# Derivative of 1/x²: Formula, Proof and Examples

> The derivative of 1/x² is -2/x³. See the power rule and quotient rule proofs, the chain rule for 1/u², a graph and examples like 1/(3x - 2)².

Source: https://practicecalculus.com/derivatives/1-over-x-squared
Updated: 2026-11-06

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

## At a glance

- **Derivative:** $-\dfrac{2}{x^3}$ for $x \ne 0$
- **As a power:** $\dfrac{d}{dx}\,x^{-2} = -2x^{-3}$
- **Chain rule form:** $\dfrac{d}{dx}\,\dfrac{1}{u^2} = -\dfrac{2u'}{u^3}$
- **Second derivative:** $\dfrac{6}{x^4}$
- **Integral version:** $\int \frac{1}{x^2}\,dx = -\frac{1}{x} + C$

## Key takeaways

- The derivative of $\frac{1}{x^2}$ is $-\frac{2}{x^3}$.
- Rewrite $\frac{1}{x^2}$ as $x^{-2}$ before differentiating. The exponent goes from $-2$ to $-3$.
- The derivative is positive for $x < 0$ and negative for $x > 0$, so the graph rises then falls.
- For $\frac{1}{u^2}$, use the chain rule: $-\frac{2u'}{u^3}$.

The **derivative of $\dfrac{1}{x^2}$ is $-\dfrac{2}{x^3}$**. Move the power to the top as $x^{-2}$, and the power rule finishes the job in one step.

$$\frac{d}{dx}\,\frac{1}{x^2} = -\frac{2}{x^3}, \qquad x \ne 0$$

## Graph of 1/x² and its derivative

```graph
title: 'Graph of y = 1/x² and its derivative y = -2/x³'
caption: 'On the left, $\frac{1}{x^2}$ rises toward the asymptote, so the derivative is positive. On the right it falls, so the derivative is negative.'
x: [-4, 4]
y: [-5, 5]
functions:
  - expr: 1/x^2
    label: y = 1/x²
  - expr: -2/x^3
    label: y = -2/x³ (derivative)
    dashed: true
points:
  - [1, 1, '(1, 1), slope -2']
  - [-1, 1, '(-1, 1), slope 2']
```

The graph of $\frac{1}{x^2}$ is symmetric about the $y$-axis, so its slopes at $x$ and $-x$ are opposites. That is why the derivative is an odd function.

## Proof with the power rule

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

This is the [power rule](/rules/power-rule) with $n = -2$. The only care needed is the subtraction: $-2 - 1 = -3$.

## Proof with the quotient rule

With top $1$ and bottom $x^2$, the [quotient rule](/rules/quotient-rule) gives

$$\frac{0 \cdot x^2 - 1 \cdot 2x}{(x^2)^2} = \frac{-2x}{x^4} = -\frac{2}{x^3}$$

Same answer, more work. Rewriting as a power is the faster habit.

## Proof from the limit definition

$$\frac{\frac{1}{(x + h)^2} - \frac{1}{x^2}}{h} = \frac{x^2 - (x + h)^2}{h\,x^2(x + h)^2} = \frac{-2x - h}{x^2(x + h)^2}$$

As $h \to 0$ this becomes $\frac{-2x}{x^4} = -\frac{2}{x^3}$.

## The chain rule version: derivative of 1/u²

$$\frac{d}{dx}\,u^{-2} = -2u^{-3}\cdot u' = -\frac{2u'}{u^3}$$

**Example 1.** $\frac{d}{dx}\,\frac{1}{(3x - 2)^2} = -\frac{2 \cdot 3}{(3x - 2)^3} = -\frac{6}{(3x - 2)^3}$.

**Example 2.** $\frac{d}{dx}\,\frac{1}{(x^2 + 4)^2} = -\frac{2 \cdot 2x}{(x^2 + 4)^3} = -\frac{4x}{(x^2 + 4)^3}$.

**Example 3.** $\frac{d}{dx}\,\frac{1}{\cos^2 x} = -\frac{2(-\sin x)}{\cos^3 x} = \frac{2\sin x}{\cos^3 x}$. Since $\frac{1}{\cos^2 x} = \sec^2 x$, this matches the [derivative of sec²x](/derivatives/sec-squared-x), $2\sec^2 x\tan x$.

## More worked examples

**Example 4: a physics shape.** Gravity and light intensity follow $F = \frac{k}{r^2}$. The rate of change with distance is $\frac{dF}{dr} = -\frac{2k}{r^3}$. Doubling the distance cuts the force to a quarter and the rate of change to an eighth.

**Example 5: second derivative.** $\frac{d}{dx}\left(-2x^{-3}\right) = 6x^{-4} = \frac{6}{x^4}$. It is positive everywhere, so both branches are concave up.

## How to remember it

Think of $\frac{1}{x^2}$ as a power that happens to sit in the denominator. Move it up as $x^{-2}$, apply the power rule like any other power, then move the result back down if you want a fraction. The same three steps work for $\frac{1}{x^3}$, $\frac{1}{x^{10}}$ or $\frac{1}{\sqrt{x}}$. A quick sign check helps too: for positive $x$ the curve is falling, so the derivative has to be negative there, and $-\frac{2}{x^3}$ is.

## Common mistakes

1. **Writing $-2x^{-1}$.** Going down one from $-2$ lands at $-3$.
2. **Flipping the derivative of the bottom.** $\frac{1}{2x}$ is not the derivative of $\frac{1}{x^2}$.
3. **Losing the minus sign.** The exponent $-2$ comes down as a negative coefficient.

Going backward, $\int \frac{1}{x^2}\,dx = -\frac{1}{x} + C$ is on the [integral of 1/x²](/integrals/1-over-x-squared) page. Compare with the [derivative of 1/x](/derivatives/1-over-x).

## Frequently asked questions

### What is the derivative of 1/x²?

It is $-\frac{2}{x^3}$. Write $\frac{1}{x^2} = x^{-2}$ and use the power rule: $-2x^{-3}$.

### What is the derivative of 3/x²?

It is $3 \cdot \left(-\frac{2}{x^3}\right) = -\frac{6}{x^3}$.

### What is the derivative of 1/(x + 1)²?

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

### Is the derivative of 1/x² equal to 1/(2x)?

No. Taking the derivative of the bottom and flipping does not work. The correct answer is $-\frac{2}{x^3}$.