Derivative of ln(tan x): Answer, Proof, Simplification

Chain rule, then clean up

Outer function $\ln u$, inner function $u = \tan x$. Differentiating the outer gives $\frac{1}{u}$ and the inner derivative is $\sec^{2}x$, so the first form of the answer is a quotient of trig functions.

Convert everything to sines and cosines and the quotient collapses. $\sec^{2}x = \frac{1}{\cos^{2}x}$ and $\frac{1}{\tan x} = \frac{\cos x}{\sin x}$, so one power of $\cos x$ cancels.

Why the simplified form is worth the extra line

$\ln\left(\tan x\right)$ needs $\tan x > 0$, so take the interval $\left(0,\frac{\pi}{2}\right)$. Written as $\frac{2}{\sin 2x}$, the derivative is transparent: $\sin 2x$ runs from $0$ up to $1$ and back to $0$ across that interval.

So the derivative is never smaller than $2$, hits exactly $2$ at $x = \frac{\pi}{4}$, and blows up at both ends of the interval. The blow-up at $x = \frac{\pi}{2}$ is invisible in $\frac{\sec^{2}x}{\tan x}$, where both parts run to infinity at once and the quotient's behaviour is not readable off the page, and nothing in that form shows the minimum value of $2$.

The derivative is also positive everywhere on the interval, which confirms that $\ln\left(\tan x\right)$ is increasing throughout, running from $-\infty$ at the left end to $+\infty$ at the right.

The mistakes students make

Three of the four below produce a wrong answer outright. The other produces a correct answer in a form that costs marks whenever the question goes on to ask about behaviour.

• Reporting $\frac{1}{\tan x}$, that is $\cot x$, after applying the logarithm rule and forgetting to multiply by the inner derivative $\sec^{2}x$.
• Cancelling inside $\frac{\sec^{2}x}{\tan x}$ down to $\sec x$. The cancellation that is legal is $\frac{\sec x}{\tan x} = \csc x$, which leaves $\sec x\csc x$, not $\sec x$. The discarded $\csc x$ is the whole difference, and $\sec x\csc x$ is just $\frac{1}{\sin x\cos x}$ again.
• Leaving the answer as $\frac{\sec^{2}x}{\tan x}$. It is correct, but it hides that the derivative never drops below $2$ and blows up at both ends of $\left(0,\frac{\pi}{2}\right)$, which is exactly what a follow-up part will ask about.
• Reading $\ln\left(\tan x\right)$ as $\ln x\cdot\tan x$ and using the product rule, which produces $\frac{\tan x}{x}+\ln x\sec^{2}x$, an answer for a different function entirely.