AP Calculus AB and BC

Derivative of x^2 ln x: Answer, Proof, and Mistakes

Q: What is the derivative of x^{2}\ln x?

It is x(2\ln x+1), equivalently 2x\ln x+x, for x>0.

Q: Why does the product rule give x and not \frac{1}{x}?

Because the second term is x^{2}\cdot\frac{1}{x}, the untouched factor x^{2} times the derivative of \ln x. Those cancel down to x.

Q: Where does x^{2}\ln x reach its minimum?

At x = \frac{1}{\sqrt{e}}, where 2\ln x+1 = 0. The minimum value there is -\frac{1}{2e}.

The derivative of x^2 ln x is x(2 ln x + 1), which expands to 2x ln x + x, for x > 0. In prime notation, if f(x) = x^2 ln x then f'(x) = x(2 ln x + 1). The product rule gives 2x ln x plus x^2 times 1/x, and that second term collapses to x.

\frac{d}{dx}\left[x^2\ln x\right] = x(2\ln x+1)

How to differentiate x^2 ln x

Set $u = x^{2}$ and $v = \ln x$ , so $u' = 2x$ and $v' = \frac{1}{x}$ , then apply the product rule $u'v+uv'$ .

\frac{d}{dx}\left(x^{2}\ln x\right) = 2x\ln x+x^{2}\cdot\frac{1}{x} = 2x\ln x+x

The step worth slowing down for is $x^{2}\cdot\frac{1}{x} = x$ . Cancelling one power of $x$ leaves $x$ , not $1$ and not $\frac{1}{x}$ .

f'(x) = x(2\ln x+1)

What the factored form shows

On the domain $x>0$ the factor $x$ is positive, so the derivative changes sign only when the bracket does.

2\ln x+1 = 0 \quad\Longrightarrow\quad \ln x = -\frac{1}{2} \quad\Longrightarrow\quad x = \frac{1}{\sqrt{e}}

To the left of $\frac{1}{\sqrt{e}}\approx 0.607$ the derivative is negative and to the right it is positive, so that input gives the minimum of $x^{2}\ln x$ , with value $-\frac{1}{2e}$ .

Product rule questions with a logarithm factor sit in Unit 2, and this one also sets up the integration by parts version, $\int x^{2}\ln x\,dx$ , later in the course.

Common mistakes with the derivative of x^2 ln x

Multiplying derivatives and answering $2x\cdot\frac{1}{x} = 2$ . The product rule adds two terms rather than multiplying two derivatives.
Writing $2x\ln x+\frac{1}{x}$ , forgetting to multiply $\frac{1}{x}$ by the factor $x^{2}$ that stays put.
Expanding the factored answer wrong: $x(2\ln x+1) = 2x\ln x+x$ , not $2x\ln x+1$ .
Treating $\ln x$ as if it had a derivative of $\ln x$ , which loses the whole point of the logarithm rule.

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 $x^{2}\ln x$ ?

It is $x(2\ln x+1)$ , equivalently $2x\ln x+x$ , for $x>0$ .

Why does the product rule give $x$ and not $\frac{1}{x}$ ?

Because the second term is $x^{2}\cdot\frac{1}{x}$ , the untouched factor $x^{2}$ times the derivative of $\ln x$ . Those cancel down to $x$ .

Where does $x^{2}\ln x$ reach its minimum?

At $x = \frac{1}{\sqrt{e}}$ , where $2\ln x+1 = 0$ . The minimum value there is $-\frac{1}{2e}$ .