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

> The derivative of 1/x is -1/x². See the limit definition proof, the power rule with x⁻¹, the chain rule for 1/u and worked examples like 1/(2x + 5).

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

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

## At a glance

- **Derivative:** $-\dfrac{1}{x^2}$ for $x \ne 0$
- **As a power:** $\dfrac{d}{dx}\,x^{-1} = -x^{-2}$
- **Chain rule form:** $\dfrac{d}{dx}\,\dfrac{1}{u} = -\dfrac{u'}{u^2}$
- **Sign:** Always negative, so $\frac{1}{x}$ decreases on each side of $0$
- **Integral version:** $\int \frac{1}{x}\,dx = \ln\lvert x\rvert + C$

## Key takeaways

- The derivative of $\frac{1}{x}$ is $-\frac{1}{x^2}$.
- Rewrite $\frac{1}{x}$ as $x^{-1}$ and the power rule gives $-x^{-2}$.
- For a reciprocal of anything, $\frac{d}{dx}\frac{1}{u} = -\frac{u'}{u^2}$.
- Do not mix this up with $\ln x$: it is $\ln x$ whose derivative is $\frac{1}{x}$.

The **derivative of $\dfrac{1}{x}$ is $-\dfrac{1}{x^2}$**. The minus sign tells you the curve is always going downhill, on both sides of the vertical asymptote.

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

## Graph of 1/x and its derivative

```graph
title: 'Graph of y = 1/x and its derivative y = -1/x²'
caption: 'Both branches of $\frac{1}{x}$ fall from left to right, so the derivative is negative everywhere it exists.'
x: [-4, 4]
y: [-5, 5]
functions:
  - expr: 1/x
    label: y = 1/x
  - expr: -1/x^2
    label: y = -1/x² (derivative)
    dashed: true
points:
  - [1, 1, '(1, 1), slope -1']
```

Near $x = 0$ the branches are nearly vertical, and $-\frac{1}{x^2}$ is a huge negative number. Far from $0$ the curve flattens, and $-\frac{1}{x^2}$ approaches $0$.

## Proof from the limit definition

$$f'(x) = \lim_{h \to 0}\frac{\frac{1}{x + h} - \frac{1}{x}}{h}$$

Combine the fractions on top over the common denominator $x(x + h)$:

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

Divide by $h$ and let $h \to 0$:

$$f'(x) = \lim_{h \to 0}\frac{-1}{x(x + h)} = -\frac{1}{x^2}$$

## Proof with the power rule

Since $\frac{1}{x} = x^{-1}$, the [power rule](/rules/power-rule) gives $-1 \cdot x^{-2} = -\frac{1}{x^2}$. The [quotient rule](/rules/quotient-rule) agrees: $\frac{0 \cdot x - 1 \cdot 1}{x^2} = -\frac{1}{x^2}$.

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

$$\frac{d}{dx}\,\frac{1}{u} = -\frac{1}{u^2}\cdot u' = -\frac{u'}{u^2}$$

| Function | Inside $u$ | Derivative |
| --- | --- | --- |
| $\frac{1}{2x + 5}$ | $2x + 5$ | $-\frac{2}{(2x + 5)^2}$ |
| $\frac{1}{x^2 + 1}$ | $x^2 + 1$ | $-\frac{2x}{(x^2 + 1)^2}$ |
| $\frac{1}{\sin x}$ | $\sin x$ | $-\frac{\cos x}{\sin^2 x}$ |
| $\frac{1}{e^x + 1}$ | $e^x + 1$ | $-\frac{e^x}{(e^x + 1)^2}$ |

The third row is the same as $-\csc x\cot x$, the [derivative of csc x](/derivatives/csc-x).

## Worked examples

**Example 1: a sum.** $\frac{d}{dx}\left(x + \frac{1}{x}\right) = 1 - \frac{1}{x^2}$. Setting this to $0$ gives $x = \pm 1$, where $x + \frac{1}{x}$ has its turning points.

**Example 2: a tangent line.** At $x = 2$, $y = \frac{1}{2}$ and the slope is $-\frac{1}{4}$:

$$y - \frac{1}{2} = -\frac{1}{4}(x - 2) \quad\Longrightarrow\quad y = -\frac{x}{4} + 1$$

**Example 3: second derivative.** $\frac{d}{dx}\left(-x^{-2}\right) = 2x^{-3} = \frac{2}{x^3}$. It is positive for $x > 0$ and negative for $x < 0$, so the right branch is concave up and the left branch concave down.

## Common mistakes

1. **Answering $\ln x$ or $\frac{1}{x^2}$.** The derivative is $-\frac{1}{x^2}$. The function $\ln x$ is the antiderivative, not the derivative.
2. **Losing the minus sign.** $x^{-1}$ brings down $-1$.
3. **Writing $x^0$.** Lowering $-1$ by one gives $-2$, not $0$.
4. **Forgetting $u'$.** $\frac{d}{dx}\frac{1}{3x}$ is $-\frac{3}{(3x)^2} = -\frac{1}{3x^2}$, not $-\frac{1}{9x^2}$.

The integral side is on the [integral of 1/x](/integrals/1-over-x) page, and the curve itself is on the [reciprocal function](/functions/reciprocal-function) page.

## Frequently asked questions

### What is the derivative of 1/x?

The derivative of $\frac{1}{x}$ is $-\frac{1}{x^2}$, for every $x \ne 0$.

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

Write it as $\frac{1}{2}x^{-1}$. The derivative is $-\frac{1}{2}x^{-2} = -\frac{1}{2x^2}$.

### What is the derivative of 5/x?

It is $5 \cdot \left(-\frac{1}{x^2}\right) = -\frac{5}{x^2}$.

### Which function has derivative 1/x?

The natural log: $\frac{d}{dx}\ln x = \frac{1}{x}$. That is why $\int \frac{1}{x}\,dx = \ln\lvert x\rvert + C$.