# Derivative of x ln x: Formula, Steps and Examples

> The derivative of x ln x is ln x + 1. See the product rule steps, the minimum at x = 1/e, the second derivative, a graph and similar examples.

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

$$ \frac{d}{dx}\,x\ln x = \ln x + 1 $$

## At a glance

- **Derivative:** $\ln x + 1$ for $x > 0$
- **Second derivative:** $\dfrac{1}{x}$
- **Minimum:** $x = \frac{1}{e}$, value $-\frac{1}{e} \approx -0.368$
- **Limit at 0:** $\lim_{x \to 0^+} x\ln x = 0$
- **Integral:** $\int x\ln x\,dx = \frac{x^2}{2}\ln x - \frac{x^2}{4} + C$

## Key takeaways

- The derivative of $x\ln x$ is $\ln x + 1$, found with the product rule.
- Setting $\ln x + 1 = 0$ gives a minimum at $x = \frac{1}{e}$, with value $-\frac{1}{e}$.
- The second derivative is $\frac{1}{x} > 0$, so the graph is concave up everywhere on its domain.
- Run backwards, the result shows $\int(\ln x + 1)\,dx = x\ln x + C$, which is the key step in integrating $\ln x$.

The **derivative of $x\ln x$ is $\ln x + 1$**. It is a product, so the [product rule](/rules/product-rule) does all the work.

$$\frac{d}{dx}\,x\ln x = \ln x + 1, \qquad x > 0$$

## Step by step with the product rule

The product rule says $(fg)' = f'g + fg'$. Take $f = x$ and $g = \ln x$:

1. $f' = 1$.
2. $g' = \frac{1}{x}$, from the [derivative of ln x](/derivatives/ln-x).
3. Combine: $1\cdot\ln x + x\cdot\frac{1}{x} = \ln x + 1$.

The $x$ in front cancels the $\frac{1}{x}$ from the log, leaving a clean answer.

## Graph of x ln x and its derivative

```graph
title: 'Graph of y = x ln x and its derivative y = ln x + 1'
caption: 'The function dips to a minimum of $-\frac{1}{e}$ at $x = \frac{1}{e}$, exactly where its derivative crosses zero.'
x: [-0.3, 3]
y: [-2, 3.5]
functions:
  - expr: x*ln(x)
    label: y = x ln x
    domain: [0.005, 3]
  - expr: ln(x) + 1
    label: y = ln x + 1 (derivative)
    dashed: true
    domain: [0.15, 3]
points:
  - [0.3679, -0.3679, 'minimum (1/e, −1/e)']
  - [1, 0, '(1, 0)']
```

## Finding the minimum

Set the derivative to zero:

$$\ln x + 1 = 0 \quad\Longrightarrow\quad \ln x = -1 \quad\Longrightarrow\quad x = e^{-1} = \frac{1}{e}$$

For $x < \frac{1}{e}$ the derivative is negative, and for $x > \frac{1}{e}$ it is positive, so this is a minimum. Its value is

$$\frac{1}{e}\ln\frac{1}{e} = \frac{1}{e}\cdot(-1) = -\frac{1}{e} \approx -0.368$$

The second derivative confirms it: $\frac{d^2}{dx^2}x\ln x = \frac{1}{x}$, which is positive for every $x > 0$, so the curve is concave up everywhere.

## What happens near x = 0

The graph seems to start at the origin, even though $x\ln x$ is not defined at $0$. That is because $\lim_{x \to 0^+} x\ln x = 0$. Write it as $\frac{\ln x}{1/x}$ and apply [L'Hôpital's rule](/rules/lhopitals-rule):

$$\lim_{x \to 0^+}\frac{\ln x}{1/x} = \lim_{x \to 0^+}\frac{1/x}{-1/x^2} = \lim_{x \to 0^+}(-x) = 0$$

Meanwhile the slope $\ln x + 1$ goes to $-\infty$, so the curve leaves the origin almost vertically.

## Related examples

**Example 1.** $\frac{d}{dx}(x\ln x - x) = \ln x + 1 - 1 = \ln x$. This shows $x\ln x - x$ is an antiderivative of $\ln x$; see the [integral of ln x](/integrals/ln-x).

**Example 2.** $\frac{d}{dx}x^2\ln x = 2x\ln x + x^2\cdot\frac{1}{x} = 2x\ln x + x = x(2\ln x + 1)$.

**Example 3.** $\frac{d}{dx}\frac{\ln x}{x} = \frac{\frac{1}{x}\cdot x - \ln x}{x^2} = \frac{1 - \ln x}{x^2}$, by the [quotient rule](/rules/quotient-rule). Its maximum is at $x = e$.

**Example 4.** $\frac{d}{dx}x\ln(3x) = \ln(3x) + 1$, since $\frac{d}{dx}\ln(3x) = \frac{1}{x}$ too.

## Common mistakes

1. **Multiplying the derivatives.** $1\cdot\frac{1}{x} = \frac{1}{x}$ is not the product rule. You need both terms.
2. **Forgetting the domain.** $x\ln x$ only exists for $x > 0$, so the minimum search only happens there.
3. **Mixing it up with $x^x$.** $\frac{d}{dx}x^x = x^x(\ln x + 1)$. The factor $\ln x + 1$ appears because $x^x = e^{x\ln x}$; see the [derivative of xˣ](/derivatives/x-to-the-x).

To go the other way, the [integral of x ln x](/integrals/x-ln-x) uses integration by parts. Practise more product rule problems on [product and quotient rules](/practice-problems/product-and-quotient-rules).

## Frequently asked questions

### What is the derivative of x ln x?

The derivative of $x\ln x$ is $\ln x + 1$.

### What is the second derivative of x ln x?

Differentiate $\ln x + 1$ to get $\frac{1}{x}$.

### Where is the minimum of x ln x?

At $x = \frac{1}{e} \approx 0.368$, where $\ln x + 1 = 0$. The minimum value is $\frac{1}{e}\ln\frac{1}{e} = -\frac{1}{e}$.

### What is the derivative of x ln x − x?

It is $(\ln x + 1) - 1 = \ln x$. That is why $\int\ln x\,dx = x\ln x - x + C$.