The Power Rule

The pattern

From first principles, $x^2$x² differentiates to 2x and $x^3$x³ differentiates to $3x^2$3x². Look at where the numbers went. In both, the power came down to the front as a multiplier, and the power on x dropped by one: 2 became 1, and 3 became 2.

The same thing happens for every whole-number power n. Expanding ${\left(x+h\right)}^n$(x + h)ⁿ gives $x^n+n{x}^{n-1}h$xⁿ + nxⁿ⁻¹h, then terms in $h^2$h², $h^3$h³ and higher powers of h. Subtract $x^n$xⁿ and divide by h: the chord gradient is $n{x}^{n-1}$nxⁿ⁻¹ plus terms that each still carry an h, and those tend to 0. So $x^n$xⁿ differentiates to $n{x}^{n-1}$nxⁿ⁻¹. This is the power rule.

It agrees with what is known about lines. x is $x^1$x¹, and the rule gives $1\times x^0=1$1 × x⁰ = 1, the gradient of y = x. The constant 1 is $x^0$x⁰, and the rule gives $0\times x^{-1}=0$0 × x⁻¹ = 0, the gradient of a level line.

A cube of side x = 2 grown by $\Delta x=0.6$Δx = 0.6. The extra volume, $2.6^3-8=9.576$2.6³ − 8 = 9.576, is three slabs of $x^2\Delta x=7.2$x²Δx = 7.2 in all, three rods of $x\Delta x^2=2.16$xΔx² = 2.16 in all, and a corner of $\Delta x^3=0.216$Δx³ = 0.216. Divided by $\Delta x$Δx it is 12 + 3.6 + 0.36 = 15.96. Drag $\Delta x$Δx to 0: the rods and the corner vanish, and the three slabs leave $3x^2=12$3x² = 12.

Two moves, in order

The rule is two moves. First, bring the power down to the front, where it multiplies: $x^3$x³ becomes $3x^3$3x³. The power on x has not changed yet, so $3x^3$3x³ is not the answer. Second, lower the power by one: $3x^3$3x³ becomes $3x^2$3x². So $x^3$x³ differentiates to $3x^2$3x².

The same two moves take $x^5$x⁵ to $5x^5$5x⁵ and then to $5x^4$5x⁴, and $x^{10}$x¹⁰ to $10x^9$10x⁹. At x = 1 the derivative of $x^5$x⁵ is 5, and at x = 2 it is $5\times 2^4=80$5 × 2⁴ = 80. Check against chords with h = 0.001: $\left(1.001^5-1\right)/0.001=5.01001$(1.001⁵ − 1) / 0.001 = 5.01001 and $\left(2.001^5-32\right)/0.001=80.08004$(2.001⁵ − 32) / 0.001 = 80.08004.

The curve $y=x^3$y = x³ with its tangents at (1, 1) and (−1, −1). The derivative $3x^2$3x² is 3 at x = 1 and at x = −1, so the two tangents are parallel, both with gradient 3.

A negative power, from first principles

Before using the rule on negative powers, test it on one. $x^{-1}$x⁻¹ is $\frac{1}{x}$1/x. Its rise over a step h is $\frac{1}{x+h}-\frac{1}{x}$1/(x + h) − 1/x, which over the common denominator x(x + h) is $-h/\left(x\left(x+h\right)\right)$−h / (x(x + h)). Divide by h: $-1/\left(x\left(x+h\right)\right)$−1 / (x(x + h)). As h tends to 0 this tends to $-\frac{1}{x^2}$−1/x², which is $-x^{-2}$−x⁻².

The two moves give the same: bring −1 down to get $-1\times x^{-1}$−1 × x⁻¹, then lower the power by one, from −1 to −2, to get $-x^{-2}$−x⁻². First principles and the rule agree.

Negative powers

So negative powers take the same two moves. For $x^{-2}$x⁻², bring −2 down: $-2x^{-2}$−2x⁻². Then lower the power by one: −2 − 1 = −3, so the answer is $-2x^{-3}$−2x⁻³, which is $-\frac{2}{x^3}$−2/x³. Lowering a negative power by one moves it further below 0.

Check at x = 1, where $-2x^{-3}=-2$−2x⁻³ = −2. With h = 0.001 the chord gradient is $\left(\frac{1}{1.001^2}-1\right)/0.001=-1.997004$(1/1.001² − 1) / 0.001 = −1.997004, and with h = −0.001 it is −2.003004. Likewise $\frac{1}{x^3}=x^{-3}$1/x³ = x⁻³ differentiates to $-3x^{-4}$−3x⁻⁴, which is −3 at x = 1; the chord with h = 0.001 gives −2.99401.

