AP Calculus AB and BC

Derivative of sqrt(ln x): Chain Rule Proof

Q: Which rule gives the derivative of sqrt(ln x)?

The chain rule. The root on the outside contributes \frac{1}{2\sqrt{\ln x}}, the log inside contributes \frac{1}{x}, and multiplying the two gives \frac{1}{2x\sqrt{\ln x}}.

Q: What is the domain of sqrt(ln x)?

The function is defined for x \geq 1, since \ln x must be zero or positive. The derivative is defined for x>1, because x=1 makes the denominator zero.

Q: Is sqrt(ln x) the same as ln(sqrt x)?

No. Their derivatives are \frac{1}{2x\sqrt{\ln x}} and \frac{1}{2x}, and even their domains differ: x\geq 1 against x>0. Composition order changes the function.

Only inputs above x = 1 are in play, since ln x sits under a root in the denominator and has to be positive there. On that domain the derivative of sqrt(ln x) is 1 divided by 2x sqrt(ln x). The square root is the outer function, so the chain rule pairs one half over sqrt(ln x) with the inner derivative 1/x.

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

The root is on the outside

Reading the function from the outside in: the last operation performed is the square root, so $\sqrt{u}$ is the outer function and $u = \ln x$ is the inner one. The outer derivative is $\frac{1}{2\sqrt{u}}$ and the inner derivative is $\frac{1}{x}$ .

\frac{d}{dx}\sqrt{\ln x} = \frac{1}{2\sqrt{\ln x}}\cdot\frac{1}{x} = \frac{1}{2x\sqrt{\ln x}}

Both factors are fractions with $1$ on top, so multiplying them just gathers everything into one denominator. Nothing cancels and nothing simplifies further: the root stays in the answer, which is the signature of a root on the outside.

Domain, and the contrast with ln(sqrt x)

Two conditions stack up. The logarithm needs $x>0$ , and the square root needs $\ln x \geq 0$ , which means $x \geq 1$ . At $x=1$ the denominator $2x\sqrt{\ln x}$ is zero, so the function has a vertical tangent there and the derivative formula holds for $x>1$ . Each of $1.5$ , $3$ and $8$ clears that bar comfortably.

Swapping the order of the two operations gives a completely different function. In $\ln\sqrt{x}$ the log is on the outside, the log property flattens it to $\frac{1}{2}\ln x$ , and no root survives in the derivative.

\frac{d}{dx}\sqrt{\ln x} = \frac{1}{2x\sqrt{\ln x}} \qquad \frac{d}{dx}\ln\sqrt{x} = \frac{1}{2x}

The mistakes students make

The first two are chain rule failures and the third is a domain failure that costs a mark on justification questions.

Answering $\frac{1}{2\sqrt{\ln x}}$ and never multiplying by the inner derivative. That misses the factor $\frac{1}{x}$ , so at $x=3$ it is three times the true value.
Answering $\frac{1}{2x}$ , the derivative of $\ln\sqrt{x}$ . Same two operations, opposite order, different function, different answer.
Evaluating the derivative at inputs like $x=0.5$ . There $\ln x$ is negative, $\sqrt{\ln x}$ is not a real number, and the function does not exist to be differentiated.

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

Which rule gives the derivative of sqrt(ln x)?

The chain rule. The root on the outside contributes $\frac{1}{2\sqrt{\ln x}}$ , the log inside contributes $\frac{1}{x}$ , and multiplying the two gives $\frac{1}{2x\sqrt{\ln x}}$ .

What is the domain of sqrt(ln x)?

The function is defined for $x \geq 1$ , since $\ln x$ must be zero or positive. The derivative is defined for $x>1$ , because $x=1$ makes the denominator zero.

Is sqrt(ln x) the same as ln(sqrt x)?

No. Their derivatives are $\frac{1}{2x\sqrt{\ln x}}$ and $\frac{1}{2x}$ , and even their domains differ: $x\geq 1$ against $x>0$ . Composition order changes the function.