², h³, ...)
The xⁿ cancels with -xⁿ. The term n·xⁿ⁻¹·h survives when you divide by h. The higher-order terms (h², h³, ...) all have h as a factor, so they vanish when you take the limit as h → 0.
What's left? n·xⁿ⁻¹.
That's the power rule. The algebra always works out the same way, no matter what n is.
The Power Rule in Action: Examples
Example 1: f(x) = x⁵
- f'(x) = 5x⁴
Example 2: f(x) = x
- Rewrite as f(x) = x¹
- f'(x) = 1·x⁰ = 1
The derivative of x is always 1. This makes intuitive sense: the function f(x) = x is a straight line with slope 1. The derivative is constant.
Example 3: f(x) = 7
- Rewrite as f(x) = 7x⁰ = 7·1 = 7 (constant function)
- f'(x) = 0·x⁻¹ = 0
The derivative of any constant is zero. Constants don't change, so their rate of change is zero.
Example 4: f(x) = x⁻¹ = 1/x
- f'(x) = -1·x⁻² = -1/x²
Example 5: f(x) = x⁻² = 1/x²
- f'(x) = -2·x⁻³ = -2/x³
Example 6: f(x) = √x = x^(1/2)
- f'(x) = (1/2)·x^(-1/2) = 1/(2√x)
Example 7: f(x) = ∛x = x^(1/3)
- f'(x) = (1/3)·x^(-2/3) = 1/(3∛(x²))
Example 8: f(x) = 1/√x = x^(-1/2)
- f'(x) = (-1/2)·x^(-3/2) = -1/(2x^(3/2)) = -1/(2x√x)
Example 9: f(x) = x^π (yes, irrational exponents work too)
- f'(x) = π·x^(π-1)
The rule works for any real number exponent.
Why Negative and Fractional Exponents Work
The power rule extends beyond positive integers because the limit definition doesn't care what n is. The binomial expansion (x+h)ⁿ generalizes to any real n via the binomial series:
(x+h)ⁿ = xⁿ [1 + (n/1)·(h/x) + (n(n-1)/2)·(h/x)² + ...]
When n is not a positive integer, the series is infinite. But the first term still dominates when h is small, and the limit still picks out n·xⁿ⁻¹.
For negative exponents:
- f(x) = x⁻ⁿ = 1/xⁿ
- Using the chain rule (which we'll get to later) or direct computation, you find f'(x) = -n·x⁻ⁿ⁻¹
For fractional exponents:
- f(x) = x^(p/q)
- f'(x) = (p/q)·x^(p/q - 1)
The algebra is messier, but the pattern holds.
Polynomials: Where the Power Rule Shines
A polynomial is a sum of power functions:
f(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₂x² + a₁x + a₀
To differentiate a polynomial, differentiate each term separately (this is called linearity of the derivative):
f'(x) = n·aₙxⁿ⁻¹ + (n-1)·aₙ₋₁xⁿ⁻² + ... + 2a₂x + a₁
The constant term a₀ disappears because its derivative is zero.
Example: f(x) = 3x⁴ - 5x³ + 2x² - 7x + 11
Differentiate term by term:
- d/dx [3x⁴] = 12x³
- d/dx [-5x³] = -15x²
- d/dx [2x²] = 4x
- d/dx [-7x] = -7
- d/dx [11] = 0
So: f'(x) = 12x³ - 15x² + 4x - 7
That's it. One line.
What About Constants? The Constant Multiple Rule
If you have a constant times a function, the derivative is that constant times the derivative of the function:
d/dx [c·f(x)] = c·f'(x)
This is why we could handle 3x⁴ by taking 3 times the derivative of x⁴.
d/dx [3x⁴] = 3 · d/dx [x⁴] = 3 · 4x³ = 12x³
The constant multiplier just tags along through the differentiation.
When the Power Rule Doesn't Apply (Yet)
The power rule works when the base is the variable and the exponent is a constant. But it doesn't directly handle:
- Functions in the exponent: f(x) = 2ˣ
- This is an exponential function, not a power function. The derivative is 2ˣ·ln(2), not x·2^(x-1).
- Variable in both base and exponent: f(x) = xˣ
- This requires logarithmic differentiation. The derivative is xˣ·(1 + ln x).
- Composite functions: f(x) = (3x + 5)⁴
- You need the chain rule. The power rule applies to the outer function, but you have to account for the inner function (3x + 5).
The power rule is powerful, but it's not the whole story. You'll need other tools (product rule, quotient rule, chain rule) to handle more complex functions. But those tools often use the power rule as a component.
The Power Rule and Physics: Velocity and Acceleration
In physics, if position is a function of time, x(t), then:
- Velocity is the derivative of position: v(t) = dx/dt
- Acceleration is the derivative of velocity: a(t) = dv/dt = d²x/dt²
If an object's position is given by x(t) = 5t², then:
- Velocity: v(t) = d/dt [5t²] = 10t
- Acceleration: a(t) = d/dt [10t] = 10
The acceleration is constant at 10 units/time². This describes motion under constant acceleration—like free fall near Earth's surface (where a ≈ 9.8 m/s²).
