AP Calculus AB and BC

Derivative of ln(sqrt x): Answer and Shortcut

Q: How do you find the derivative of ln(sqrt x)?

Rewrite \ln\sqrt{x} as \frac{1}{2}\ln x first, and the derivative is immediate: \frac{1}{2x}, for x>0.

Q: Do you need the chain rule for ln(sqrt x)?

Not if you rewrite. \ln\sqrt{x} = \frac{1}{2}\ln x is a constant multiple of \ln x. The chain rule does work and gives the same \frac{1}{2x}, but it is the longer path.

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

No. \ln\sqrt{x} has derivative \frac{1}{2x}, while \sqrt{\ln x} has derivative \frac{1}{2x\sqrt{\ln x}}. Reversing the order of composition changes both the function and its domain.

Rewriting first is what makes this easy: the log property turns ln(sqrt x) into one half of ln x, and half of ln x differentiates to 1/(2x), valid for every x > 0. No chain rule is needed once the rewrite is done.

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

Rewrite before differentiating

Write the root as a power, $\sqrt{x} = x^{1/2}$ , then use $\ln\left(x^{p}\right) = p\ln x$ to bring the exponent out in front. The composition disappears and what remains is a constant multiple of $\ln x$ .

\ln\sqrt{x} = \ln\left(x^{1/2}\right) = \frac{1}{2}\ln x

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

The chain rule route reaches the same place with more work: $\frac{1}{\sqrt{x}}\cdot\frac{1}{2\sqrt{x}} = \frac{1}{2x}$ , since $\sqrt{x}\cdot\sqrt{x} = x$ . Use it as a check, but rewriting first is faster and leaves less to go wrong.

The same two operations in the other order

$\ln\sqrt{x}$ and $\sqrt{\ln x}$ are built from the same pair of operations, and that is the end of the resemblance. The log on the outside can be flattened by a log property; the root on the outside cannot be flattened by anything.

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

The domains part company too. $\ln\sqrt{x}$ accepts every $x>0$ , which is why $x=0.5$ is a legitimate sample input here, while $\sqrt{\ln x}$ needs $x\geq 1$ and rejects that input outright.

The mistakes students make

All three come from working through the composition instead of rewriting first.

Answering $\frac{1}{\sqrt{x}}$ , the outer derivative on its own, with the inner derivative $\frac{1}{2\sqrt{x}}$ left out.
Getting $\frac{1}{\sqrt{x}}\cdot\frac{1}{2\sqrt{x}}$ and then writing $\frac{1}{2\sqrt{x}}$ , as though only one root cancels. The two roots multiply to $x$ , giving $\frac{1}{2x}$ .
Pulling the exponent out as $2$ rather than $\frac{1}{2}$ and answering $\frac{2}{x}$ . The exponent on $x$ under the root is $\frac{1}{2}$ , not $2$ .

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

How do you find the derivative of ln(sqrt x)?

Rewrite $\ln\sqrt{x}$ as $\frac{1}{2}\ln x$ first, and the derivative is immediate: $\frac{1}{2x}$ , for $x>0$ .

Do you need the chain rule for ln(sqrt x)?

Not if you rewrite. $\ln\sqrt{x} = \frac{1}{2}\ln x$ is a constant multiple of $\ln x$ . The chain rule does work and gives the same $\frac{1}{2x}$ , but it is the longer path.

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

No. $\ln\sqrt{x}$ has derivative $\frac{1}{2x}$ , while $\sqrt{\ln x}$ has derivative $\frac{1}{2x\sqrt{\ln x}}$ . Reversing the order of composition changes both the function and its domain.