# Derivative of Cube Root of x: Formula, Proof and Examples

> The derivative of the cube root of x is 1/(3∛(x²)). See the power rule and limit proofs, the vertical tangent at 0 and chain rule examples.

Source: https://practicecalculus.com/derivatives/cube-root-x
Updated: 2026-11-06

$$ \frac{d}{dx}\,\sqrt[3]{x} = \frac{1}{3\sqrt[3]{x^2}} $$

## At a glance

- **Derivative:** $\dfrac{1}{3\sqrt[3]{x^2}}$ for $x \ne 0$
- **As a power:** $\dfrac{d}{dx}\,x^{1/3} = \tfrac{1}{3}x^{-2/3}$
- **Chain rule form:** $\dfrac{d}{dx}\sqrt[3]{u} = \dfrac{u'}{3\sqrt[3]{u^2}}$
- **At x = 0:** Not differentiable, the tangent is vertical
- **Domain:** All real numbers, unlike $\sqrt{x}$

## Key takeaways

- The derivative of $\sqrt[3]{x}$ is $\frac{1}{3}x^{-2/3} = \frac{1}{3\sqrt[3]{x^2}}$.
- The cube root is defined for negative $x$, and the formula works there too.
- The derivative is always positive, so $\sqrt[3]{x}$ always increases.
- At $x = 0$ the slope is infinite, so the derivative does not exist there.

The **derivative of $\sqrt[3]{x}$ is $\dfrac{1}{3\sqrt[3]{x^2}}$**, for every $x \ne 0$. Write the cube root as the power $x^{1/3}$ and the power rule does the rest.

$$\frac{d}{dx}\,\sqrt[3]{x} = \frac{1}{3}x^{-2/3} = \frac{1}{3\sqrt[3]{x^2}}$$

## Graph of ∛x and its derivative

```graph
title: 'Graph of y = ∛x and its derivative'
caption: 'The cube root passes through the origin with a vertical tangent. Its derivative is positive on both sides and spikes near $x = 0$.'
x: [-6, 6]
y: [-2.5, 3]
functions:
  - expr: x^(1/3)
    label: y = ∛x
  - expr: 1/(3abs(x)^(2/3))
    label: y = 1/(3∛(x²)) (derivative)
    dashed: true
points:
  - [1, 1, '(1, 1), slope 1/3']
  - [-1, -1, '(-1, -1), slope 1/3']
```

Notice that the derivative is the same at $x = 1$ and $x = -1$. Since $x^2$ is the same for $x$ and $-x$, the derivative is an even function.

## Proof with the power rule

Using the [power rule](/rules/power-rule) with $n = \frac{1}{3}$:

$$\frac{d}{dx}\,x^{1/3} = \frac{1}{3}x^{1/3 - 1} = \frac{1}{3}x^{-2/3}$$

A negative exponent moves the power to the bottom, and $x^{2/3} = \sqrt[3]{x^2}$.

## Proof from the limit definition

Let $a = \sqrt[3]{x + h}$ and $b = \sqrt[3]{x}$. Then $a^3 - b^3 = h$. The difference of cubes $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ gives

$$\frac{a - b}{h} = \frac{a - b}{a^3 - b^3} = \frac{1}{a^2 + ab + b^2}$$

As $h \to 0$, $a \to b$, so the limit is $\frac{1}{3b^2} = \frac{1}{3\sqrt[3]{x^2}}$. This needs $b \ne 0$, which is exactly why $x = 0$ fails.

## Why the derivative fails at 0

At $x = 0$ the difference quotient is $\frac{\sqrt[3]{h}}{h} = h^{-2/3}$, which goes to $+\infty$ from both sides. The curve is continuous but has a vertical tangent, a standard example of a function that is continuous but not [differentiable](/learn/calculus-1/the-derivative/differentiability).

## Chain rule and other roots

For a cube root of an expression, the [chain rule](/rules/chain-rule) gives

$$\frac{d}{dx}\sqrt[3]{u} = \frac{u'}{3\sqrt[3]{u^2}}$$

**Example 1.** $\frac{d}{dx}\sqrt[3]{5x - 1} = \frac{5}{3\sqrt[3]{(5x - 1)^2}}$.

**Example 2.** $\frac{d}{dx}\,6\sqrt[3]{x} = 6 \cdot \frac{1}{3}x^{-2/3} = 2x^{-2/3}$.

**Example 3.** Any root works the same way: $\frac{d}{dx}\sqrt[n]{x} = \frac{1}{n}x^{\frac{1}{n} - 1}$. For $n = 4$ this is $\frac{1}{4}x^{-3/4}$.

**Example 4: a tangent line.** At $x = 8$, $\sqrt[3]{8} = 2$ and the slope is $\frac{1}{3 \cdot 4} = \frac{1}{12}$. So $\sqrt[3]{9} \approx 2 + \frac{1}{12} \approx 2.083$. The true value is about $2.080$.

## Common mistakes

1. **Writing $x^{-1/3}$.** The new exponent is $\frac{1}{3} - 1 = -\frac{2}{3}$.
2. **Putting the 3 in the wrong place.** The answer is $\frac{1}{3}x^{-2/3}$, not $3x^{-2/3}$.
3. **Assuming negative $x$ is off limits.** Cube roots of negatives are real, so the derivative exists for every $x$ except $0$.

Compare with the [derivative of √x](/derivatives/sqrt-x), which is only defined for $x > 0$.

## Frequently asked questions

### What is the derivative of the cube root of x?

It is $\frac{1}{3\sqrt[3]{x^2}}$, which is the same as $\frac{1}{3}x^{-2/3}$.

### Is the cube root of x differentiable at 0?

No. The function is continuous at $0$, but the difference quotient $\frac{h^{1/3}}{h} = h^{-2/3}$ grows without bound, so the tangent at the origin is vertical.

### What is the derivative of x^(2/3)?

By the power rule, $\frac{2}{3}x^{-1/3} = \frac{2}{3\sqrt[3]{x}}$. Unlike $x^{1/3}$, this one changes sign at $0$, giving a sharp cusp.

### What is the derivative of the cube root of (x² + 1)?

By the chain rule, $\frac{1}{3}(x^2 + 1)^{-2/3} \cdot 2x = \frac{2x}{3\sqrt[3]{(x^2 + 1)^2}}$.