A reciprocal has to be written as a negative power first. $\frac{1}{x^2}$1/x² is $x^{-2}$x⁻², so its derivative is $-2x^{-3}=-\frac{2}{x^3}$−2x⁻³ = −2/x³. It is not $\frac{1}{2x}$1/(2x), which comes from differentiating the denominator and keeping the fraction.

The curve $y=\frac{1}{x}$y = 1/x with its tangents at (1, 1) and (2, 0.5). The derivative $-x^{-2}$−x⁻² is −1 at x = 1 and $-\frac{1}{4}$−1/4 at x = 2, so the curve falls more gently further out.

Roots are powers

A root is a fractional power. $\sqrt{x}$√x is $x^{\frac{1}{2}}$x^(1/2), and the same two moves apply. Bring $\frac{1}{2}$1/2 down: $\left(\frac{1}{2}\right)x^{\frac{1}{2}}$(1/2)x^(1/2). Lower the power by one: $\frac{1}{2}-1=-\frac{1}{2}$1/2 − 1 = −1/2. So $\sqrt{x}$√x differentiates to $\left(\frac{1}{2}\right)x^{-\frac{1}{2}}$(1/2)x^(−1/2). Written back with the root, $x^{-\frac{1}{2}}$x^(−1/2) is $\frac{1}{\sqrt{x}}$1/√x, so the derivative is $\frac{1}{2\sqrt{x}}$1/(2√x).

Check at x = 4, where $\frac{1}{2\sqrt{4}}=\frac{1}{4}=0.25$1/(2√4) = 1/4 = 0.25. The chord with h = 0.01 gives $\left(\sqrt{4.01}-2\right)÷0.01=0.249844$(√4.01 − 2) ÷ 0.01 = 0.249844.

The cube root runs the same way. $\sqrt[3]{x}$∛x is $x^{\frac{1}{3}}$x^(1/3), which gives $\left(\frac{1}{3}\right)x^{-\frac{2}{3}}$(1/3)x^(−2/3), that is $\frac{1}{3\sqrt[3]{x^2}}$1/(3∛(x²)). At x = 8, $\sqrt[3]{64}=4$∛64 = 4, so the derivative is $\frac{1}{12}=0.083333$1/12 = 0.083333; the chord with h = 0.01 gives 0.083299. At x = 0 both formulas divide by 0: $\sqrt{x}$√x and $\sqrt[3]{x}$∛x stand vertical there and have no derivative.

The curve $y=\sqrt{x}$y = √x with its tangent at (4, 2), the line y = 0.25x + 1. The gradient there is $\frac{1}{2\sqrt{4}}=\frac{1}{4}$1/(2√4) = 1/4.

Rewrite first, then the two moves

Any power of x, written as a root or a fraction, is rewritten as $x^n$xⁿ first. $\frac{1}{\sqrt{x}}$1/√x is $x^{-\frac{1}{2}}$x^(−1/2), which differentiates to $-\left(\frac{1}{2}\right)x^{-\frac{3}{2}}$−(1/2)x^(−3/2); at x = 4 that is $-\frac{1}{16}=-0.0625$−1/16 = −0.0625, and the chord with h = 0.001 gives −0.062488.

$x\sqrt{x}$x√x is $x^1\times x^{\frac{1}{2}}=x^{\frac{3}{2}}$x¹ × x^(1/2) = x^(3/2), which differentiates to $\left(\frac{3}{2}\right)x^{\frac{1}{2}}=\left(\frac{3}{2}\right)\sqrt{x}$(3/2)x^(1/2) = (3/2)√x; at x = 4 that is 3, and the chord with h = 0.001 gives 3.000187. $\frac{1}{x^4}$1/x⁴ is $x^{-4}$x⁻⁴, which differentiates to $-4x^{-5}$−4x⁻⁵.

The usual mistakes

Bringing the power down and not lowering it: $x^5$x⁵ to $5x^5$5x⁵. The second move is part of the rule; the answer is $5x^4$5x⁴.

Lowering the power without bringing it down: $x^5$x⁵ to $x^4$x⁴. The 5 has to come to the front first.

Dropping the minus sign of a negative power: $x^{-2}$x⁻² to $2x^{-3}$2x⁻³. The power −2 comes to the front with its sign, giving $-2x^{-3}$−2x⁻³.

Lowering a negative power the wrong way: $x^{-2}$x⁻² to $-2x^{-1}$−2x⁻¹. One less than −2 is −3.

Stopping after the first move on a root: $\sqrt{x}$√x to $\frac{\sqrt{x}}{2}$√x/2. The power must also drop, from $\frac{1}{2}$1/2 to $-\frac{1}{2}$−1/2, giving $\frac{1}{2\sqrt{x}}$1/(2√x). And $\frac{2}{\sqrt{x}}$2/√x turns the half upside down: the power comes down as $\frac{1}{2}$1/2, not 2.