Q.Prove that the function $f$ given by $f(x) = \log \sin x$ is increasing on $\left(0, \frac{\pi}{2}\right)$ and decreasing on $\left(\frac{\pi}{2}, \pi\right)$.

Concept: Monotonic Function Analysis — we check the sign of $f'(x)$ on each interval.

Step 1: Differentiate $f(x) = \log(\sin x)$ using the chain rule:

Step 2: On $\left(0, \frac{\pi}{2}\right)$, $\sin x > 0$ and $\cos x > 0$, so $\cot x > 0$. Hence $f'(x) > 0$, so $f$ is increasing here. …

To determine where a function is increasing or decreasing, we look at the sign of its derivative. If $f'(x) > 0$ on an interval, the function is increasing there; if $f'(x) < 0$, it is decreasing. This is the Monotonic Function Analysis — a direct application of the first derivative test.

Here, $f(x) = \log \sin x$. The domain is $(0, \pi)$ because $\sin x$ must be positive for the logarithm to be defined. Within this domain, $\sin x$ is positive on $(0, \pi)$, so we are safe.

Let’s work through the steps.

1. Find the derivative. Using the chain rule:

So $f'(x) = \cot x$.

2. Analyse the sign of $\cot x$ on $(0, \pi)$.

   Recall that $\cot x = \frac{\cos x}{\sin x}$. On $(0, \pi)$, $\sin x > 0$, so the sign of $\cot x$ is the same as the sign of $\cos x$.

   • On $(0, \pi/2)$: $\cos x > 0$, so $\cot x > 0$. Hence $f'(x) > 0$, meaning $f$ is increasing on this interval.
   • At $x = \pi/2$: $\cos(\pi/2) = 0$, so $\cot(\pi/2) = 0$. This is a critical point.
   • On $(\pi/2, \pi)$: $\cos x < 0$, so $\cot x < 0$. Hence $f'(x) < 0$, meaning $f$ is decreasing on this interval.

3. Interpret the result. …

Method: Monotonicity of $\log(\text{trig function})$ via the Chain Rule

Use this method whenever the function is a logarithm applied to a trigonometric (or other) function — the derivative always reduces to a ratio of the inner function's derivative to the inner function itself, and its sign then follows from the signs of the inner function and its derivative on each sub-interval.

Steps

Step 1: Confirm the domain first.

$\log(g(x))$ is defined only where $g(x)>0$; state this domain restriction before differentiating (for $g(x)=\sin x$, this is where $\sin x>0$, i.e. $(0,\pi)$ and its periodic repeats).

Step 2: Differentiate using the chain rule for logarithms.

For $g(x)=\sin x$, $g'(x)=\cos x$, so $f'(x) = \dfrac{\cos x}{\sin x} = \cot x$.

Step 3: Split the required interval at the point(s) where the sign of $f'(x)$ can change. …

Common Mistakes

Mistake 1: Forgetting to restrict the domain to where $\sin x>0$ before analysing monotonicity.

Why it's wrong: $\log\sin x$ is undefined wherever $\sin x\le 0$, so any monotonicity claim outside $(0,\pi)$ (or its periodic repeats) is meaningless. Correct approach: always state the domain first, then work only within it.

Mistake 2: Writing the derivative as $\cot x$ but not actually checking its sign separately on each part of the interval.

Why it's wrong: $\cot x$ genuinely changes sign at $x=\tfrac{\pi}{2}$ within $(0,\pi)$, so stopping at "the derivative is $\cot x$" without a sign analysis leaves the monotonicity question unanswered. Correct approach: explicitly test the sign of $\cos x$ (equivalently $\cot x$, since $\sin x>0$ throughout) on each sub-interval separately. …