AP Calculus AB and BC

Derivative of x^(ln x): Answer, Proof, Mistakes

Q: What is d/dx of x^ln x?

It is \frac{2\ln x \cdot x^{\ln x}}{x} for x>0. Equivalently, y' = y\cdot\frac{2\ln x}{x} with y = x^{\ln x}.

Q: Why does taking logs help with x^ln x?

Because \ln\left(x^{\ln x}\right) = \left(\ln x\right)^{2}. A messy variable exponent becomes a simple square, and one chain rule finishes the job.

Q: Where is the minimum of x^(ln x)?

At x=1, where 2\ln x = 0. The derivative changes from negative to positive there, and the minimum value is 1^{0}=1.

This one has a minimum at x = 1, and the derivative shows it: 2 ln x times x^(ln x), all divided by x, for x > 0. Taking logs turns the function into ln y = (ln x)^2, a perfect square that differentiates in one chain rule step, and the factor 2 ln x changes sign exactly at x = 1.

\frac{d}{dx}\left[x^{\ln x}\right] = \frac{2\ln x \cdot x^{\ln x}}{x}

Taking logs collapses the exponent

Set $y = x^{\ln x}$ and take the natural log of both sides. Because the exponent is itself $\ln x$ , the rule $\ln\left(a^{b}\right)=b\ln a$ produces $\ln x$ multiplied by $\ln x$ , that is a square.

\ln y = \ln x\cdot\ln x = \left(\ln x\right)^{2}

Now differentiate. The right side is a chain rule on the square, giving $2\ln x$ times the derivative of $\ln x$ , and the left side gives $\frac{y'}{y}$ .

\frac{y'}{y} = 2\ln x\cdot\frac{1}{x} \quad\Longrightarrow\quad y' = \frac{2\ln x \cdot x^{\ln x}}{x}

The minimum at x = 1

On the domain $x>0$ both $x$ and $x^{\ln x}$ are positive, so the sign of the derivative is decided entirely by the factor $2\ln x$ . That factor is negative below $1$ and positive above $1$ .

y'=0 \quad\Longleftrightarrow\quad \ln x = 0 \quad\Longleftrightarrow\quad x = 1

The function falls then rises, so $x=1$ gives a minimum. Its value is $1^{\ln 1} = 1^{0} = 1$ . That splits the values checked here in two: $x=0.6$ falls on the decreasing side, while $x=1.5$ and $x=3$ sit on the increasing side.

The mistakes students make

The log step is short here, which is exactly why the errors below slip through unchecked.

Answering $\ln x\cdot x^{\ln x-1}$ from the power rule. The exponent $\ln x$ is not a constant, so the power rule does not apply at all.
Differentiating $\left(\ln x\right)^{2}$ as $2\ln x$ and losing the inner $\frac{1}{x}$ . That produces $2\ln x\cdot x^{\ln x}$ , which is $x$ times too large.
Reporting $\frac{2\ln x}{x}$ as the final derivative. That is $\frac{y'}{y}$ , and it still needs multiplying by $y = x^{\ln x}$ .

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 d/dx of x^ln x?

It is $\frac{2\ln x \cdot x^{\ln x}}{x}$ for $x>0$ . Equivalently, $y' = y\cdot\frac{2\ln x}{x}$ with $y = x^{\ln x}$ .

Why does taking logs help with x^ln x?

Because $\ln\left(x^{\ln x}\right) = \left(\ln x\right)^{2}$ . A messy variable exponent becomes a simple square, and one chain rule finishes the job.

Where is the minimum of x^(ln x)?

At $x=1$ , where $2\ln x = 0$ . The derivative changes from negative to positive there, and the minimum value is $1^{0}=1$ .