AP Calculus AB and BC

Derivative of ln(x^2): Answer, Proof, Mistakes

Q: What is the derivative of ln(x^2)?

It is \frac{2}{x}. The chain rule gives \frac{1}{x^2}\cdot 2x = \frac{2}{x}, and rewriting \ln(x^2) = 2\ln|x| gives the same \frac{2}{x}.

Q: Is the derivative of ln(x^2) the same as the derivative of 2 ln x?

Yes, the formula is the same, \frac{2}{x}. The only difference is the domain: \ln(x^2) and 2\ln|x| are defined for all x \neq 0, while 2\ln x needs x > 0.

Q: Why does the x^2 disappear from the answer?

The chain rule puts x^2 in the denominator and 2x in the numerator. The common factor of x cancels, leaving \frac{2}{x}.

The derivative of ln(x^2) with respect to x is 2/x. By the chain rule with inner function x^2, the derivative is (1/x^2) times 2x, which simplifies to 2/x. You get the same answer by rewriting ln(x^2) as 2 ln|x| and differentiating to 2 times 1/x. It holds for every x except 0.

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

The proof: chain rule on ln(x^2)

The outer function is $\ln u$ and the inner is $u = x^2$ . Differentiate the outer to $\frac{1}{u}$ , keep the inner, then multiply by the inner derivative $u' = 2x$ .

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

The $x^2$ in the denominator and the $2x$ from the chain rule cancel down to $\frac{2}{x}$ . Leaving the answer as $\frac{2x}{x^2}$ is not wrong, but the graders expect the simplified $\frac{2}{x}$ .

The shortcut: rewrite ln(x^2) as 2 ln|x|

A log of a power comes down as a coefficient: $\ln(x^2) = 2\ln|x|$ . The absolute value matters because $x^2$ is positive for every $x \neq 0$ , so the domain of $\ln(x^2)$ is all $x \neq 0$ , wider than the domain of $\ln x$ .

\frac{d}{dx}\left[2\ln|x|\right] = 2\cdot\frac{1}{x} = \frac{2}{x}

Both routes agree, and the derivative $\frac{2}{x}$ is itself defined for all $x \neq 0$ , matching the domain of the original function.

Common mistakes

Answering $\frac{1}{x^2}$ by differentiating $\ln$ and stopping. That skips the chain rule factor $2x$ from the inner function $x^2$ .
Confusing $\ln(x^2)$ with $(\ln x)^2$ . The second is a square of a log and differentiates by the chain rule to $\frac{2\ln x}{x}$ , a different function.
Leaving $\frac{2x}{x^2}$ unsimplified. It equals $\frac{2}{x}$ , and simplifying avoids the false impression the answer still depends on $x^2$ .
Writing $2\ln x$ instead of $2\ln|x|$ and then claiming the domain is only $x > 0$ . The original $\ln(x^2)$ is defined for negative $x$ too.

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 ln(x^2)?

It is $\frac{2}{x}$ . The chain rule gives $\frac{1}{x^2}\cdot 2x = \frac{2}{x}$ , and rewriting $\ln(x^2) = 2\ln|x|$ gives the same $\frac{2}{x}$ .

Is the derivative of ln(x^2) the same as the derivative of 2 ln x?

Yes, the formula is the same, $\frac{2}{x}$ . The only difference is the domain: $\ln(x^2)$ and $2\ln|x|$ are defined for all $x \neq 0$ , while $2\ln x$ needs $x > 0$ .

Why does the x^2 disappear from the answer?

The chain rule puts $x^2$ in the denominator and $2x$ in the numerator. The common factor of $x$ cancels, leaving $\frac{2}{x}$ .