AP Calculus AB and BC

Derivative of x ln x: Answer, Product Rule, Mistakes

The derivative of x ln x with respect to x is ln x plus 1. The product rule gives ln x plus x times one over x, and that second term simplifies to 1. Setting the derivative to zero shows the function has a minimum at x equals one over e.

ddx[xln⁡x]=ln⁡x+1\frac{d}{dx}\left[x\ln x\right] = \ln x + 1

How to differentiate x ln x

Use the product rule with f(x)=xf(x) = x and g(x)=ln⁡xg(x) = \ln x.

ddx(xln⁡x)=1⋅ln⁡x+x⋅1x\frac{d}{dx}\left(x\ln x\right) = 1\cdot\ln x + x\cdot\frac{1}{x}

The second term collapses because x⋅1x=1x\cdot\frac{1}{x} = 1, which is what makes this derivative so clean.

ddx(xln⁡x)=ln⁡x+1\frac{d}{dx}\left(x\ln x\right) = \ln x + 1

The minimum at 1 over e

Setting ln⁡x+1=0\ln x + 1 = 0 gives ln⁡x=−1\ln x = -1, so x=e−1x = e^{-1}. The derivative is negative to the left and positive to the right, so this is a minimum by the First Derivative Test.

The minimum value is 1eln⁡1e=−1e\frac{1}{e}\ln\frac{1}{e} = -\frac{1}{e}, roughly −0.368-0.368. Note the function is negative on 0<x<10 < x < 1 even though it goes to 00 at both ends of that interval.

Where the derivative of x ln x shows up on the AP exam

Topic 2.8 on both exams. The function is a standard optimization and curve sketching example, and its behaviour as x→0+x \to 0^{+} is a common L'Hopital exercise: xln⁡x→0x\ln x \to 0, because the xx factor beats the diverging logarithm.

Common mistakes with the derivative of x ln x

  • Answering 1x\frac{1}{x}, multiplying the derivatives instead of using the product rule.
  • Answering ln⁡x+x\ln x + x, failing to simplify x⋅1xx\cdot\frac{1}{x}.
  • Answering ln⁡x\ln x, dropping the second term entirely.
  • Forgetting the domain. ln⁡x\ln x requires x>0x > 0, so the function and its derivative do not exist for negative xx.

Check yourself, not just the answer

Type derivatives and get graded on mathematical equivalence, with rule-level hints when you miss, in the Derivative Practice Checker.

Frequently asked questions

What is the derivative of x ln x?

It is ln⁡x+1\ln x + 1.

Why does the second term become 1?

The product rule gives x⋅1xx\cdot\frac{1}{x}, and that simplifies to 11.

Where does x ln x have a minimum?

At x=1ex = \frac{1}{e}, where the derivative is zero and changes from negative to positive. The minimum value is −1e-\frac{1}{e}.

What is the integral of x ln x?

It is x22ln⁡x−x24+C\frac{x^{2}}{2}\ln x - \frac{x^{2}}{4} + C, by integration by parts.