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Exponential and log derivatives

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.

Exponential and log derivatives Updated 2 min read

At a glance

\frac{d}{dx}\,x\ln x = \ln x + 1
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
On this page
  1. Step by step with the product rule
  2. Graph of x ln x and its derivative
  3. Finding the minimum
  4. What happens near x = 0
  5. Related examples
  6. Common mistakes

The derivative of x\ln x is \ln x + 1. It is a product, so the 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.
  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 of y = x ln x and its derivative y = ln x + 1
The function dips to a minimum of -\frac{1}{e} at x = \frac{1}{e}, exactly where its derivative crosses zero.

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:

\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.

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.

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. 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.

To go the other way, the integral of x ln x uses integration by parts. Practise more product rule problems on product and quotient rules.

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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.

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