Where a Derivative Fails to Exist

What a derivative needs

The derivative of f at a is the limit of the chord gradient $\left(f\left(a+h\right)-f\left(a\right)\right)/h$(f(a + h) − f(a)) / h as h tends to 0. Like any limit, it exists only when the values from the right (h positive) and from the left (h negative) settle on the same number. A derivative has to be a single, finite number.

So a derivative can fail in a few ways. At a corner the two sides settle on two different numbers. At a cusp they grow without bound, one positive and one negative. At a vertical tangent they grow without bound with the same sign. At a jump, one side runs off because the curve itself is broken.

A corner: y = |x| at 0

For f(x) = |x| at 0, the chord gradient is $\left(|h|-0\right)/h=|h|÷h$(|h| − 0) / h = |h| ÷ h. For every positive h, |h| = h and the quotient is 1. For every negative h, |h| = −h and the quotient is −1. So with h = 0.1 it is 1, and with h = −0.1 it is −1, and with h = 0.001 and h = −0.001 it is still 1 and −1.

From the right the chord gradients settle on 1, and from the left on −1. Two different one-sided limits mean there is no limit, so |x| has no derivative at 0. The graph shows why: to the right of 0 it is the line y = x, with gradient 1, and to the left the line y = −x, with gradient −1, and they meet in a corner.

Away from 0 nothing goes wrong. At x = 2 the curve is the line y = x, and the chord from 2 to 3 has gradient 3 − 2 = 1, its derivative there. At x = −3 the derivative is −1.

Two chords from the origin, to the points at x = h and x = −h, with both gradients printed. On y = |x| they stay at 1 and −1 however small h becomes. The second button draws the cusp, where they grow without bound in opposite directions; the third draws $y=\frac{x^2}{2}$y = x²/2, where both close on 0, the derivative there.

A cusp

The curve $y=\sqrt[3]{x^2}$y = ∛(x²), the cube root of $x^2$x², comes down to the origin from both sides and meets itself in a sharp point, called a cusp. At 0 the chord gradient is $\sqrt[3]{h^2}/h$∛(h²) / h, which simplifies to $1÷\sqrt[3]{h}$1 ÷ ∛h.

From the right: with h = 1 it is 1, with $h=\frac{1}{8}$h = 1/8 it is 2, with h = 0.001 it is 10, and with h = 0.000001 it is 100. From the left the values are the same with a minus sign: −1, −2, −10 and −100. The gradients on the right run to $+\infty$+∞ and those on the left to $-\infty$−∞, so neither side settles on a number, and there is no derivative at 0.

Chord gradients of $y=\sqrt[3]{x^2}$y = ∛(x²) from the origin. Each row is one size of step, from 1 down to $10^{-6}\left(\text{one millionth}\right)$10⁻⁶ (one millionth), taken to the right of 0 and to the left. Each time the step is divided by 1000, the gradient is multiplied by 10, with opposite signs on the two sides.

The cusp of $y=\sqrt[3]{x^2}$y = ∛(x²) at the origin, with the chords to $\left(\frac{1}{8},\frac{1}{4}\right)$(1/8, 1/4) and $\left(-\frac{1}{8},\frac{1}{4}\right)$(−1/8, 1/4). Their gradients are 2 and −2, and the closer the second point comes, the steeper the two chords stand.

A vertical tangent: $y=\sqrt[3]{x}$y = ∛x at 0

The curve $y=\sqrt[3]{x}$y = ∛x passes through the origin standing upright. The chord gradient at 0 is $\sqrt[3]{h}/h=1÷\sqrt[3]{h^2}$∛h / h = 1 ÷ ∛(h²). The chord to (8, 2) has gradient $2÷8=\frac{1}{4}$2 ÷ 8 = 1/4, the chord to (1, 1) has gradient 1, the chord to $\left(\frac{1}{8},\frac{1}{2}\right)$(1/8, 1/2) has gradient 4, and with h = 0.001 the gradient is 100.