If position is x(t) = t³ - 6t² + 9t, then:
- Velocity: v(t) = 3t² - 12t + 9
- Acceleration: a(t) = 6t - 12
The acceleration is no longer constant. It's a linear function of time.
The power rule makes computing these derivatives trivial. Without it, you'd have to do the limit definition every time.
The Power Rule and Optimization
One of the main applications of derivatives is finding maxima and minima. You set the derivative equal to zero and solve.
Example: A farmer has 100 meters of fencing and wants to build a rectangular pen with maximum area. One side of the pen is against a barn, so he only needs to fence three sides. What dimensions maximize the area?
Let x = width of the pen (perpendicular to barn) Let y = length of the pen (parallel to barn)
Constraint: 2x + y = 100 (total fencing) So: y = 100 - 2x
Area: A = x·y = x(100 - 2x) = 100x - 2x²
To maximize area, take the derivative and set it to zero: dA/dx = 100 - 4x = 0 4x = 100 x = 25
So: y = 100 - 2(25) = 50
The optimal dimensions are 25 meters wide by 50 meters long, giving an area of 1250 m².
The power rule made the derivative of 100x - 2x² instant: 100 - 4x. No limit computation needed.
The Power Rule and Taylor Series
When you approximate a function near a point using a Taylor series, the coefficients involve derivatives. The power rule makes computing those derivatives straightforward.
For f(x) = x³, the derivatives are:
- f(x) = x³
- f'(x) = 3x²
- f''(x) = 6x
- f'''(x) = 6
- f''''(x) = 0 (and all higher derivatives are zero)
The Taylor series around x = 0 is: f(x) = 0 + 0·x + 0·x² + (6/6)·x³ = x³
Which is just the original function. But for more complicated functions, the power rule simplifies the derivative chain enormously.
The Power Rule for Functions of Functions: A Preview of the Chain Rule
What if you have (3x + 5)⁴? You can't directly apply the power rule because the base isn't just x—it's a function of x.
You need the chain rule, which says: d/dx [f(g(x))] = f'(g(x)) · g'(x)
For (3x + 5)⁴:
- Outer function: f(u) = u⁴, so f'(u) = 4u³
- Inner function: g(x) = 3x + 5, so g'(x) = 3
- Combine: d/dx [(3x+5)⁴] = 4(3x+5)³ · 3 = 12(3x+5)³
The power rule applies to the outer function, then you multiply by the derivative of the inner function.
This generalizes to any composition. The power rule is the backbone, the chain rule is the connector.
Why the Power Rule Is So Damn Useful
Because it's:
- Fast. One step. No limit computation.
- General. Works for any exponent—positive, negative, fractional, irrational.
- Composable. Combines with other rules (product, quotient, chain) to handle complex functions.
- Everywhere. Polynomials, rational functions, radicals—they all reduce to power functions.
If you memorize one derivative rule, make it this one. Everything else builds on it.
Common Mistakes
Mistake 1: Applying the power rule to the exponent instead of the base.
Wrong: d/dx [x³] = x³·ln(x) (that's for 3ˣ, not x³) Right: d/dx [x³] = 3x²
Mistake 2: Forgetting to subtract 1 from the exponent.
Wrong: d/dx [x⁴] = 4x⁴ Right: d/dx [x⁴] = 4x³
Mistake 3: Thinking the power rule applies when the variable is in the exponent.
Wrong: d/dx [2ˣ] = x·2^(x-1) Right: d/dx [2ˣ] = 2ˣ·ln(2)
Mistake 4: Not rewriting roots and fractions as powers.
Hard way: d/dx [√x] = ??? (no direct rule for radicals) Easy way: d/dx [x^(1/2)] = (1/2)·x^(-1/2) = 1/(2√x)
The Power Rule in Reverse: Antiderivatives
If differentiation pulls the exponent down and subtracts one, integration (antidifferentiation) does the reverse: adds one to the exponent and divides.
∫ xⁿ dx = x^(n+1) / (n+1) + C (for n ≠ -1)
This is the power rule in reverse. To check: differentiate x^(n+1)/(n+1) and you get xⁿ back.
For n = -1, the antiderivative is ln|x| + C, which is a special case (because x⁰ / 0 is undefined).
But for all other powers, the reverse power rule works perfectly.
The Big Sentence
The power rule is one exponent move: pull it down, subtract one, done.
Further Reading
- Stewart, James. Calculus: Early Transcendentals (8th ed.). Cengage Learning, 2015.
- Chapter 3 covers differentiation rules, including the power rule with detailed examples.
- Spivak, Michael. Calculus (4th ed.). Publish or Perish, 2008.
- Rigorous treatment of the derivative and proof of the power rule using the binomial theorem.
- Strang, Gilbert. Calculus (3rd ed.). Wellesley-Cambridge Press, 2017.
- Clear presentation of the power rule and its applications to polynomials and optimization.
- Thomas, George B., et al. Thomas' Calculus (14th ed.). Pearson, 2017.
- Comprehensive examples of the power rule applied to physics, engineering, and economics.
- Apostol, Tom M. Calculus, Volume 1 (2nd ed.). Wiley, 1991.
- Proof-oriented approach to derivatives with careful treatment of the power rule for all real exponents.
Part 3 of the Calculus Derivatives series.
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