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The power rule

Basic Calculus · Derivatives

The Power Rule

Bring the power down in front, then lower it by one. That single move differentiates every polynomial you will ever meet.

Differentiate

f(x) = 3x⁴

The rule: xⁿ becomes nxⁿ⁻¹.

Power down in front, power minus one behind.

x² → 2x

the slope machine of the parabola

x³ → 3x²

same move, one power up

x⁵ → 5x⁴, x → 1

x is x¹, so it becomes 1·x⁰ = 1

Coefficients ride along; lone constants vanish.

3x⁴ → 3 · 4x³ = 12x³

the 3 multiplies what the rule produces

7 → 0

y = 7 is a flat line — slope zero everywhere

A constant's derivative is 0, not the constant. If it has no x, it contributes nothing to the slope.

Sums: differentiate term by term.

f(x) = 2x³ − 5x + 7

three terms, three little jobs

2x³ → 6x², −5x → −5, 7 → 0

each term on its own

f′(x) = 6x² − 5

reassemble

Asked for the x that gives a stated slope? Set f′ equal to it and solve. For f(x) = x³ + 3x², where is the slope 24? f′(x) = 3x² + 6x = 24, so x² + 2x − 8 = 0, (x + 4)(x − 2) = 0, and x = 2 or x = −4.

Roots and fractions: rewrite as powers first.

The rule works for every power, negative and fractional too. Differentiate 8√x and 3x³.

8√x = 8x^1/2

a square root is the power ½

→ 8 · ½ x^−1/2 = 4√x

½ − 1 = −½; a negative power goes downstairs

3x³ = 3x⁻³

a power on the bottom becomes negative

→ 3 · (−3)x⁻⁴ = −9x⁴

−3 − 1 = −4, and the −3 comes down

Lowering a negative power makes it MORE negative

x⁻³ becomes x⁻⁴, not x⁻². And 1x³ is never differentiated as “13x²” — rewrite first, then apply the rule.

Tidy first: expand products, split fractions.

Differentiate (x − 2)(3x + 1), and find g′(2) for g(x) = x³ + 4x.

(x − 2)(3x + 1) = 3x² − 5x − 2

expand — never differentiate each bracket and multiply

→ 6x − 5

term by term

x³ + 4x = x² + 4x⁻¹

divide each top term by x

g′(x) = 2x − 4x² → g′(2) = 4 − 1 = 3

plug in last

Unknown coefficients work the same way. If f(x) = ax³ + x and f′(1) = 7, then f′(x) = 3ax² + 1, so 3a + 1 = 7 and a = 2. Two conditions on f′ give two equations for two unknowns.

Variations: a number under the bar, a root of a power, a root on the bottom.

A number below the bar. Differentiate x⁴8 and 52x.

x⁴8 = ⅛x⁴ → ⅛ · 4x³ = ½x³

a number below is just a coefficient

52x = (52)x⁻¹

only the x moves up; the 2 stays below the 5

→ (52)(−1)x⁻² = −52x²

power down, then minus one

What is different: the coefficient is a fraction. 52x is not 10x⁻¹ — the 2 divides, it does not multiply.

A root of a power. Find f′(8) for f(x) = ³√x².

³√x² = x^2/3

power over root

f′(x) = ⅔x^−1/3 = 23³√x

23 − 1 = −13

f′(8) = 23 × 2 = 13

³√8 = 2

What is different: subtracting 1 from a fraction crosses zero and makes the power negative, so x ends up below the bar.

A root on the bottom. Find g′(4) for g(x) = 6√x.

6√x = 6x^−1/2

up from the bottom: the sign flips

g′(x) = 6 · (−½)x^−3/2 = −3x^−3/2

−½ − 1 = −32

g′(4) = −3 ÷ 4^3/2 = −3 ÷ 8 = −38

4^3/2 = (√4)³ = 8

What is different: two signs to track — the negative power and the negative result. The curve is falling at x = 4.

The three classic slips.

Slip	Looks like
power not lowered	3x⁴ → 12x⁴
power not brought down	3x⁴ → 3x³
constant kept alive	… + 7 → … + 7
negative power raised	x⁻² → −2x⁻¹

On the exam

Exam derivatives hide the power rule behind a root, a 1x, a product or a fraction — rewrite as a sum of powers first, then differentiate. Questions then ask for f′ at a point, or the x (or coefficient) that gives a stated slope. Quick check: the derivative of a degree-n polynomial has degree n − 1.

Check: it must agree with slopes you can see.

y = x² f′(1) = 2 matches the tangent drawn in the slope lesson ✓

y = x² f′(0) = 0 flat at the vertex, as every parabola is ✓

Remember the flow

Power down front · Subtract one · Coefficients ride · Lone constants die

Term by term, every polynomial falls in seconds.

Now you try

Practice problem

Find the slope at x = 2

f(x) = x⁴ − 3x² + 5x − 2

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