From the left the values are the same, and positive: the chord to $\left(-\frac{1}{8},-\frac{1}{2}\right)$(−1/8, −1/2) also has gradient 4. Both sides grow without bound, so the chords close on the vertical line x = 0. That is the tangent at the origin, and a vertical line has no gradient, because its run is 0 and a rise divided by 0 is not a number. So $\sqrt[3]{x}$∛x has no derivative at 0, though its tangent there is a perfectly good line.

The curve $y=\sqrt[3]{x}$y = ∛x with the chords from the origin to (1, 1), gradient 1, and to $\left(\frac{1}{8},\frac{1}{2}\right)$(1/8, 1/2), gradient 4. The closer the second point, the nearer the chord stands to the vertical tangent x = 0.

A jump

Let f(x) = 1 for x < 1 and f(x) = 3 for $x\ge 1$x ≥ 1, so the graph jumps from height 1 up to 3 at x = 1, and f(1) = 3. From the right, every chord from (1, 3) runs along the level piece, so $\left(3-3\right)/h=0$(3 − 3) / h = 0 for every positive h.

From the left, the second point is on the lower piece, so the rise is 1 − 3 = −2 and the quotient is $-2/h$−2 / h. With h = −0.1 that is 20, with h = −0.01 it is 200, and with h = −0.001 it is 2000. These chords join a point at height 1 to the point at height 3, and as h shrinks they stand more and more upright. The left side settles on nothing, so there is no derivative at 1. Where the curve is broken there is no single tangent to shrink the chords onto.

A jump at x = 1: y = 1 for x < 1 and y = 3 for $x\ge 1$x ≥ 1. The hollow dot at (1, 1) is where the left piece ends without reaching x = 1; the solid dot at (1, 3) is the value f(1) = 3.

Differentiable forces continuous

If f has a derivative at a, then f is continuous at a. The reason is one line. For h not 0, $f\left(a+h\right)-f\left(a\right)=h\times \left(\left(f\left(a+h\right)-f\left(a\right)\right)/h\right)$f(a + h) − f(a) = h × ((f(a + h) − f(a)) / h). As h tends to 0, the first factor tends to 0 and the second tends to f'(a), a number. The limit of a product is the product of the limits, so f(a + h) − f(a) tends to 0 × f'(a) = 0, which says f(a + h) tends to f(a): f is continuous at a.

So a break of any kind, a jump or a hole or an asymptote, rules out a derivative there. The jump above has no derivative at 1 for exactly this reason.

The reverse fails

A continuous function need not have a derivative. y = |x| is continuous at 0: from both sides it tends to 0, and |0| = 0. Yet it has no derivative there. The cusp of $\sqrt[3]{x^2}$∛(x²) and the vertical tangent of $\sqrt[3]{x}$∛x are unbroken too, and neither has a derivative at 0.

So continuity is needed for a derivative, but it is not enough. To decide whether a derivative exists at a point, first check that the curve is unbroken there, and then work out the chord gradient from each side: the derivative exists only if both sides settle on the same finite number.

The usual mistakes

Saying |x| has derivative 0 at 0, by averaging −1 and 1. The two sides disagree, and a derivative is one limit, not the average of two.

Saying the derivative of |x| at 0 is 1. That is the right-hand side only; the left-hand side gives −1.

Giving "very large" or "infinity" as the derivative of $\sqrt[3]{x}$∛x at 0. The chord gradients grow without bound and settle on no number, so there is no derivative.

Giving the size of a jump as the derivative. A jump from 1 to 3 is a change in height of 2, not a rate of change at a point.

Taking continuity to promise a derivative. A curve can be unbroken with a corner, a cusp or a vertical tangent.

A reservoir

In the application below, a reservoir's depth follows one formula in a dry spell and a straight line once the rain starts on day 6. Both give 102 centimeters on day 6, so the depth is unbroken, but the chord gradients from the left tend to −6 and those from the right are 5: a corner, with no derivative on day 6.