The Power Rule

The pattern

From first principles, $x^{2}$ differentiates to 2x and $x^{3}$ differentiates to $3x^{2}$. 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 $\left(x + h\right)^{n}$ gives $x^{n} + nx^{n-1}h$, then terms in $h^{2}$, $h^{3}$ and higher powers of h. Subtract $x^{n}$ and divide by h: the chord gradient is $nx^{n-1}$ plus terms that each still carry an h, and those tend to 0. So $x^{n}$ differentiates to $nx^{n-1}$. This is the power rule.

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

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

Two moves, in order

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

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

The curve $y = x^{3}$ with its tangents at (1, 1) and (−1, −1). The derivative $3x^{2}$ 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. $x^{-1}$ is $\frac{1}{x}$. Its rise over a step h is $\frac{1}{{x + h}} - \frac{1}{x}$, which over the common denominator x(x + h) is $-h / \left(x\left(x + h\right)\right)$. Divide by h: $-1 / \left(x\left(x + h\right)\right)$. As h tends to 0 this tends to $- \frac{1}{x^{2}}$, which is $-x^{-2}$.

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

Negative powers

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

Check at x = 1, where $-2x^{-3} = -2$. With h = 0.001 the chord gradient is $\left(\frac{1}{1.001^{2}} - 1\right) / 0.001 = -1.997004$, and with h = −0.001 it is −2.003004. Likewise $\frac{1}{x^{3}} = x^{-3}$ differentiates to $-3x^{-4}$, 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. $\frac{1}{x^{2}}$ is $x^{-2}$, so its derivative is $-2x^{-3} = - \frac{2}{x^{3}}$. It is not $\frac{1}{{2x}}$, which comes from differentiating the denominator and keeping the fraction.

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

Roots are powers

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

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

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

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

Rewrite first, then the two moves

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

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

The usual mistakes

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

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

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

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

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