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

The proof: chain rule on ln(sin x)

The outer function is $\ln u$ and the inner function is $u = \sin x$. Differentiate the outer to $\frac{1}{u}$, leave the inner untouched inside it, then multiply by the inner derivative $u' = \cos x$.

The ratio $\frac{\cos x}{\sin x}$ is the definition of $\cot x$, so a two step chain rule (Topic 3.1) collapses into one named function.

Where ln(sin x) is defined

A logarithm accepts positive inputs only, so this function needs $\sin x > 0$. That gives the intervals $(0, \pi)$, $(2\pi, 3\pi)$, and every shift of those by $2\pi$. Outside them the function does not exist, so neither does a slope.

On $(0, \pi)$ the graph climbs from $-\infty$ to a maximum of $\ln 1 = 0$ at $x = \frac{\pi}{2}$, then falls back to $-\infty$. The derivative agrees: $\cot x$ is positive on $\left(0, \frac{\pi}{2}\right)$, zero at $\frac{\pi}{2}$, and negative on $\left(\frac{\pi}{2}, \pi\right)$.

The same fact read backwards: the integral of cot x

Since differentiating $\ln(\sin x)$ produces $\cot x$, an antiderivative of $\cot x$ is a logarithm of sine. The absolute value in the standard form covers the intervals where $\sin x$ is negative.

The substitution behind it is $u = \sin x$ with $du = \cos x\,dx$, which turns $\int \frac{\cos x}{\sin x}\,dx$ into $\int \frac{1}{u}\,du$.

Common mistakes

• Answering $\csc x$. That is $\frac{1}{\sin x}$, the outer part alone, with the inner derivative $\cos x$ never applied.
• Answering $\tan x$ by flipping the ratio. Cosine lands on top, so the quotient is $\frac{\cos x}{\sin x}$.
• Confusing $\ln(\sin x)$ with $\sin(\ln x)$, whose derivative is $\frac{\cos(\ln x)}{x}$.
• Using the formula where $\sin x \leq 0$, for example at $x = 4$. The original function is undefined there.